[{"Name":"Introduction to Electric Dipole","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Intro To Electric Dipoles (Potential And Energy)","Duration":"11m 23s","ChapterTopicVideoID":21438,"CourseChapterTopicPlaylistID":99476,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/21438.jpeg","UploadDate":"2020-04-21T09:34:33.3930000","DurationForVideoObject":"PT11M23S","Description":null,"MetaTitle":"Intro To Electric Dipoles (Potential And Energy): Video + Workbook | Proprep","MetaDescription":"Electric Dipole - Introduction to Electric Dipole. Watch the video made by an expert in the field. Download the workbook and maximize your learning.","Canonical":"https://www.proprep.uk/general-modules/all/physics-2-electricity-and-magnetism/electric-dipole/introduction-to-electric-dipole/vid22324","VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.980","Text":"Hello. In this lesson,"},{"Start":"00:01.980 ","End":"00:05.770","Text":"we\u0027re going to be speaking about the electric dipole."},{"Start":"00:05.770 ","End":"00:09.150","Text":"An electric dipole, some kind of body."},{"Start":"00:09.150 ","End":"00:11.880","Text":"It could be 2 point charges or anything else,"},{"Start":"00:11.880 ","End":"00:16.635","Text":"where 1 charge is positive and the other charge is negative."},{"Start":"00:16.635 ","End":"00:22.725","Text":"Usually the charges themselves magnitude are equal."},{"Start":"00:22.725 ","End":"00:24.750","Text":"Between the 2 charges,"},{"Start":"00:24.750 ","End":"00:27.225","Text":"there\u0027s a distance d. Now,"},{"Start":"00:27.225 ","End":"00:30.435","Text":"we assume that our distance d between the 2 charges"},{"Start":"00:30.435 ","End":"00:34.020","Text":"is something that\u0027s constant and that it isn\u0027t changing."},{"Start":"00:34.020 ","End":"00:38.760","Text":"Electric dipoles are very common even in everyday life and in nature."},{"Start":"00:38.760 ","End":"00:42.750","Text":"An example is the water molecule."},{"Start":"00:42.750 ","End":"00:47.840","Text":"If this is our oxygen with a charge of negative 2,"},{"Start":"00:47.840 ","End":"00:49.445","Text":"and then on top of it,"},{"Start":"00:49.445 ","End":"00:56.102","Text":"we have our hydrogen atoms where each one has 1 positive charge."},{"Start":"00:56.102 ","End":"00:58.947","Text":"Here we can see that we have a dipole."},{"Start":"00:58.947 ","End":"01:01.880","Text":"Here\u0027s our positive charge of plus 2,"},{"Start":"01:01.880 ","End":"01:04.520","Text":"and here\u0027s our negative charge of negative 2."},{"Start":"01:04.520 ","End":"01:09.304","Text":"Now, even though the total charge of our water molecule"},{"Start":"01:09.304 ","End":"01:14.480","Text":"of this dipole over here is going to be equal to 0 because they cancel each other out."},{"Start":"01:14.480 ","End":"01:19.310","Text":"We can still see that there\u0027s different charges within our system."},{"Start":"01:19.310 ","End":"01:27.185","Text":"The way that our dipoles interact is slightly different to just regular charges."},{"Start":"01:27.185 ","End":"01:29.810","Text":"That\u0027s what we\u0027re going to learn in this lesson."},{"Start":"01:29.810 ","End":"01:37.780","Text":"The first thing that we need to discuss is the potential of the electric dipole."},{"Start":"01:37.780 ","End":"01:44.300","Text":"The potential of an electric dipole in some point in space is going to equal"},{"Start":"01:44.300 ","End":"01:48.005","Text":"the potential from our positive point charge"},{"Start":"01:48.005 ","End":"01:52.075","Text":"plus the potential from our negative point charge."},{"Start":"01:52.075 ","End":"01:58.880","Text":"We\u0027ll get that the potential somewhere in space is going to be equal to kq"},{"Start":"01:58.880 ","End":"02:07.260","Text":"divided by r1 plus k multiplied by minus q divided by r2."},{"Start":"02:07.260 ","End":"02:11.255","Text":"Here, if we\u0027re looking at this point in space over here,"},{"Start":"02:11.255 ","End":"02:14.450","Text":"and we want to know what our potential is equal to."},{"Start":"02:14.450 ","End":"02:19.190","Text":"Our r1 is going to be this vector,"},{"Start":"02:19.190 ","End":"02:25.255","Text":"this is r1, and our r2 is going to be this vector."},{"Start":"02:25.255 ","End":"02:30.845","Text":"What we have going on over here is great if we\u0027re relatively close to our dipole."},{"Start":"02:30.845 ","End":"02:34.430","Text":"But what happens if we\u0027re far away from our dipole?"},{"Start":"02:34.430 ","End":"02:39.245","Text":"If this is our dipole and we\u0027re located extremely far away,"},{"Start":"02:39.245 ","End":"02:43.244","Text":"what is our potential at this point?"},{"Start":"02:43.244 ","End":"02:46.565","Text":"As we can see when we\u0027re extremely far away,"},{"Start":"02:46.565 ","End":"02:53.420","Text":"what this is going to look like is some point with net charge 0."},{"Start":"02:53.420 ","End":"02:57.785","Text":"But we know that there is going to be some potential because at the end,"},{"Start":"02:57.785 ","End":"02:59.330","Text":"even though they cancel each other out,"},{"Start":"02:59.330 ","End":"03:01.580","Text":"we do have a positive and a negative charge."},{"Start":"03:01.580 ","End":"03:05.480","Text":"This is going to affect our potential."},{"Start":"03:05.480 ","End":"03:11.185","Text":"The first thing that we\u0027re going to do is we\u0027re going to draw an axis."},{"Start":"03:11.185 ","End":"03:18.215","Text":"We can draw a tiny little axis over here where the origin is exactly at the midpoint,"},{"Start":"03:18.215 ","End":"03:20.790","Text":"at d divided by 2."},{"Start":"03:20.790 ","End":"03:23.835","Text":"The midpoint between the 2 charges."},{"Start":"03:23.835 ","End":"03:29.644","Text":"Then what we do is our radius is going to be from this origin,"},{"Start":"03:29.644 ","End":"03:31.625","Text":"from d divided by 2,"},{"Start":"03:31.625 ","End":"03:35.325","Text":"all the way to this point over here."},{"Start":"03:35.325 ","End":"03:38.435","Text":"This is our r vector."},{"Start":"03:38.435 ","End":"03:42.470","Text":"Now, the way that we mathematically define when"},{"Start":"03:42.470 ","End":"03:47.210","Text":"we\u0027re far away from our dipole is when the magnitude of"},{"Start":"03:47.210 ","End":"03:57.235","Text":"our r vector is significantly larger than d. The distance between the 2 charges."},{"Start":"03:57.235 ","End":"04:03.525","Text":"What is the potential at this point far away from our dipole?"},{"Start":"04:03.525 ","End":"04:12.075","Text":"Our potential is given as k multiplied by p vector.r vector."},{"Start":"04:12.075 ","End":"04:19.948","Text":"This is a dot product divided by the magnitude of our r vector^3."},{"Start":"04:19.948 ","End":"04:23.250","Text":"So p is something which is new."},{"Start":"04:23.250 ","End":"04:30.800","Text":"Our p is equal to our charge multiplied by d,"},{"Start":"04:30.800 ","End":"04:34.465","Text":"the distance between the 2 charges."},{"Start":"04:34.465 ","End":"04:37.950","Text":"Now notice that the d is a vector."},{"Start":"04:37.950 ","End":"04:41.345","Text":"Let\u0027s see. When we\u0027re dealing with the electric field,"},{"Start":"04:41.345 ","End":"04:43.550","Text":"we always go from positive to negative."},{"Start":"04:43.550 ","End":"04:51.850","Text":"However, our dipole moment is always going to point from negative to positive."},{"Start":"04:51.850 ","End":"04:55.206","Text":"This is our d vector,"},{"Start":"04:55.206 ","End":"04:57.810","Text":"from negative to positive."},{"Start":"04:57.810 ","End":"05:00.425","Text":"What is our p? What does this mean?"},{"Start":"05:00.425 ","End":"05:02.570","Text":"It\u0027s our dipole moment."},{"Start":"05:02.570 ","End":"05:08.285","Text":"Now, another way of writing out the potential is that we can write that this is equal to"},{"Start":"05:08.285 ","End":"05:14.840","Text":"k multiplied by the dipole moment.r-hat instead of r vector."},{"Start":"05:14.840 ","End":"05:16.790","Text":"Then the only difference is,"},{"Start":"05:16.790 ","End":"05:21.205","Text":"is that it\u0027s divided by r^2 rather than r^3."},{"Start":"05:21.205 ","End":"05:26.104","Text":"Because we have a dot product here and here between 2 vectors,"},{"Start":"05:26.104 ","End":"05:29.940","Text":"we\u0027re going to get a scalar quantity,"},{"Start":"05:29.940 ","End":"05:33.090","Text":"which is great because our potential is a scalar quantity."},{"Start":"05:33.090 ","End":"05:37.865","Text":"We can see that the potential of our dipole"},{"Start":"05:37.865 ","End":"05:43.265","Text":"is going to be or is going to behave like 1 divided by r^2."},{"Start":"05:43.265 ","End":"05:48.200","Text":"However, the electric potential of a single point charge,"},{"Start":"05:48.200 ","End":"05:49.925","Text":"which is called a monopole,"},{"Start":"05:49.925 ","End":"05:54.830","Text":"behaves like 1 divided by r. We can see that"},{"Start":"05:54.830 ","End":"05:59.840","Text":"the potential of a dipole decreases to 0 a lot quicker."},{"Start":"05:59.840 ","End":"06:02.825","Text":"Because of this difference in size between"},{"Start":"06:02.825 ","End":"06:06.080","Text":"the dipole potential and the monopole potential,"},{"Start":"06:06.080 ","End":"06:09.965","Text":"we\u0027re only going to use this equation over here when we\u0027re dealing with"},{"Start":"06:09.965 ","End":"06:14.075","Text":"a dipole whose net charge exactly cancels out."},{"Start":"06:14.075 ","End":"06:16.513","Text":"If there\u0027s negative q it\u0027s plus q,"},{"Start":"06:16.513 ","End":"06:19.345","Text":"and negative 2q it\u0027s plus 2q."},{"Start":"06:19.345 ","End":"06:24.875","Text":"Otherwise, if we have some net charge which is different to 0,"},{"Start":"06:24.875 ","End":"06:28.310","Text":"either a negative charge or a positive charge. It doesn\u0027t really matter."},{"Start":"06:28.310 ","End":"06:33.150","Text":"Then we\u0027ll use the equation for potential to deal with a monopole."},{"Start":"06:33.150 ","End":"06:37.250","Text":"We\u0027ll consider it as some point charge."},{"Start":"06:37.250 ","End":"06:40.010","Text":"Only if the charge is exactly cancel out,"},{"Start":"06:40.010 ","End":"06:44.450","Text":"we\u0027ll use this dipole equation for our potential."},{"Start":"06:44.450 ","End":"06:46.474","Text":"Let\u0027s give a little example."},{"Start":"06:46.474 ","End":"06:52.325","Text":"Let\u0027s imagine that this axis over here is our z-axis."},{"Start":"06:52.325 ","End":"06:58.620","Text":"We want to find the potential somewhere on the z-axis and now height z."},{"Start":"06:59.090 ","End":"07:03.450","Text":"Let\u0027s imagine that this height z is much larger than our"},{"Start":"07:03.450 ","End":"07:07.800","Text":"d. This diagram is not drawn to scale."},{"Start":"07:07.800 ","End":"07:10.295","Text":"We can say therefore,"},{"Start":"07:10.295 ","End":"07:12.545","Text":"from our example along the z-axis,"},{"Start":"07:12.545 ","End":"07:18.350","Text":"that our r vector is going to be equal to 0, 0, z."},{"Start":"07:18.350 ","End":"07:20.450","Text":"It only has components on the z-axis,"},{"Start":"07:20.450 ","End":"07:24.260","Text":"which means that it\u0027s z in the z-direction."},{"Start":"07:24.260 ","End":"07:26.390","Text":"It\u0027s the same way of writing the same thing."},{"Start":"07:26.390 ","End":"07:34.015","Text":"Then our dipole moment is going to be equal to q,"},{"Start":"07:34.015 ","End":"07:35.955","Text":"one of the charges."},{"Start":"07:35.955 ","End":"07:40.050","Text":"Q multiplied by d vector."},{"Start":"07:40.050 ","End":"07:42.695","Text":"Our d vector is given by the magnitude,"},{"Start":"07:42.695 ","End":"07:43.745","Text":"so that\u0027s d,"},{"Start":"07:43.745 ","End":"07:47.270","Text":"that\u0027s the distance multiplied by the direction."},{"Start":"07:47.270 ","End":"07:49.955","Text":"We can see that it\u0027s going in the z-direction."},{"Start":"07:49.955 ","End":"07:53.400","Text":"We have d going in the positive z-direction."},{"Start":"07:53.930 ","End":"07:57.570","Text":"Now let\u0027s find out therefore,"},{"Start":"07:57.570 ","End":"08:00.780","Text":"what our potential for this dipole is."},{"Start":"08:00.780 ","End":"08:05.480","Text":"This is going to be equal to k, and then in brackets,"},{"Start":"08:05.480 ","End":"08:12.935","Text":"the dipole moment, which is qdz-hat dot product with our r vector."},{"Start":"08:12.935 ","End":"08:18.790","Text":"Z in the z-hat direction."},{"Start":"08:18.790 ","End":"08:23.090","Text":"Then that\u0027s going to be divided by r^3."},{"Start":"08:23.090 ","End":"08:26.840","Text":"It\u0027s just the magnitude of our r vector^3."},{"Start":"08:26.840 ","End":"08:31.650","Text":"The magnitude of our r vector is simply z and then cubed."},{"Start":"08:31.650 ","End":"08:36.815","Text":"Now we can see that our z.z is going to be equal to 1."},{"Start":"08:36.815 ","End":"08:43.235","Text":"We\u0027ll be left with kqd multiplied by 1 divided by z^3."},{"Start":"08:43.235 ","End":"08:47.790","Text":"Of course, there\u0027s a z over here, I forgot."},{"Start":"08:47.790 ","End":"08:50.473","Text":"Then this can cancel out,"},{"Start":"08:50.473 ","End":"08:58.725","Text":"so we\u0027ll be left with kqd divided by z^3 because of this z over here."},{"Start":"08:58.725 ","End":"09:00.822","Text":"This is the potential."},{"Start":"09:00.822 ","End":"09:03.200","Text":"Now, if we were being asked,"},{"Start":"09:03.200 ","End":"09:08.075","Text":"what energy would I need to bring some charge"},{"Start":"09:08.075 ","End":"09:14.865","Text":"q tag from infinity until this point over here?"},{"Start":"09:14.865 ","End":"09:20.310","Text":"In infinity our energy is equal to 0 or our potential is equal to 0."},{"Start":"09:20.310 ","End":"09:23.330","Text":"The energy that we would need to bring this q tag from"},{"Start":"09:23.330 ","End":"09:27.860","Text":"infinity until this point over here is going to be"},{"Start":"09:27.860 ","End":"09:35.375","Text":"equal to our charge q tag multiplied by the potential at this point over here,"},{"Start":"09:35.375 ","End":"09:41.855","Text":"which is equal to kqd divided by z^2."},{"Start":"09:41.855 ","End":"09:45.560","Text":"Now something interesting to note is say we wanted to check what"},{"Start":"09:45.560 ","End":"09:49.355","Text":"the potential was at this point over here, where it\u0027s just on the,"},{"Start":"09:49.355 ","End":"09:50.600","Text":"let\u0027s say x-axis,"},{"Start":"09:50.600 ","End":"09:53.240","Text":"but this is the same for the y-axis,"},{"Start":"09:53.240 ","End":"09:57.635","Text":"is specifically our dipole moment is going in the z-direction."},{"Start":"09:57.635 ","End":"10:00.995","Text":"If our dipole moment only has a z component."},{"Start":"10:00.995 ","End":"10:05.121","Text":"If we wanted to measure the potential on either the x or y-axis,"},{"Start":"10:05.121 ","End":"10:08.405","Text":"we could say that our r vector is equal to,"},{"Start":"10:08.405 ","End":"10:11.455","Text":"let\u0027s just say x, 0, 0."},{"Start":"10:11.455 ","End":"10:14.630","Text":"Then we can see that the dot product between"},{"Start":"10:14.630 ","End":"10:20.105","Text":"the dipole moment and our r vector would be equal to,"},{"Start":"10:20.105 ","End":"10:23.530","Text":"the dipole moment here would still be the same, q."},{"Start":"10:23.530 ","End":"10:29.495","Text":"The charge multiplied by d in the z-direction.x,"},{"Start":"10:29.495 ","End":"10:32.895","Text":"some x in the x-direction."},{"Start":"10:32.895 ","End":"10:37.170","Text":"We can see that we will get 0."},{"Start":"10:37.170 ","End":"10:41.360","Text":"Then there will be no potential on the x or y plane"},{"Start":"10:41.360 ","End":"10:44.690","Text":"if our dipole moment is only in the z-direction."},{"Start":"10:44.690 ","End":"10:46.760","Text":"But what does that actually mean?"},{"Start":"10:46.760 ","End":"10:49.520","Text":"That we get p.r=0."},{"Start":"10:49.520 ","End":"10:54.290","Text":"That means that if I wanted to bring some charge q tag from"},{"Start":"10:54.290 ","End":"10:59.795","Text":"infinity up until this point over here on my x or y-axis,"},{"Start":"10:59.795 ","End":"11:02.780","Text":"the energy that I\u0027d have to put in,"},{"Start":"11:02.780 ","End":"11:05.720","Text":"in order to do that would be equal to 0."},{"Start":"11:05.720 ","End":"11:08.874","Text":"Because the energy is equal to the charge,"},{"Start":"11:08.874 ","End":"11:12.260","Text":"q tag that I want to bring from infinity to here,"},{"Start":"11:12.260 ","End":"11:16.715","Text":"multiply it by my potential for the electric dipole,"},{"Start":"11:16.715 ","End":"11:20.083","Text":"which would be equal to in this case 0."},{"Start":"11:20.083 ","End":"11:24.090","Text":"That\u0027s the end of this lesson."}],"ID":22324},{"Watched":false,"Name":"Dipole Electric Field","Duration":"7m 55s","ChapterTopicVideoID":21439,"CourseChapterTopicPlaylistID":99476,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.650","Text":"Hello. In this lesson,"},{"Start":"00:01.650 ","End":"00:05.490","Text":"we\u0027re going to speak about the electric field due to a dipole."},{"Start":"00:05.490 ","End":"00:06.930","Text":"In the last lesson,"},{"Start":"00:06.930 ","End":"00:11.640","Text":"we defined what an electric dipole is and that\u0027s 2 charges."},{"Start":"00:11.640 ","End":"00:14.040","Text":"Here specifically, there are 2 point charges"},{"Start":"00:14.040 ","End":"00:16.980","Text":"where 1 has a plus q charge and the other 1 has"},{"Start":"00:16.980 ","End":"00:20.730","Text":"a negative q charge and between the 2 charges there\u0027s"},{"Start":"00:20.730 ","End":"00:25.530","Text":"a distance d. Then we also defined a new term,"},{"Start":"00:25.530 ","End":"00:27.420","Text":"which was this p vector,"},{"Start":"00:27.420 ","End":"00:29.025","Text":"which was a dipole moment,"},{"Start":"00:29.025 ","End":"00:31.245","Text":"which is defined as q,"},{"Start":"00:31.245 ","End":"00:33.565","Text":"one of the charges of the dipole,"},{"Start":"00:33.565 ","End":"00:36.365","Text":"multiplied by the d vector,"},{"Start":"00:36.365 ","End":"00:42.005","Text":"where the d vector is the distance between the 2 charges plus"},{"Start":"00:42.005 ","End":"00:45.485","Text":"the direction that our vector is going from and it always goes"},{"Start":"00:45.485 ","End":"00:50.640","Text":"from a negative charge until a positive charge."},{"Start":"00:50.870 ","End":"00:56.765","Text":"In most questions that you\u0027ll be asked about when you\u0027re dealing with a dipole,"},{"Start":"00:56.765 ","End":"01:00.950","Text":"they\u0027ll ask you what the potential is far away from the dipole."},{"Start":"01:00.950 ","End":"01:03.890","Text":"It\u0027s always going to be far away from the dipole because otherwise,"},{"Start":"01:03.890 ","End":"01:07.870","Text":"you can just find the potential as 2-point charges if you\u0027re close."},{"Start":"01:07.870 ","End":"01:10.070","Text":"We\u0027re always going to be far away from the dipole,"},{"Start":"01:10.070 ","End":"01:14.120","Text":"where our r vector or a distance away from the dipole"},{"Start":"01:14.120 ","End":"01:18.640","Text":"is significantly bigger than the distance between the 2 charges."},{"Start":"01:18.640 ","End":"01:21.740","Text":"In this lesson, we\u0027re going to see how to find"},{"Start":"01:21.740 ","End":"01:26.515","Text":"the electric field far away from the dipole due to the dipole."},{"Start":"01:26.515 ","End":"01:31.640","Text":"This is the equation for the electric field due to a dipole."},{"Start":"01:31.640 ","End":"01:33.680","Text":"It\u0027s a bit weird as we can see,"},{"Start":"01:33.680 ","End":"01:34.819","Text":"it looks a bit complicated,"},{"Start":"01:34.819 ","End":"01:37.685","Text":"but all we have to do is learn how to play"},{"Start":"01:37.685 ","End":"01:42.070","Text":"around with it and how to sub in all of our values."},{"Start":"01:42.070 ","End":"01:44.520","Text":"Let\u0027s write out this equation."},{"Start":"01:44.520 ","End":"01:50.630","Text":"We have the electric field due to a dipole is equal to our k constant multiplied by,"},{"Start":"01:50.630 ","End":"01:52.310","Text":"so these are square brackets, sorry,"},{"Start":"01:52.310 ","End":"01:54.480","Text":"the bottom part somehow cut out,"},{"Start":"01:54.480 ","End":"01:56.735","Text":"so 3 multiplied by,"},{"Start":"01:56.735 ","End":"01:58.745","Text":"so first, we do the brackets,"},{"Start":"01:58.745 ","End":"02:00.505","Text":"this is very important."},{"Start":"02:00.505 ","End":"02:10.470","Text":"We have our dipole moment dot-product with our r-hat and then we multiply this,"},{"Start":"02:10.470 ","End":"02:15.880","Text":"we\u0027ll get a scalar quantity over here in the r-hat direction."},{"Start":"02:15.880 ","End":"02:19.865","Text":"We have a scalar quantity inside"},{"Start":"02:19.865 ","End":"02:26.090","Text":"the brackets and then we get the direction as well, with the i hat,"},{"Start":"02:26.090 ","End":"02:29.365","Text":"and then we minus our dipole moment again,"},{"Start":"02:29.365 ","End":"02:32.795","Text":"and then we divide all of this by r^3,"},{"Start":"02:32.795 ","End":"02:36.110","Text":"which is the magnitude of our distance"},{"Start":"02:36.110 ","End":"02:40.225","Text":"from the center of the dipole until our point in space."},{"Start":"02:40.225 ","End":"02:43.910","Text":"Let\u0027s now give a quick little worked example."},{"Start":"02:43.910 ","End":"02:49.384","Text":"Let\u0027s say that we want to see what our electric field is going to be at this point."},{"Start":"02:49.384 ","End":"02:56.090","Text":"Let\u0027s say that it\u0027s very far away from our dipole,"},{"Start":"02:56.090 ","End":"03:04.970","Text":"so our r is going to be bigger than d and that this point is somewhere along the z-axis,"},{"Start":"03:04.970 ","End":"03:08.425","Text":"and it\u0027s at some arbitrary height z."},{"Start":"03:08.425 ","End":"03:15.140","Text":"First, let\u0027s define for this point over here where we want to find our electric field,"},{"Start":"03:15.140 ","End":"03:17.855","Text":"what our dipole moment is equal to."},{"Start":"03:17.855 ","End":"03:19.975","Text":"Its p is equal to q,"},{"Start":"03:19.975 ","End":"03:24.090","Text":"the charge, multiplied by our d vector."},{"Start":"03:24.090 ","End":"03:27.320","Text":"Our d vector is the vector going"},{"Start":"03:27.320 ","End":"03:30.440","Text":"from the negative charge to the positive charge so it\u0027s in"},{"Start":"03:30.440 ","End":"03:34.520","Text":"the positive z-direction and its magnitude is of course"},{"Start":"03:34.520 ","End":"03:41.370","Text":"d. That is our dipole moment."},{"Start":"03:41.370 ","End":"03:45.290","Text":"Then we have our r vector which is going from the center of"},{"Start":"03:45.290 ","End":"03:50.615","Text":"the dipole until our point in space where we\u0027re measuring our electric field."},{"Start":"03:50.615 ","End":"03:55.715","Text":"Our r vector is simply going to be equal to"},{"Start":"03:55.715 ","End":"04:01.100","Text":"z in the z-direction because we said that we\u0027re measuring at some height z."},{"Start":"04:01.100 ","End":"04:06.545","Text":"Now notice that when we\u0027re dealing with our electric field,"},{"Start":"04:06.545 ","End":"04:12.980","Text":"the equation uses our r-hat rather than our r vector so all"},{"Start":"04:12.980 ","End":"04:19.780","Text":"we have to do is we have to convert our r vector into a unit vector r-hat."},{"Start":"04:19.780 ","End":"04:22.520","Text":"That is simply going to be equal to,"},{"Start":"04:22.520 ","End":"04:25.780","Text":"so it\u0027s going to be 0, 0,"},{"Start":"04:25.780 ","End":"04:30.170","Text":"z, 0 in the x-direction,"},{"Start":"04:30.170 ","End":"04:35.570","Text":"0 in the y-direction, and z in the z-direction divided by the magnitude."},{"Start":"04:35.570 ","End":"04:41.430","Text":"That\u0027s going to be equal to the square root z^2."},{"Start":"04:41.600 ","End":"04:44.485","Text":"If we had x and y components,"},{"Start":"04:44.485 ","End":"04:47.497","Text":"it would be the square root of x^2 plus y^2 plus z^2,"},{"Start":"04:47.497 ","End":"04:49.310","Text":"but we only have the z component."},{"Start":"04:49.310 ","End":"04:52.745","Text":"That\u0027s going to be equal to 0,"},{"Start":"04:52.745 ","End":"04:56.900","Text":"0, z divided by z."},{"Start":"04:56.900 ","End":"05:03.395","Text":"Now we can see that that\u0027s simply going to be equal to 0, 0,"},{"Start":"05:03.395 ","End":"05:05.330","Text":"1 in the z-direction,"},{"Start":"05:05.330 ","End":"05:08.630","Text":"which is simply equal to z-hat,"},{"Start":"05:08.630 ","End":"05:11.635","Text":"or 1 times z-hat, it\u0027s the same thing."},{"Start":"05:11.635 ","End":"05:14.390","Text":"This is our r-hat and now we can just plug this"},{"Start":"05:14.390 ","End":"05:17.020","Text":"in to our equation for our electric field."},{"Start":"05:17.020 ","End":"05:20.645","Text":"We can see that our electric field is equal to"},{"Start":"05:20.645 ","End":"05:27.710","Text":"k multiplied by 3 times our dipole moment,"},{"Start":"05:27.710 ","End":"05:31.445","Text":"which is qdz-hat,"},{"Start":"05:31.445 ","End":"05:36.320","Text":"dot-product with our r-hat, which is z-hat."},{"Start":"05:36.320 ","End":"05:38.990","Text":"Then after the brackets,"},{"Start":"05:38.990 ","End":"05:40.390","Text":"we have our r-hat,"},{"Start":"05:40.390 ","End":"05:45.220","Text":"which we already said our r-hat is z-hat,"},{"Start":"05:46.280 ","End":"05:49.910","Text":"and then minus our dipole moment,"},{"Start":"05:49.910 ","End":"05:55.210","Text":"which is going to again be equal to qdz-hat."},{"Start":"05:55.210 ","End":"06:05.485","Text":"Then we close off r^2 brackets and we divide everything by the magnitude of our r vector."},{"Start":"06:05.485 ","End":"06:09.440","Text":"Here we\u0027ve divided by the magnitude of our r vector."},{"Start":"06:09.440 ","End":"06:12.920","Text":"I\u0027m actually going to give a little note here."},{"Start":"06:12.920 ","End":"06:19.070","Text":"Our r over here is simply equal to the magnitude of our i vector,"},{"Start":"06:19.070 ","End":"06:21.920","Text":"not r-hat because then it would just be divided by 1^3."},{"Start":"06:21.920 ","End":"06:28.850","Text":"Our z.z is simply going to be equal to 1."},{"Start":"06:28.850 ","End":"06:33.305","Text":"What we\u0027re going to have is we\u0027re going to have k multiplied by"},{"Start":"06:33.305 ","End":"06:38.700","Text":"3 times qd times 1 in the z-direction so"},{"Start":"06:38.700 ","End":"06:44.385","Text":"3 times qd in the z-direction minus"},{"Start":"06:44.385 ","End":"06:51.900","Text":"qd in the z-direction divided by z^3."},{"Start":"06:51.900 ","End":"06:57.330","Text":"Therefore, once we minus everything,"},{"Start":"06:57.330 ","End":"07:06.340","Text":"we\u0027ll get that E is equal to 2kqd divided by z^3,"},{"Start":"07:07.670 ","End":"07:11.580","Text":"and of course, our E field is in the z-direction."},{"Start":"07:11.580 ","End":"07:19.435","Text":"What we can see is that the electric field due to a dipole behaves like 1 divided by z^3."},{"Start":"07:19.435 ","End":"07:22.805","Text":"However, the electric field due to a monopole or"},{"Start":"07:22.805 ","End":"07:27.810","Text":"a point charge behaves like 1 divided by z^2."},{"Start":"07:28.940 ","End":"07:34.985","Text":"What we can see is that the electric field and the potential for a dipole,"},{"Start":"07:34.985 ","End":"07:41.705","Text":"the power is 1 power bigger and then 1 divided by that."},{"Start":"07:41.705 ","End":"07:46.085","Text":"We can see that our potential and our electric field"},{"Start":"07:46.085 ","End":"07:48.380","Text":"dissipates quicker when we\u0027re dealing with"},{"Start":"07:48.380 ","End":"07:51.610","Text":"a dipole than when we\u0027re dealing with a monopole."},{"Start":"07:51.610 ","End":"07:54.670","Text":"That\u0027s the end of this lesson."}],"ID":22325},{"Watched":false,"Name":"Exercise 1","Duration":"46m 26s","ChapterTopicVideoID":21440,"CourseChapterTopicPlaylistID":99476,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.875","Text":"Hello, in this lesson,"},{"Start":"00:01.875 ","End":"00:04.760","Text":"we\u0027re going to be answering the following question."},{"Start":"00:04.760 ","End":"00:06.839","Text":"Here we have 2 charges,"},{"Start":"00:06.839 ","End":"00:12.120","Text":"q and negative q that are placed at x is equal a,"},{"Start":"00:12.120 ","End":"00:16.935","Text":"and x is equal negative a over here."},{"Start":"00:16.935 ","End":"00:20.489","Text":"We\u0027ll go straight into question number 1,"},{"Start":"00:20.489 ","End":"00:23.895","Text":"which is to calculate the force acting on a third charge,"},{"Start":"00:23.895 ","End":"00:28.320","Text":"Q, which is placed at some random point x, y, 0."},{"Start":"00:28.320 ","End":"00:32.369","Text":"Now, it\u0027s important to note this line over here"},{"Start":"00:32.369 ","End":"00:36.585","Text":"and this angle of 45 degrees is irrelevant to the next question."},{"Start":"00:36.585 ","End":"00:41.030","Text":"So just ignore everything that\u0027s in red right now and just assume that"},{"Start":"00:41.030 ","End":"00:47.730","Text":"this point Q is that some random place along the x, y plane."},{"Start":"00:47.730 ","End":"00:50.720","Text":"For question number 1,"},{"Start":"00:50.720 ","End":"00:53.615","Text":"we\u0027re trying to find the force on this charge Q."},{"Start":"00:53.615 ","End":"00:57.620","Text":"What we\u0027re going to do is we\u0027re going to use Coulomb\u0027s Law,"},{"Start":"00:57.620 ","End":"01:00.609","Text":"and then we\u0027re going to use superposition."},{"Start":"01:00.609 ","End":"01:07.760","Text":"Coulomb\u0027s Law says that the electric field for a point charge is equal"},{"Start":"01:07.760 ","End":"01:15.865","Text":"to kq divided by r squared in the radial direction."},{"Start":"01:15.865 ","End":"01:18.770","Text":"We\u0027re going to have an electric field for"},{"Start":"01:18.770 ","End":"01:22.115","Text":"this positive charge that we\u0027re going to calculate."},{"Start":"01:22.115 ","End":"01:25.520","Text":"Then we\u0027re also going to find the electric field of"},{"Start":"01:25.520 ","End":"01:31.860","Text":"this negative charge at this exact point over here."},{"Start":"01:31.860 ","End":"01:36.410","Text":"Once we found these 2 electric fields for these 2 charges,"},{"Start":"01:36.410 ","End":"01:37.985","Text":"we\u0027re going to superimpose them."},{"Start":"01:37.985 ","End":"01:39.440","Text":"We\u0027re just going to add them up."},{"Start":"01:39.440 ","End":"01:43.835","Text":"Then we have the electric field at this point,"},{"Start":"01:43.835 ","End":"01:45.965","Text":"and then we just multiply it by Q,"},{"Start":"01:45.965 ","End":"01:48.079","Text":"and then we have my capital Q,"},{"Start":"01:48.079 ","End":"01:51.140","Text":"and then we have the force at this point."},{"Start":"01:51.140 ","End":"01:57.124","Text":"Now, it\u0027s important to note that r is a vector pointing from our charge,"},{"Start":"01:57.124 ","End":"01:59.960","Text":"either positive q or a negative q,"},{"Start":"01:59.960 ","End":"02:03.919","Text":"to a charge Q over here."},{"Start":"02:03.919 ","End":"02:10.382","Text":"We\u0027re going to be using a vector r that looks like so."},{"Start":"02:10.382 ","End":"02:18.640","Text":"This is r plus and this over here is r minus."},{"Start":"02:18.680 ","End":"02:26.200","Text":"Right now we want to find out what our vectors r plus and r minus are."},{"Start":"02:26.210 ","End":"02:31.070","Text":"Here we can see we have this red line that\u0027s pointing to our Q,"},{"Start":"02:31.070 ","End":"02:32.735","Text":"which is somewhere in space."},{"Start":"02:32.735 ","End":"02:34.790","Text":"Let\u0027s call this r line,"},{"Start":"02:34.790 ","End":"02:37.520","Text":"so it\u0027s actually an r vector."},{"Start":"02:37.520 ","End":"02:39.410","Text":"Let\u0027s call it r vector,"},{"Start":"02:39.410 ","End":"02:44.210","Text":"and of course r plus and r minus are also both vectors."},{"Start":"02:44.210 ","End":"02:50.690","Text":"Now let\u0027s define this vector from the origin pointing to this charge over here,"},{"Start":"02:50.690 ","End":"02:56.909","Text":"Q and let\u0027s call this vector r tag."},{"Start":"02:57.160 ","End":"03:03.725","Text":"What we can see is that in order to get this r vector,"},{"Start":"03:03.725 ","End":"03:05.420","Text":"so let\u0027s write this over here,"},{"Start":"03:05.420 ","End":"03:11.314","Text":"what we\u0027re going to do is we\u0027re going to add up our r tag vector,"},{"Start":"03:11.314 ","End":"03:16.294","Text":"plus our r plus vector."},{"Start":"03:16.294 ","End":"03:19.519","Text":"That we can see if we add this vector plus this vector,"},{"Start":"03:19.519 ","End":"03:24.760","Text":"we\u0027re just going to get our r vector from the origin to our point Q."},{"Start":"03:24.760 ","End":"03:28.875","Text":"What I want is I want my r plus vector."},{"Start":"03:28.875 ","End":"03:38.610","Text":"I can write that my r plus vector is equal to my r vector minus my r tag vector."},{"Start":"03:38.610 ","End":"03:42.715","Text":"My r vector is x, y, 0."},{"Start":"03:42.715 ","End":"03:44.720","Text":"Now, we can see that we\u0027re just working on this x,"},{"Start":"03:44.720 ","End":"03:48.380","Text":"y plane, so my z is always going to be equal to 0."},{"Start":"03:48.380 ","End":"03:51.634","Text":"In order to save some space and some writing,"},{"Start":"03:51.634 ","End":"03:53.615","Text":"I\u0027m just going to work in 2D."},{"Start":"03:53.615 ","End":"03:55.430","Text":"Instead of writing x, y 0,"},{"Start":"03:55.430 ","End":"03:56.960","Text":"I\u0027m just going to write x y."},{"Start":"03:56.960 ","End":"04:01.155","Text":"We have x, y minus my r tag vector."},{"Start":"04:01.155 ","End":"04:05.075","Text":"My r tag vector is pointing from the origin until my charge,"},{"Start":"04:05.075 ","End":"04:11.300","Text":"so I know that my charge Q is located at a in the x direction,"},{"Start":"04:11.300 ","End":"04:15.760","Text":"and at 0 in the y direction."},{"Start":"04:16.600 ","End":"04:20.975","Text":"Then we know that when we\u0027re doing vector subtraction,"},{"Start":"04:20.975 ","End":"04:23.710","Text":"we subtract each component separately."},{"Start":"04:23.710 ","End":"04:28.629","Text":"We have the x component minus the x component and the y component minus the y component."},{"Start":"04:28.629 ","End":"04:34.950","Text":"So I have x minus a and then y minus 0, so just y."},{"Start":"04:34.950 ","End":"04:38.679","Text":"This is my r plus vector."},{"Start":"04:38.679 ","End":"04:43.823","Text":"Then we can do a similar thing with the r negative vector."},{"Start":"04:43.823 ","End":"04:47.029","Text":"I\u0027m just going to substitute in my r minus"},{"Start":"04:47.029 ","End":"04:50.569","Text":"vector and that\u0027s just going to be equal to my r vector."},{"Start":"04:50.569 ","End":"04:57.109","Text":"Again, minus this r tag vector over here,"},{"Start":"04:57.109 ","End":"04:59.569","Text":"but for the minus."},{"Start":"04:59.569 ","End":"05:02.459","Text":"Let\u0027s do it in green."},{"Start":"05:02.459 ","End":"05:07.395","Text":"Over here, let\u0027s call this r tag,"},{"Start":"05:07.395 ","End":"05:10.590","Text":"tag, so minus r tag, tag."},{"Start":"05:10.590 ","End":"05:15.455","Text":"Then I\u0027m going to have my r vector is still x,"},{"Start":"05:15.455 ","End":"05:20.070","Text":"y, and then minus my r tag tag,"},{"Start":"05:20.070 ","End":"05:25.434","Text":"so my location of the negative Q charge is at negative a,"},{"Start":"05:25.434 ","End":"05:28.245","Text":"so negative a in the x direction,"},{"Start":"05:28.245 ","End":"05:31.129","Text":"and then again I have 0 in the y direction."},{"Start":"05:31.129 ","End":"05:34.200","Text":"Then when I just do this,"},{"Start":"05:34.200 ","End":"05:36.210","Text":"I\u0027ll have x minus, minus a,"},{"Start":"05:36.210 ","End":"05:38.595","Text":"so that\u0027s x plus a,"},{"Start":"05:38.595 ","End":"05:40.994","Text":"and then y minus 0,"},{"Start":"05:40.994 ","End":"05:43.449","Text":"which is just y."},{"Start":"05:43.960 ","End":"05:47.269","Text":"Now for our Coulomb\u0027s law,"},{"Start":"05:47.269 ","End":"05:53.810","Text":"we have to plug in the size or the magnitude of our r vector."},{"Start":"05:53.810 ","End":"05:58.339","Text":"As we know the magnitude of an r vector is written like this,"},{"Start":"05:58.339 ","End":"06:01.655","Text":"so let\u0027s just write the equation for both of them."},{"Start":"06:01.655 ","End":"06:06.965","Text":"This is the r plus vector and this will represent the r minus vector."},{"Start":"06:06.965 ","End":"06:09.690","Text":"Then we just use Pythagoras,"},{"Start":"06:09.690 ","End":"06:14.340","Text":"so it\u0027s equal to the x component squared, so x."},{"Start":"06:14.340 ","End":"06:17.539","Text":"In 1 side we have minus and 1 we have plus."},{"Start":"06:17.539 ","End":"06:21.410","Text":"The minus a represents the r plus vector,"},{"Start":"06:21.410 ","End":"06:23.970","Text":"so we\u0027ll put minus over here,"},{"Start":"06:23.970 ","End":"06:27.585","Text":"and plus a represents the r minus vector,"},{"Start":"06:27.585 ","End":"06:31.110","Text":"so we\u0027ll put the plus over here a,"},{"Start":"06:31.110 ","End":"06:33.050","Text":"and then this is squared,"},{"Start":"06:33.050 ","End":"06:36.830","Text":"and then we add plus y^2,"},{"Start":"06:36.830 ","End":"06:40.189","Text":"the y components, which were both vectors are the same."},{"Start":"06:40.189 ","End":"06:45.170","Text":"Then we take all of this and we take the square root of it."},{"Start":"06:45.170 ","End":"06:52.259","Text":"The next thing we can see is that we have this r vector equation over here."},{"Start":"06:52.480 ","End":"06:55.880","Text":"Instead of trying to work it out,"},{"Start":"06:55.880 ","End":"07:04.155","Text":"instead of writing Coulomb\u0027s law as e is equal to kq divided by r^2 r hat,"},{"Start":"07:04.155 ","End":"07:07.610","Text":"we can just write it and it\u0027s exactly the same thing,"},{"Start":"07:07.610 ","End":"07:12.910","Text":"kq divided by r^3, r vector."},{"Start":"07:12.910 ","End":"07:15.920","Text":"I\u0027m reminding you that r vector is"},{"Start":"07:15.920 ","End":"07:20.864","Text":"just the magnitude of the r vector in the r hat direction."},{"Start":"07:20.864 ","End":"07:26.720","Text":"We can see that if we substitute into the r vector direction the r hat,"},{"Start":"07:26.720 ","End":"07:28.640","Text":"so we come back to this equation."},{"Start":"07:28.640 ","End":"07:31.564","Text":"We can see that these 2 equations are the same."},{"Start":"07:31.564 ","End":"07:32.840","Text":"But for us right now,"},{"Start":"07:32.840 ","End":"07:35.314","Text":"because we have r vector,"},{"Start":"07:35.314 ","End":"07:41.320","Text":"it\u0027s easier for us to use this instead of trying to calculate our r hat."},{"Start":"07:41.320 ","End":"07:44.539","Text":"Let\u0027s write out the E field."},{"Start":"07:44.539 ","End":"07:49.190","Text":"Let\u0027s write it out first for the positive and then we\u0027ll add in"},{"Start":"07:49.190 ","End":"07:54.185","Text":"the details for the negative E field or the E field for the negative charge."},{"Start":"07:54.185 ","End":"08:01.399","Text":"First of all, we have kq and for the positive charge,"},{"Start":"08:01.399 ","End":"08:02.719","Text":"our q is positive,"},{"Start":"08:02.719 ","End":"08:06.679","Text":"so we can have a positive coefficient over here, kq,"},{"Start":"08:06.679 ","End":"08:14.200","Text":"and then we\u0027ll divide it by our r plus vector."},{"Start":"08:14.200 ","End":"08:17.850","Text":"We have to use the magnitude over here."},{"Start":"08:18.330 ","End":"08:21.335","Text":"We already calculated the magnitude,"},{"Start":"08:21.335 ","End":"08:26.964","Text":"so we have x minus"},{"Start":"08:26.964 ","End":"08:34.120","Text":"a for the r plus vector squared plus y^2."},{"Start":"08:34.120 ","End":"08:40.190","Text":"Then we have to take this, all of this."},{"Start":"08:40.190 ","End":"08:43.069","Text":"First of all it\u0027s to the power of a half of the magnitude,"},{"Start":"08:43.069 ","End":"08:45.995","Text":"but then we have to cube it over here because of this equation."},{"Start":"08:45.995 ","End":"08:47.975","Text":"We have 1/2,"},{"Start":"08:47.975 ","End":"08:51.140","Text":"so we have this to the power of a half and then cubed."},{"Start":"08:51.140 ","End":"08:54.790","Text":"It\u0027s going to be to the power of 3/2."},{"Start":"08:54.790 ","End":"08:58.845","Text":"Then we add in our r vector."},{"Start":"08:58.845 ","End":"09:07.515","Text":"That is going to be x minus a for the r plus, and then y."},{"Start":"09:07.515 ","End":"09:11.194","Text":"Now if we want to incorporate into this same equation,"},{"Start":"09:11.194 ","End":"09:13.245","Text":"our E minus,"},{"Start":"09:13.245 ","End":"09:16.525","Text":"the electric field for the negative charge."},{"Start":"09:16.525 ","End":"09:18.975","Text":"We\u0027re just going to change some details."},{"Start":"09:18.975 ","End":"09:21.560","Text":"First of all it\u0027s a negative q over here,"},{"Start":"09:21.560 ","End":"09:26.075","Text":"so the coefficient of kq will be minus."},{"Start":"09:26.075 ","End":"09:31.290","Text":"Then we know that our r vector is x plus a,"},{"Start":"09:31.290 ","End":"09:36.290","Text":"so we\u0027ll just put a plus in here if we\u0027re going to the minus."},{"Start":"09:36.290 ","End":"09:39.905","Text":"Then we know that the magnitude over here."},{"Start":"09:39.905 ","End":"09:45.840","Text":"In this, we also have x plus a^2, and that\u0027s it."},{"Start":"09:45.840 ","End":"09:50.870","Text":"Now we have our E plus and E minus fields incorporated into here."},{"Start":"09:51.650 ","End":"09:57.140","Text":"Now what we\u0027ll find out is the total electric field at this point."},{"Start":"09:57.140 ","End":"10:01.130","Text":"Then we\u0027re just going to multiply that electric field by this charge Q,"},{"Start":"10:01.130 ","End":"10:02.750","Text":"and then we get the force,"},{"Start":"10:02.750 ","End":"10:04.765","Text":"which is what we\u0027re trying to find."},{"Start":"10:04.765 ","End":"10:06.764","Text":"Let\u0027s write it like so."},{"Start":"10:06.764 ","End":"10:13.200","Text":"The total electric field is simply going to be equal to E plus,"},{"Start":"10:13.200 ","End":"10:16.930","Text":"plus our E minus."},{"Start":"10:16.970 ","End":"10:23.764","Text":"Our E plus is simply kq in the x minus"},{"Start":"10:23.764 ","End":"10:32.935","Text":"a y direction divided by x minus a^2 plus y^2,"},{"Start":"10:32.935 ","End":"10:36.076","Text":"and all of this to the power of 3/2."},{"Start":"10:36.076 ","End":"10:39.124","Text":"Let\u0027s just scroll over here."},{"Start":"10:39.124 ","End":"10:40.850","Text":"Then we\u0027re adding the E minus,"},{"Start":"10:40.850 ","End":"10:47.434","Text":"so then we have negative kq in the x plus a,"},{"Start":"10:47.434 ","End":"10:53.659","Text":"y direction divided by x"},{"Start":"10:53.659 ","End":"11:01.639","Text":"plus a^2 plus y^2 to the power of 3/2."},{"Start":"11:01.639 ","End":"11:04.235","Text":"Now we can simplify this a little."},{"Start":"11:04.235 ","End":"11:08.495","Text":"We can see that this is equal to kq,"},{"Start":"11:08.495 ","End":"11:10.865","Text":"which is a common multiple."},{"Start":"11:10.865 ","End":"11:14.740","Text":"Then we have this over here."},{"Start":"11:14.780 ","End":"11:19.355","Text":"Let\u0027s do the x direction first."},{"Start":"11:19.355 ","End":"11:21.935","Text":"Let\u0027s do it like this."},{"Start":"11:21.935 ","End":"11:27.500","Text":"We have x minus a divided by"},{"Start":"11:27.500 ","End":"11:35.940","Text":"x minus a^2 plus y^2 to the power of 3/2."},{"Start":"11:35.940 ","End":"11:41.225","Text":"Then we also have over here x plus a in the x direction."},{"Start":"11:41.225 ","End":"11:49.499","Text":"Then we can have minus and then we have x plus a."},{"Start":"11:49.499 ","End":"11:59.860","Text":"We can just write this as minus x plus a."},{"Start":"11:59.860 ","End":"12:01.465","Text":"Don\u0027t change that to a plus."},{"Start":"12:01.465 ","End":"12:06.774","Text":"Divided by, here we can see it\u0027s x plus a^2"},{"Start":"12:06.774 ","End":"12:12.025","Text":"plus y^2 to the power of 3/2."},{"Start":"12:12.025 ","End":"12:14.905","Text":"All of this is in the x hat direction."},{"Start":"12:14.905 ","End":"12:19.120","Text":"Then we\u0027re going to add in the y direction."},{"Start":"12:19.120 ","End":"12:22.045","Text":"The y direction is this,"},{"Start":"12:22.045 ","End":"12:26.665","Text":"y in the y direction,"},{"Start":"12:26.665 ","End":"12:33.955","Text":"divided by x minus a^2 plus y^2."},{"Start":"12:33.955 ","End":"12:36.925","Text":"All of this is to the power of 3/2."},{"Start":"12:36.925 ","End":"12:41.650","Text":"Then again, we have this minus coefficient which goes on over here."},{"Start":"12:41.650 ","End":"12:48.249","Text":"Minus y divided by x plus a^2"},{"Start":"12:48.249 ","End":"12:56.080","Text":"plus y^2 to the power of 3/2 in the y direction."},{"Start":"12:56.080 ","End":"13:02.680","Text":"This is our total electric field."},{"Start":"13:02.680 ","End":"13:07.119","Text":"Then in order to find the force,"},{"Start":"13:07.119 ","End":"13:13.390","Text":"this is simply equal to q multiplied by this total electric field."},{"Start":"13:13.390 ","End":"13:16.464","Text":"In that case, if you want,"},{"Start":"13:16.464 ","End":"13:19.555","Text":"so I can just erase this."},{"Start":"13:19.555 ","End":"13:25.555","Text":"I\u0027ll just add in an f over here and multiply this by q."},{"Start":"13:25.555 ","End":"13:33.890","Text":"Then what we get is that this is our final answer."},{"Start":"13:34.260 ","End":"13:38.305","Text":"This was our answer to Question 1."},{"Start":"13:38.305 ","End":"13:42.970","Text":"Now, let\u0027s answer Question 2."},{"Start":"13:42.970 ","End":"13:50.649","Text":"Question 2 is that we assume that the distance between Q and the origin,"},{"Start":"13:50.649 ","End":"13:52.719","Text":"this distance over here,"},{"Start":"13:52.719 ","End":"13:56.680","Text":"is much larger than the distance between the other 2 charges."},{"Start":"13:56.680 ","End":"14:00.430","Text":"The distance between the other 2 charges is just 2a."},{"Start":"14:00.430 ","End":"14:05.724","Text":"What we\u0027re in actual fact doing is assuming that"},{"Start":"14:05.724 ","End":"14:11.740","Text":"the magnitude of our r vector is much bigger than 2a."},{"Start":"14:13.250 ","End":"14:18.159","Text":"Then we\u0027re being told over here in this diagram that"},{"Start":"14:18.159 ","End":"14:22.779","Text":"the angle between Q and the x-axis is 45 degrees, see diagram."},{"Start":"14:22.779 ","End":"14:24.459","Text":"Now we\u0027re using this diagram."},{"Start":"14:24.459 ","End":"14:27.820","Text":"Now what we\u0027re going to do is we\u0027re going to use our answer to"},{"Start":"14:27.820 ","End":"14:32.420","Text":"1 to calculate the force on Q."},{"Start":"14:33.810 ","End":"14:37.135","Text":"Because here we\u0027re given an angle,"},{"Start":"14:37.135 ","End":"14:40.690","Text":"it\u0027s a good idea and a good clue for you to change"},{"Start":"14:40.690 ","End":"14:44.560","Text":"from your Cartesian coordinates to polar coordinates."},{"Start":"14:44.560 ","End":"14:49.210","Text":"Give you x and y in terms of r and θ."},{"Start":"14:49.210 ","End":"15:00.745","Text":"We know that x=r cosine(θ) and y=r sine(θ)."},{"Start":"15:00.745 ","End":"15:05.695","Text":"Of course, rθ specifically in this question is 45 degrees."},{"Start":"15:05.695 ","End":"15:09.670","Text":"But I\u0027m going to write out the whole equation as general as"},{"Start":"15:09.670 ","End":"15:14.079","Text":"possible using just a standard θ and then at the end,"},{"Start":"15:14.079 ","End":"15:17.350","Text":"I\u0027m going to substitute in 45 degrees so that we see"},{"Start":"15:17.350 ","End":"15:22.160","Text":"that we get the specific answer for this case over here."},{"Start":"15:22.620 ","End":"15:28.480","Text":"I\u0027ve written out the electric field that we calculated in Question 1."},{"Start":"15:28.480 ","End":"15:31.795","Text":"It\u0027s just, we take out the capital Q,"},{"Start":"15:31.795 ","End":"15:33.685","Text":"and this is our electric field."},{"Start":"15:33.685 ","End":"15:37.360","Text":"If we\u0027re assuming that the magnitude of r is much greater than 2a,"},{"Start":"15:37.360 ","End":"15:40.539","Text":"that means that we can also"},{"Start":"15:40.539 ","End":"15:48.535","Text":"write that the magnitude of r vector is much greater than a."},{"Start":"15:48.535 ","End":"15:52.750","Text":"Our r vector is made up of x and y components."},{"Start":"15:52.750 ","End":"15:55.615","Text":"If we\u0027re saying that our r is much greater than a,"},{"Start":"15:55.615 ","End":"16:01.255","Text":"then we\u0027re also saying that our x under y is much greater than a."},{"Start":"16:01.255 ","End":"16:04.820","Text":"We can just write this."},{"Start":"16:05.010 ","End":"16:09.280","Text":"Now what we want to test is we want to see if,"},{"Start":"16:09.280 ","End":"16:13.130","Text":"if we cancel out this a everywhere,"},{"Start":"16:13.130 ","End":"16:16.820","Text":"i.e., we\u0027re using the zeroth order."},{"Start":"16:16.820 ","End":"16:20.380","Text":"If we\u0027re going to get some electric field."},{"Start":"16:20.380 ","End":"16:23.319","Text":"Because if our x is much larger than a,"},{"Start":"16:23.319 ","End":"16:24.550","Text":"let\u0027s say, if x is,"},{"Start":"16:24.550 ","End":"16:27.430","Text":"let\u0027s say a million and a is 1."},{"Start":"16:27.430 ","End":"16:29.559","Text":"A million divided by everything,"},{"Start":"16:29.559 ","End":"16:34.780","Text":"or a million plus 1 divided by the same denominator doesn\u0027t really make a difference."},{"Start":"16:34.780 ","End":"16:38.335","Text":"We can assume that a is approximately equal to 0."},{"Start":"16:38.335 ","End":"16:41.900","Text":"Then let\u0027s see what happens when we cross it out."},{"Start":"16:41.900 ","End":"16:45.870","Text":"This a will disappear and this a will disappear."},{"Start":"16:45.870 ","End":"16:48.930","Text":"Similarly, this a in the denominator and this a in"},{"Start":"16:48.930 ","End":"16:54.250","Text":"the denominator and here and here is also going to cross out."},{"Start":"16:54.270 ","End":"17:00.310","Text":"Now we see that we have x divided by x^2 plus y^2 to the power of"},{"Start":"17:00.310 ","End":"17:07.015","Text":"3/2 minus x divided by x^2 plus y^2 to the power of 3/2."},{"Start":"17:07.015 ","End":"17:12.280","Text":"The electric field in the x direction is going to completely cancel out."},{"Start":"17:12.280 ","End":"17:16.929","Text":"Then we have y divided by x^2 plus y^2 to the power of 3/2 minus"},{"Start":"17:16.929 ","End":"17:22.120","Text":"y divided by x^2 plus y^2 to the power of 3/2 in the y direction."},{"Start":"17:22.120 ","End":"17:27.710","Text":"Now we can see that the y components of the electric field also cancels out."},{"Start":"17:28.380 ","End":"17:36.670","Text":"That means it doesn\u0027t make sense that we get this E field that is equal to 0."},{"Start":"17:36.670 ","End":"17:38.499","Text":"Because we know that we have charges,"},{"Start":"17:38.499 ","End":"17:40.465","Text":"we\u0027re going to have some E field."},{"Start":"17:40.465 ","End":"17:47.560","Text":"That means that we can\u0027t use this first-order."},{"Start":"17:47.560 ","End":"17:51.250","Text":"We\u0027re going to have to go to higher orders in order to get"},{"Start":"17:51.250 ","End":"17:57.170","Text":"some tiny E field at this point over here."},{"Start":"17:58.200 ","End":"18:03.850","Text":"Now we\u0027re going to move and try and do a first-order approximation."},{"Start":"18:03.850 ","End":"18:07.540","Text":"This is the idea of using the Taylor series approximation."},{"Start":"18:07.540 ","End":"18:12.400","Text":"Let\u0027s just look at this function over here."},{"Start":"18:12.400 ","End":"18:18.385","Text":"x plus a divided by x plus a^2 plus y^2 to the power of 3/2."},{"Start":"18:18.385 ","End":"18:20.650","Text":"When I do this approximation,"},{"Start":"18:20.650 ","End":"18:26.390","Text":"I\u0027m just going to use the first-order approximation on the denominator."},{"Start":"18:26.730 ","End":"18:29.650","Text":"This is something that is allowed,"},{"Start":"18:29.650 ","End":"18:33.955","Text":"you\u0027re allowed to use approximations just on parts of the function."},{"Start":"18:33.955 ","End":"18:37.360","Text":"We\u0027re going to approximate the denominator."},{"Start":"18:37.360 ","End":"18:43.609","Text":"What we have is 1 divided by x"},{"Start":"18:43.609 ","End":"18:51.010","Text":"plus a^2 plus y^2 and all of this is to the power of 3/2."},{"Start":"18:51.010 ","End":"18:56.230","Text":"Now what I\u0027m going to do is I\u0027m going to open up this brackets."},{"Start":"18:56.230 ","End":"19:06.190","Text":"What I\u0027m going to have is x^2 plus 2ax plus a^2 plus y^2."},{"Start":"19:06.190 ","End":"19:12.070","Text":"All of this to the power of 3/2."},{"Start":"19:12.070 ","End":"19:20.350","Text":"Now I\u0027m going to remember that r^2=x^2 plus y^2."},{"Start":"19:20.350 ","End":"19:21.760","Text":"This is a definition."},{"Start":"19:21.760 ","End":"19:23.470","Text":"You\u0027ve already seen this before."},{"Start":"19:23.470 ","End":"19:26.380","Text":"Here I have x^2 and y^2."},{"Start":"19:26.380 ","End":"19:33.295","Text":"I\u0027m going to rewrite this over here."},{"Start":"19:33.295 ","End":"19:40.224","Text":"I have 1 divided by x^2 plus y^2 is r^2 plus"},{"Start":"19:40.224 ","End":"19:47.725","Text":"2ax plus a^2 to the power of 3/2."},{"Start":"19:47.725 ","End":"19:53.375","Text":"Then I\u0027m also going to remember that x=r cos θ."},{"Start":"19:53.375 ","End":"19:59.170","Text":"Then I\u0027m going to write this as r^2 plus 2a and"},{"Start":"19:59.170 ","End":"20:06.160","Text":"then x(r) cos of θ plus a^2."},{"Start":"20:06.160 ","End":"20:11.165","Text":"All of this to the power of 3/2."},{"Start":"20:11.165 ","End":"20:18.804","Text":"Now a common practice is to take the common factor out,"},{"Start":"20:18.804 ","End":"20:22.929","Text":"where we choose the common factor to be the largest value."},{"Start":"20:22.929 ","End":"20:27.459","Text":"Here we\u0027re being told that r is much bigger than a so r is"},{"Start":"20:27.459 ","End":"20:34.000","Text":"the largest value so we\u0027re going to take r^2 out."},{"Start":"20:34.000 ","End":"20:37.236","Text":"We\u0027re going to have r^2 over here and all"},{"Start":"20:37.236 ","End":"20:40.599","Text":"this is multiplied by so now we just substitute this in."},{"Start":"20:40.599 ","End":"20:43.334","Text":"We have 1 plus,"},{"Start":"20:43.334 ","End":"20:49.419","Text":"here we\u0027re going to have 2a cosine of Theta divided"},{"Start":"20:49.419 ","End":"20:56.110","Text":"by r. If I multiply this value by r^2 I\u0027ll get back to 2ar cosine of Theta,"},{"Start":"20:56.110 ","End":"21:01.390","Text":"plus a divided by r^2."},{"Start":"21:01.390 ","End":"21:06.740","Text":"All of this is of course to the power of 3/2."},{"Start":"21:07.740 ","End":"21:11.140","Text":"Just add another bracket in there."},{"Start":"21:11.140 ","End":"21:20.814","Text":"If we know that a is much smaller than r and if we divide both sides by r,"},{"Start":"21:20.814 ","End":"21:25.435","Text":"we\u0027ll get that a divided by r is much smaller than 1."},{"Start":"21:25.435 ","End":"21:31.840","Text":"What we have over here where we have 2a cosine of Theta divided by r we know that,"},{"Start":"21:31.840 ","End":"21:36.280","Text":"that is going to be much smaller than 1 and definitely a divided"},{"Start":"21:36.280 ","End":"21:41.755","Text":"by r which is a very small number squared is going to be even smaller."},{"Start":"21:41.755 ","End":"21:46.194","Text":"Because I\u0027m working with a first-order approximation,"},{"Start":"21:46.194 ","End":"21:52.225","Text":"anything where I have a divided by r is considered first-order."},{"Start":"21:52.225 ","End":"21:55.090","Text":"Something like this I won\u0027t cancel out because I have"},{"Start":"21:55.090 ","End":"21:58.630","Text":"2a cosine of Theta divided by r which is first-order."},{"Start":"21:58.630 ","End":"22:02.530","Text":"However, a divided by r^2 is"},{"Start":"22:02.530 ","End":"22:07.164","Text":"already in the second-order realm for second-order approximations."},{"Start":"22:07.164 ","End":"22:11.590","Text":"Because I\u0027m only using an accuracy of first-order I can cross"},{"Start":"22:11.590 ","End":"22:15.910","Text":"out all the values that are of second-order or lower,"},{"Start":"22:15.910 ","End":"22:20.680","Text":"or so if I had a divided by r^3 so I would also cross that out."},{"Start":"22:20.680 ","End":"22:26.169","Text":"Now it\u0027s important to note if I use the first-order approximation and again I"},{"Start":"22:26.169 ","End":"22:32.200","Text":"get that my total E field when I use the first-order approximation is equal to 0,"},{"Start":"22:32.200 ","End":"22:37.179","Text":"like we saw in my zeroth-order approximation then I will have to use"},{"Start":"22:37.179 ","End":"22:40.329","Text":"my second-order approximation in which case I won\u0027t be able to"},{"Start":"22:40.329 ","End":"22:44.665","Text":"cancel out things such as a divided by r^2."},{"Start":"22:44.665 ","End":"22:46.810","Text":"If I\u0027m on my second-order approximation,"},{"Start":"22:46.810 ","End":"22:53.584","Text":"if I would have a divided by r^3 for instance so that I could cross out."},{"Start":"22:53.584 ","End":"22:57.520","Text":"It\u0027s important to remember that."},{"Start":"22:57.770 ","End":"23:00.089","Text":"Now, let\u0027s carry on."},{"Start":"23:00.089 ","End":"23:05.479","Text":"I\u0027ll scroll down to get a little bit more space and let\u0027s just rub this out."},{"Start":"23:05.479 ","End":"23:09.170","Text":"Now, let\u0027s carry on with this."},{"Start":"23:09.420 ","End":"23:16.315","Text":"First of all I can see that over here I have r^2 and it is also to the power of 3/2."},{"Start":"23:16.315 ","End":"23:21.475","Text":"R^2 to the power of 3/2 is like r^3,"},{"Start":"23:21.475 ","End":"23:25.525","Text":"but of course it\u0027s in the denominator so it\u0027s 1 divided by r^3."},{"Start":"23:25.525 ","End":"23:30.459","Text":"Then what I want to do is what I want to write what\u0027s in this brackets"},{"Start":"23:30.459 ","End":"23:33.010","Text":"but instead of writing it in the denominator I want"},{"Start":"23:33.010 ","End":"23:36.130","Text":"to write it as if it\u0027s in the numerator."},{"Start":"23:36.130 ","End":"23:45.385","Text":"What I\u0027m going to do is I\u0027m going to write 1 plus 2a cosine of Theta divided by r,"},{"Start":"23:45.385 ","End":"23:51.715","Text":"came not crossing anything out yet plus a divided by r^2."},{"Start":"23:51.715 ","End":"23:56.604","Text":"Then because all of this in the denominator was to the power of 3/2"},{"Start":"23:56.604 ","End":"24:01.660","Text":"so in the numerator it\u0027s to the power of negative 3/2."},{"Start":"24:01.660 ","End":"24:05.499","Text":"This is the exact same thing as what\u0027s written here just in a"},{"Start":"24:05.499 ","End":"24:09.559","Text":"slightly more simple to understand way."},{"Start":"24:10.140 ","End":"24:12.820","Text":"Now I\u0027ll cross this out,"},{"Start":"24:12.820 ","End":"24:15.729","Text":"because it\u0027s a very small thing,"},{"Start":"24:15.729 ","End":"24:18.549","Text":"and it doesn\u0027t match our first-order approximation it\u0027s"},{"Start":"24:18.549 ","End":"24:22.420","Text":"already second-order and I can say it\u0027s approaching 0."},{"Start":"24:22.420 ","End":"24:25.284","Text":"Now we\u0027re going to use this equation,"},{"Start":"24:25.284 ","End":"24:30.055","Text":"this approximation that we use often so please remember it and write it down."},{"Start":"24:30.055 ","End":"24:37.105","Text":"It says that if we have 1 plus x to the power of some thing, so Alpha."},{"Start":"24:37.105 ","End":"24:44.595","Text":"Where x has to be significantly smaller than 1,"},{"Start":"24:44.595 ","End":"24:48.314","Text":"so we have 1 plus a very small number to the power of Alpha."},{"Start":"24:48.314 ","End":"24:51.149","Text":"We can say that this is approximately equal to"},{"Start":"24:51.149 ","End":"24:54.530","Text":"if we\u0027re going to first order approximations,"},{"Start":"24:54.530 ","End":"24:57.430","Text":"1 plus Alpha x."},{"Start":"24:57.430 ","End":"25:02.180","Text":"If we were using the zeroth-order we\u0027ll just keep this 1,"},{"Start":"25:02.400 ","End":"25:08.720","Text":"and if we going to the first-order we have plus Alpha x."},{"Start":"25:09.210 ","End":"25:16.089","Text":"Using this equation over here I can therefore say that this is"},{"Start":"25:16.089 ","End":"25:22.930","Text":"approximately equal to so let\u0027s just say that this is equal to our x over here."},{"Start":"25:22.930 ","End":"25:26.965","Text":"We know that this is significantly smaller than 1 because we know that"},{"Start":"25:26.965 ","End":"25:30.805","Text":"a divided by r is significantly smaller than 1."},{"Start":"25:30.805 ","End":"25:33.894","Text":"We saw that just a few moments ago."},{"Start":"25:33.894 ","End":"25:40.150","Text":"Then, my 1 divided by r^3 remains and then it is going to be"},{"Start":"25:40.150 ","End":"25:47.125","Text":"multiplied by this 1 plus Alpha x so plus,"},{"Start":"25:47.125 ","End":"25:50.920","Text":"so our alpha is negative 3/2,"},{"Start":"25:50.920 ","End":"25:55.150","Text":"that\u0027s our power and then our x is this,"},{"Start":"25:55.150 ","End":"26:02.810","Text":"so 2a cosine of Theta divided by r."},{"Start":"26:03.570 ","End":"26:08.860","Text":"Then of course we can cancel out these 2s and"},{"Start":"26:08.860 ","End":"26:14.810","Text":"there we have our first-order approximation for the denominator."},{"Start":"26:16.980 ","End":"26:22.210","Text":"This approximation is of course for the denominator which"},{"Start":"26:22.210 ","End":"26:26.800","Text":"is x plus a^2 plus y^2 to the power of 3/2."},{"Start":"26:26.800 ","End":"26:29.635","Text":"However, we also have this denominator,"},{"Start":"26:29.635 ","End":"26:34.975","Text":"which is within x minus a instead of x plus a."},{"Start":"26:34.975 ","End":"26:36.969","Text":"Another way that we can write this,"},{"Start":"26:36.969 ","End":"26:41.019","Text":"so if we have 1 divided by x minus"},{"Start":"26:41.019 ","End":"26:47.980","Text":"a^2 plus y^2 to the power of 3/2."},{"Start":"26:47.980 ","End":"26:51.370","Text":"The exact way to write this is just,"},{"Start":"26:51.370 ","End":"27:01.915","Text":"we can write x plus minus a^2 plus y^2 to the power of 3/2."},{"Start":"27:01.915 ","End":"27:04.795","Text":"This means the exact same thing."},{"Start":"27:04.795 ","End":"27:08.620","Text":"Then, using this we can clearly see that we have"},{"Start":"27:08.620 ","End":"27:12.895","Text":"the exact same equation that we had over here."},{"Start":"27:12.895 ","End":"27:16.810","Text":"The only difference is that instead of having this a,"},{"Start":"27:16.810 ","End":"27:21.650","Text":"we\u0027ll substitute everywhere we see this a with just minus a."},{"Start":"27:22.680 ","End":"27:26.619","Text":"Therefore, that means that this will be approximately equal"},{"Start":"27:26.619 ","End":"27:30.550","Text":"to the same as we got here for the x plus a so 1"},{"Start":"27:30.550 ","End":"27:38.680","Text":"divided by r^3 and then we have 1 plus and then we have negative 3 divided by 2,"},{"Start":"27:38.680 ","End":"27:40.479","Text":"we\u0027ll cross it out in a second."},{"Start":"27:40.479 ","End":"27:43.900","Text":"Then, we\u0027ll have instead of multiplied by"},{"Start":"27:43.900 ","End":"27:49.660","Text":"2a cosine of Theta it will be multiplied by 2 multiplied by"},{"Start":"27:49.660 ","End":"27:55.945","Text":"negative a multiplied by cosine of Theta divided by"},{"Start":"27:55.945 ","End":"28:01.570","Text":"r. Then again we\u0027ll"},{"Start":"28:01.570 ","End":"28:07.075","Text":"get that these 2s cancel out and of course this negative and this negative cancel out,"},{"Start":"28:07.075 ","End":"28:09.805","Text":"so we\u0027ll get that this is positive."},{"Start":"28:09.805 ","End":"28:16.450","Text":"Then we\u0027ll get that this is just equal to 1 divided by r^3,"},{"Start":"28:16.450 ","End":"28:24.340","Text":"multiplied by 1 plus 3a cosine Theta"},{"Start":"28:24.340 ","End":"28:31.144","Text":"divided by r. Now here,"},{"Start":"28:31.144 ","End":"28:39.025","Text":"we have the approximation for where we have x plus a somewhere in the denominator,"},{"Start":"28:39.025 ","End":"28:44.608","Text":"and here we have the approximation for if we have x minus a."},{"Start":"28:44.608 ","End":"28:46.473","Text":"Let\u0027s scroll down."},{"Start":"28:46.473 ","End":"28:49.273","Text":"Now let\u0027s write the e-field."},{"Start":"28:49.273 ","End":"28:54.537","Text":"Let\u0027s just draw a line here to make it somewhat neater."},{"Start":"28:54.537 ","End":"29:03.108","Text":"Now we can say that the total e-field with the approximation is approximately equal to."},{"Start":"29:03.108 ","End":"29:08.526","Text":"If we remember or we have it written down in front of you,"},{"Start":"29:08.526 ","End":"29:11.597","Text":"so we have kq as a common factor."},{"Start":"29:11.597 ","End":"29:14.084","Text":"Then we can also take out,"},{"Start":"29:14.084 ","End":"29:17.642","Text":"we have this 1 divided by r^3 in both,"},{"Start":"29:17.642 ","End":"29:20.855","Text":"so we can take that out over here."},{"Start":"29:20.855 ","End":"29:24.072","Text":"Now we can do our approximation."},{"Start":"29:24.072 ","End":"29:25.743","Text":"If you remember,"},{"Start":"29:25.743 ","End":"29:27.774","Text":"we had this x minus a,"},{"Start":"29:27.774 ","End":"29:30.647","Text":"so x minus a in the numerator."},{"Start":"29:30.647 ","End":"29:35.446","Text":"This is going to be multiplied by this over here,"},{"Start":"29:35.446 ","End":"29:37.611","Text":"so this one over here."},{"Start":"29:37.611 ","End":"29:44.971","Text":"We\u0027re going to multiply this by 1 plus because we already took out the r^3,"},{"Start":"29:44.971 ","End":"29:52.609","Text":"1 plus 3a cosine of Theta divided by r because we had x minus a divided by this."},{"Start":"29:52.609 ","End":"29:55.772","Text":"Then we\u0027re going to have minus,"},{"Start":"29:55.772 ","End":"29:58.334","Text":"and then we had an x plus a."},{"Start":"29:58.334 ","End":"30:03.107","Text":"Then this x plus a was divided by this over here,"},{"Start":"30:03.107 ","End":"30:06.915","Text":"x plus a, which we saw as equal to this."},{"Start":"30:06.915 ","End":"30:10.979","Text":"We already took out the 1 divided by r^3."},{"Start":"30:10.979 ","End":"30:16.293","Text":"This will be multiplied by 1 minus 3a cosine of Theta"},{"Start":"30:16.293 ","End":"30:22.127","Text":"divided by r. All of this is of course in the x-direction."},{"Start":"30:22.127 ","End":"30:28.613","Text":"This is all in the x direction and then we have the y-direction."},{"Start":"30:28.613 ","End":"30:30.738","Text":"We can take out the y."},{"Start":"30:30.738 ","End":"30:33.622","Text":"If we go back to the e-field,"},{"Start":"30:33.622 ","End":"30:38.023","Text":"so we just remember what we\u0027re looking at."},{"Start":"30:38.023 ","End":"30:40.039","Text":"We have here y and y,"},{"Start":"30:40.039 ","End":"30:44.339","Text":"so we can take that out as a common factor."},{"Start":"30:44.339 ","End":"30:47.510","Text":"We can take that out over here."},{"Start":"30:47.510 ","End":"30:50.561","Text":"First, we had this over here,"},{"Start":"30:50.561 ","End":"30:52.218","Text":"so we have this,"},{"Start":"30:52.218 ","End":"30:56.435","Text":"so 1 plus 3a cosine of Theta divided by r,"},{"Start":"30:56.435 ","End":"30:58.870","Text":"and then we have minus."},{"Start":"30:58.870 ","End":"31:03.240","Text":"I\u0027m just reminding you I have the e-field"},{"Start":"31:03.240 ","End":"31:08.272","Text":"right in front of me so if you can\u0027t remember it,"},{"Start":"31:08.272 ","End":"31:10.860","Text":"copy this down quickly."},{"Start":"31:10.860 ","End":"31:12.936","Text":"Then we had minus y,"},{"Start":"31:12.936 ","End":"31:15.795","Text":"which we already took out,"},{"Start":"31:15.795 ","End":"31:18.791","Text":"and then we have our x plus a."},{"Start":"31:18.791 ","End":"31:24.676","Text":"That means we\u0027re using this approximation over here."},{"Start":"31:24.676 ","End":"31:32.298","Text":"We\u0027re going to have minus 1 minus 3a cosine of Theta divided by r. Then,"},{"Start":"31:32.298 ","End":"31:37.058","Text":"of course, all of this is in the y direction."},{"Start":"31:37.058 ","End":"31:41.708","Text":"Now we can just simplify this a little bit."},{"Start":"31:41.708 ","End":"31:44.775","Text":"We can open up the brackets."},{"Start":"31:44.775 ","End":"31:48.119","Text":"Here we have x multiplied by 1,"},{"Start":"31:48.119 ","End":"31:53.491","Text":"but then we have here negative x multiplied by 1."},{"Start":"31:53.491 ","End":"31:56.861","Text":"Those are going to cancel out."},{"Start":"31:56.861 ","End":"32:04.317","Text":"Then what we\u0027re going to be left with is negative a multiplied by 1."},{"Start":"32:04.317 ","End":"32:09.363","Text":"First, let\u0027s write out our kq divided by r^3."},{"Start":"32:09.363 ","End":"32:13.476","Text":"Then we saw that we have negative 2a."},{"Start":"32:13.476 ","End":"32:15.952","Text":"Let\u0027s write that out."},{"Start":"32:15.952 ","End":"32:18.295","Text":"We have negative 2a."},{"Start":"32:18.295 ","End":"32:27.451","Text":"Then we have x multiplied by 3a cosine of Theta divided by r minus x multiplied by"},{"Start":"32:27.451 ","End":"32:32.764","Text":"minus 3a cosine of Theta divided by r. We have x"},{"Start":"32:32.764 ","End":"32:38.397","Text":"times 3a times this minus x multiplied by minus,"},{"Start":"32:38.397 ","End":"32:40.803","Text":"so we have another x."},{"Start":"32:40.803 ","End":"32:45.919","Text":"Then we\u0027re going to add plus 2 multiplied by"},{"Start":"32:45.919 ","End":"32:53.080","Text":"3ax cosine of Theta divided by r. If you didn\u0027t see what I did,"},{"Start":"32:53.080 ","End":"32:58.136","Text":"please just work it out on a piece of paper."},{"Start":"32:58.136 ","End":"33:01.044","Text":"I just did some algebra."},{"Start":"33:01.044 ","End":"33:05.716","Text":"That\u0027s all we have in the x-direction."},{"Start":"33:05.716 ","End":"33:09.825","Text":"Now let\u0027s look in the y direction."},{"Start":"33:09.825 ","End":"33:13.070","Text":"let\u0027s put a plus over here."},{"Start":"33:13.070 ","End":"33:15.292","Text":"Then we have here,"},{"Start":"33:15.292 ","End":"33:16.480","Text":"1 minus 1,"},{"Start":"33:16.480 ","End":"33:19.014","Text":"so that cancels out."},{"Start":"33:19.014 ","End":"33:25.011","Text":"then we have 3a cosine of Theta divided by r minus"},{"Start":"33:25.011 ","End":"33:31.674","Text":"minus 3a cosine of Theta divided by r, so that\u0027s plus."},{"Start":"33:31.674 ","End":"33:34.726","Text":"Here we just have plus 2,"},{"Start":"33:34.726 ","End":"33:38.104","Text":"and then we can put in the y."},{"Start":"33:38.104 ","End":"33:44.778","Text":"We have plus 2 times 3ay cosine of Theta divided by r,"},{"Start":"33:44.778 ","End":"33:48.853","Text":"and this is in the y-direction."},{"Start":"33:48.853 ","End":"33:56.444","Text":"Then this is basically what we get as an approximation for"},{"Start":"33:56.444 ","End":"34:04.459","Text":"our total e-field when a charge Q is far away from the origin,"},{"Start":"34:04.459 ","End":"34:11.954","Text":"or when we saw that our r is significantly bigger than a."},{"Start":"34:11.954 ","End":"34:14.915","Text":"Now before we finish,"},{"Start":"34:14.915 ","End":"34:19.559","Text":"we have to substitute in r Thetas."},{"Start":"34:19.559 ","End":"34:26.109","Text":"We remember that x is equal to r cosine of Theta,"},{"Start":"34:26.109 ","End":"34:31.236","Text":"and that y is equal to r sine of Theta."},{"Start":"34:31.236 ","End":"34:35.918","Text":"X divided by r as cosine of Theta,"},{"Start":"34:35.918 ","End":"34:44.602","Text":"so here we have x divided by r. Y divided by r is sine of Theta,"},{"Start":"34:44.602 ","End":"34:49.137","Text":"and here we have y divided by r."},{"Start":"34:49.137 ","End":"34:54.858","Text":"We\u0027re going to have kq divided by r^3."},{"Start":"34:54.858 ","End":"35:01.443","Text":"Then we\u0027re going to have here negative 2a,"},{"Start":"35:01.443 ","End":"35:04.887","Text":"and then we have plus,"},{"Start":"35:04.887 ","End":"35:12.719","Text":"and then we have 2 times 3a cosine squared Theta."},{"Start":"35:12.719 ","End":"35:17.134","Text":"When we plug in 45 degrees,"},{"Start":"35:17.134 ","End":"35:27.307","Text":"we\u0027re just going to get that this is equal to 3a divided by 2."},{"Start":"35:27.307 ","End":"35:30.328","Text":"Then, of course,"},{"Start":"35:30.328 ","End":"35:34.060","Text":"these 2s cancel out."},{"Start":"35:34.060 ","End":"35:41.560","Text":"That\u0027s what we have in the x-direction."},{"Start":"35:41.560 ","End":"35:46.519","Text":"Then in the y-direction,"},{"Start":"35:46.519 ","End":"35:50.407","Text":"we\u0027re going to have"},{"Start":"35:50.407 ","End":"35:57.970","Text":"2 times 3a sine Theta cosine Theta."},{"Start":"35:57.970 ","End":"36:04.422","Text":"Y divided by r is sine Theta"},{"Start":"36:04.422 ","End":"36:12.226","Text":"multiplied by this cosine Theta,"},{"Start":"36:12.226 ","End":"36:12.876","Text":"which when we plug in our 45-degree angle for Theta,"},{"Start":"36:12.876 ","End":"36:13.701","Text":"what we\u0027re going to get is that this is simply equal to 3a"},{"Start":"36:13.701 ","End":"36:17.177","Text":"in the y direction."},{"Start":"36:17.177 ","End":"36:23.340","Text":"Then we can simplify it further and then what we get is that the total E field,"},{"Start":"36:23.340 ","End":"36:27.870","Text":"when we use first-order approximation is equal to kq divided by"},{"Start":"36:27.870 ","End":"36:36.105","Text":"r cubed multiplied by a in the x direction and multiplied by 3a in the y-direction."},{"Start":"36:36.105 ","End":"36:41.625","Text":"This is our E field for a question 2 and of course,"},{"Start":"36:41.625 ","End":"36:50.320","Text":"the force for q is simply going to be capital Q multiplied by this E field."},{"Start":"36:52.220 ","End":"36:57.675","Text":"We were asked to calculate the force on Q. This is the force."},{"Start":"36:57.675 ","End":"37:03.030","Text":"We just multiply our electric field by capital Q, like we did in 1."},{"Start":"37:03.030 ","End":"37:05.700","Text":"Now let\u0027s move on to question number 3."},{"Start":"37:05.700 ","End":"37:10.350","Text":"Calculate the dipole moment of charges q and negative q."},{"Start":"37:10.350 ","End":"37:15.300","Text":"The equation for the dipole moment is denoted by this P,"},{"Start":"37:15.300 ","End":"37:22.995","Text":"and it is a vector quantity and it\u0027s equal to the charge q multiplied by d vector."},{"Start":"37:22.995 ","End":"37:28.140","Text":"Where the d vector is the vector connecting the two charges and it goes,"},{"Start":"37:28.140 ","End":"37:32.040","Text":"and this is very important from negative to positive."},{"Start":"37:32.040 ","End":"37:36.674","Text":"Usually we\u0027re used to an electric field going from positive to negative."},{"Start":"37:36.674 ","End":"37:39.449","Text":"But note when we\u0027re calculating the dipole moment,"},{"Start":"37:39.449 ","End":"37:44.200","Text":"the d vector is from the negative charge to the positive charge."},{"Start":"37:45.200 ","End":"37:48.790","Text":"It\u0027s important to remember that."},{"Start":"37:49.910 ","End":"37:54.610","Text":"Here we can see that our charge is q,"},{"Start":"37:54.650 ","End":"38:01.125","Text":"and our d vector is the distance between the two charges."},{"Start":"38:01.125 ","End":"38:03.915","Text":"That is to 2a."},{"Start":"38:03.915 ","End":"38:08.970","Text":"That\u0027s the magnitude of d. The direction of d is just in"},{"Start":"38:08.970 ","End":"38:11.700","Text":"the positive x direction from negative to positive is"},{"Start":"38:11.700 ","End":"38:15.520","Text":"pointing in the positive x direction."},{"Start":"38:15.890 ","End":"38:19.260","Text":"This is the answer, you can leave it like this."},{"Start":"38:19.260 ","End":"38:22.995","Text":"This is correct. Another way to write it as just in vector form."},{"Start":"38:22.995 ","End":"38:26.910","Text":"You can write 2aq for the x components,"},{"Start":"38:26.910 ","End":"38:28.860","Text":"and then we don\u0027t have a y hat,"},{"Start":"38:28.860 ","End":"38:32.160","Text":"we have 0 in the y direction and we don\u0027t have a z hat,"},{"Start":"38:32.160 ","End":"38:34.755","Text":"we have 0 in the z direction."},{"Start":"38:34.755 ","End":"38:37.830","Text":"Both ways of writing this are completely correct."},{"Start":"38:37.830 ","End":"38:40.080","Text":"We\u0027re just going to choose this way."},{"Start":"38:40.080 ","End":"38:43.110","Text":"Now let\u0027s do question number 4."},{"Start":"38:43.110 ","End":"38:48.165","Text":"Again, we\u0027re being asked to calculate this for us on capital Q,"},{"Start":"38:48.165 ","End":"38:52.110","Text":"but this time using the equation for a dipole field."},{"Start":"38:52.110 ","End":"39:00.730","Text":"Then we\u0027re being asked how this answer compares to our answer for question 2."},{"Start":"39:01.520 ","End":"39:09.810","Text":"Number 1, our answer is going to be the same answer that we got for question 2,"},{"Start":"39:09.810 ","End":"39:13.485","Text":"I\u0027m already telling you and let\u0027s see that we get this."},{"Start":"39:13.485 ","End":"39:20.039","Text":"The electric field for a dipole is given by k multiplied by"},{"Start":"39:20.039 ","End":"39:24.929","Text":"3p dot r hat in"},{"Start":"39:24.929 ","End":"39:31.109","Text":"the r hat direction multiplied by our dipole moment over here,"},{"Start":"39:31.109 ","End":"39:35.290","Text":"divided by r cubed."},{"Start":"39:35.990 ","End":"39:43.935","Text":"Before we begin, let\u0027s see what our p dot r hat is equal to."},{"Start":"39:43.935 ","End":"39:50.489","Text":"We know that the dot product is just the magnitude of our p vector multiplied by"},{"Start":"39:50.489 ","End":"39:56.775","Text":"the magnitude of r hat vector multiplied by cosine of the angle."},{"Start":"39:56.775 ","End":"39:59.940","Text":"We know specifically in this case that the angle is 45 degrees."},{"Start":"39:59.940 ","End":"40:02.830","Text":"We\u0027ll plug that in later."},{"Start":"40:03.260 ","End":"40:07.065","Text":"In that case, what we\u0027re going to get,"},{"Start":"40:07.065 ","End":"40:09.240","Text":"let\u0027s move down here,"},{"Start":"40:09.240 ","End":"40:16.695","Text":"is that we\u0027ll have k multiplied by 3 times the magnitude of our p vector."},{"Start":"40:16.695 ","End":"40:19.425","Text":"Now we know that the magnitude of r hat is 1."},{"Start":"40:19.425 ","End":"40:21.570","Text":"That\u0027s the definition of these hat vector."},{"Start":"40:21.570 ","End":"40:24.075","Text":"It\u0027s 1 in this r direction."},{"Start":"40:24.075 ","End":"40:26.355","Text":"Just 1, we don\u0027t have to include that,"},{"Start":"40:26.355 ","End":"40:33.735","Text":"multiplied by cosine of Theta and then we have all of this in the r hat direction,"},{"Start":"40:33.735 ","End":"40:36.705","Text":"and then we have negative."},{"Start":"40:36.705 ","End":"40:38.235","Text":"We have this 3 here,"},{"Start":"40:38.235 ","End":"40:40.260","Text":"oh no, this 3, isn\u0027t multiplying this,"},{"Start":"40:40.260 ","End":"40:44.100","Text":"it\u0027s negative just our p vector."},{"Start":"40:44.100 ","End":"40:48.375","Text":"Like so and all this divided by r cubed."},{"Start":"40:48.375 ","End":"40:55.140","Text":"Then let\u0027s rewrite this."},{"Start":"40:55.140 ","End":"40:57.765","Text":"We can scroll over here."},{"Start":"40:57.765 ","End":"41:03.119","Text":"What we\u0027re going to have is k divided by r cubed and"},{"Start":"41:03.119 ","End":"41:09.165","Text":"then we have all of this multiplied by 3 and then our P,"},{"Start":"41:09.165 ","End":"41:12.435","Text":"the magnitude is just 2aq."},{"Start":"41:12.435 ","End":"41:22.229","Text":"We have 3 multiplied by 2aq multiplied by cosine of Theta and this is in"},{"Start":"41:22.229 ","End":"41:27.509","Text":"the r direction multiplied by our P vector which is simply"},{"Start":"41:27.509 ","End":"41:34.480","Text":"2aq in the x direction, x hat."},{"Start":"41:35.330 ","End":"41:39.885","Text":"What we can do is we can take out our q from both places."},{"Start":"41:39.885 ","End":"41:46.035","Text":"Let\u0027s rub it out and just put this as this common factor over here."},{"Start":"41:46.035 ","End":"41:50.130","Text":"Now we have 3 times 2a cosine of Theta."},{"Start":"41:50.130 ","End":"41:54.090","Text":"R hat minus 2a x hat."},{"Start":"41:54.090 ","End":"41:59.085","Text":"What do we want to do is we want to get our r hat in terms of x hat."},{"Start":"41:59.085 ","End":"42:07.409","Text":"We know that r hat is simply equal to r vector divided by r. What is r vector?"},{"Start":"42:07.409 ","End":"42:10.575","Text":"We\u0027re specifically working in two dimensions."},{"Start":"42:10.575 ","End":"42:15.270","Text":"R vector we can write as x in the x direction,"},{"Start":"42:15.270 ","End":"42:18.700","Text":"plus y in the y direction."},{"Start":"42:20.000 ","End":"42:23.414","Text":"Now let\u0027s scroll down."},{"Start":"42:23.414 ","End":"42:26.805","Text":"Then let\u0027s carry on over here."},{"Start":"42:26.805 ","End":"42:32.474","Text":"What we have is kq divided by r cubed multiplied by,"},{"Start":"42:32.474 ","End":"42:33.915","Text":"let\u0027s first write this out."},{"Start":"42:33.915 ","End":"42:41.114","Text":"We have negative 2a in the x direction and then we have plus 3"},{"Start":"42:41.114 ","End":"42:49.289","Text":"multiplied by 2a cosine Theta in the r hat direction,"},{"Start":"42:49.289 ","End":"42:52.530","Text":"sorry, this is divided by r as well."},{"Start":"42:52.530 ","End":"43:01.109","Text":"We have x in the x hat direction divided by r plus 3 multiplied by"},{"Start":"43:01.109 ","End":"43:09.479","Text":"2a cosine Theta y y hat divided by"},{"Start":"43:09.479 ","End":"43:15.299","Text":"r. Now I hope you remember that x divided by r is equal to"},{"Start":"43:15.299 ","End":"43:22.540","Text":"cosine of Theta and the y divided by r is equal to sine of Theta."},{"Start":"43:22.580 ","End":"43:32.100","Text":"Then what we get is we\u0027ll get kq divided by r cubed and then we have"},{"Start":"43:32.100 ","End":"43:37.154","Text":"negative 2a x hat plus 3 multiplied by"},{"Start":"43:37.154 ","End":"43:47.290","Text":"2a cosine Theta multiplied by x divided by r which is cosine Theta."},{"Start":"43:47.420 ","End":"43:56.445","Text":"What we\u0027ll have is cosine squared Theta in the x direction plus"},{"Start":"43:56.445 ","End":"44:00.435","Text":"3 times 2a and then we have"},{"Start":"44:00.435 ","End":"44:04.770","Text":"cosine Theta multiplied by y divided by r which is sine Theta."},{"Start":"44:04.770 ","End":"44:06.930","Text":"We have sine Theta,"},{"Start":"44:06.930 ","End":"44:12.340","Text":"cosine Theta in the y direction."},{"Start":"44:13.160 ","End":"44:19.260","Text":"Now we\u0027re going to plug in that Theta is equal to 45 degrees."},{"Start":"44:19.260 ","End":"44:25.240","Text":"In which case, cosine squared of 45 degrees is equal to 1/2."},{"Start":"44:25.520 ","End":"44:32.265","Text":"Then here we have sine of 45 degrees multiplied by cosine of 45 degrees."},{"Start":"44:32.265 ","End":"44:36.370","Text":"This is also equal to 1/2."},{"Start":"44:36.890 ","End":"44:39.480","Text":"Let\u0027s just do this over here."},{"Start":"44:39.480 ","End":"44:43.184","Text":"Then we can say that this is equal to kq divided by"},{"Start":"44:43.184 ","End":"44:47.320","Text":"r cubed and then we have negative 2a in the x,"},{"Start":"44:47.320 ","End":"44:52.005","Text":"direction plus this 2 and this 1\\2 will cancel out,"},{"Start":"44:52.005 ","End":"44:55.980","Text":"plus 3a in the x direction,"},{"Start":"44:55.980 ","End":"44:59.880","Text":"plus and then this 2 and this 1\\2 will also cancel out,"},{"Start":"44:59.880 ","End":"45:03.690","Text":"plus 3a in the y direction."},{"Start":"45:03.690 ","End":"45:07.050","Text":"This is simply equal to kq divided by"},{"Start":"45:07.050 ","End":"45:10.950","Text":"r cubed and then we have negative 2a plus 3a which has"},{"Start":"45:10.950 ","End":"45:18.960","Text":"just a in the x direction plus 3a in the y direction."},{"Start":"45:20.300 ","End":"45:25.120","Text":"I\u0027m just going to move this answer up."},{"Start":"45:27.290 ","End":"45:30.179","Text":"Just scroll a little bit more here."},{"Start":"45:30.179 ","End":"45:31.620","Text":"This is what we got,"},{"Start":"45:31.620 ","End":"45:33.479","Text":"just now I\u0027ve just moved it up."},{"Start":"45:33.479 ","End":"45:41.970","Text":"Therefore we know that our force is equal to Q multiplied by the E field."},{"Start":"45:41.970 ","End":"45:48.475","Text":"We\u0027ll get that this is equal to kq multiplied by capital Q divided by r cubed,"},{"Start":"45:48.475 ","End":"45:55.285","Text":"a in the x direction plus 3a in the y direction."},{"Start":"45:55.285 ","End":"45:59.145","Text":"As we can see, this is exactly what we got for question 2,"},{"Start":"45:59.145 ","End":"46:01.455","Text":"kqQ divided by r cubed,"},{"Start":"46:01.455 ","End":"46:04.989","Text":"a, in the x plus 3a in the y."},{"Start":"46:05.320 ","End":"46:08.810","Text":"This is our answer to question number 4."},{"Start":"46:08.810 ","End":"46:12.440","Text":"We got as we expected the same answer as in question 2,"},{"Start":"46:12.440 ","End":"46:20.280","Text":"just using the equation for a dipole field because we had over here a plus q and minus q."},{"Start":"46:20.280 ","End":"46:23.760","Text":"We know that we\u0027re going to have some dipole over here."},{"Start":"46:23.760 ","End":"46:26.740","Text":"That\u0027s the end of this lesson."}],"ID":22326},{"Watched":false,"Name":"Exercise 2","Duration":"18m 30s","ChapterTopicVideoID":21296,"CourseChapterTopicPlaylistID":99476,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.065","Text":"Hello. In this lesson we\u0027re going to be answering the following question."},{"Start":"00:04.065 ","End":"00:07.410","Text":"We\u0027re given a dipole of moment p,"},{"Start":"00:07.410 ","End":"00:10.260","Text":"which is p in the x-direction then 00 in"},{"Start":"00:10.260 ","End":"00:15.495","Text":"the y and z-directions and the dipole is located at the origin."},{"Start":"00:15.495 ","End":"00:18.165","Text":"Question Number 1, which is what we\u0027ll start with,"},{"Start":"00:18.165 ","End":"00:21.360","Text":"is we\u0027re being asked to calculate the magnitude of p,"},{"Start":"00:21.360 ","End":"00:24.540","Text":"such that if an electron was located at a, 0,"},{"Start":"00:24.540 ","End":"00:27.765","Text":"0 and it had a velocity of v, 0, 0,"},{"Start":"00:27.765 ","End":"00:32.620","Text":"it would eventually come to a stop at b, 0, 0."},{"Start":"00:33.230 ","End":"00:37.035","Text":"First let\u0027s draw out our system."},{"Start":"00:37.035 ","End":"00:39.750","Text":"Here\u0027s our xy-axis,"},{"Start":"00:39.750 ","End":"00:44.595","Text":"and we have a dipole of p, 0, 0."},{"Start":"00:44.595 ","End":"00:48.830","Text":"What we can see is that it only has an x-component and it\u0027s located in the origin,"},{"Start":"00:48.830 ","End":"00:53.160","Text":"so it\u0027s located something like this."},{"Start":"00:53.590 ","End":"00:59.975","Text":"We can tell that because of the x-component over here, it\u0027s positive,"},{"Start":"00:59.975 ","End":"01:04.760","Text":"so we can tell that this is going to be the positively charged particle,"},{"Start":"01:04.760 ","End":"01:07.595","Text":"and this is going to be the negatively charged particle."},{"Start":"01:07.595 ","End":"01:09.260","Text":"Because remember with dipoles,"},{"Start":"01:09.260 ","End":"01:17.030","Text":"the electric field or the dipole moment rather goes from negative to positive."},{"Start":"01:18.360 ","End":"01:23.465","Text":"Then we have an electron located a, 0, 0."},{"Start":"01:23.465 ","End":"01:30.035","Text":"Let\u0027s say that this over here is where our electron is located."},{"Start":"01:30.035 ","End":"01:32.960","Text":"Here, our x is equal to a."},{"Start":"01:32.960 ","End":"01:35.240","Text":"It has no y or z component,"},{"Start":"01:35.240 ","End":"01:37.820","Text":"and we\u0027re being told that it has"},{"Start":"01:37.820 ","End":"01:43.370","Text":"this velocity also in the x-direction and it appears in the positive x-direction,"},{"Start":"01:43.370 ","End":"01:47.740","Text":"v going in the x-direction."},{"Start":"01:48.230 ","End":"01:52.685","Text":"We\u0027re being told that it will come to a stop at b, 0, 0."},{"Start":"01:52.685 ","End":"01:57.645","Text":"It\u0027s also somewhere along the x-axis."},{"Start":"01:57.645 ","End":"02:01.560","Text":"This is at x is equal to b."},{"Start":"02:01.560 ","End":"02:05.900","Text":"Whenever we\u0027re being told that something has some velocity"},{"Start":"02:05.900 ","End":"02:10.145","Text":"and it starts at a certain point and stops at a certain point,"},{"Start":"02:10.145 ","End":"02:14.210","Text":"then my immediate reaction is to think to use"},{"Start":"02:14.210 ","End":"02:19.800","Text":"energy and energy conservation in order to answer these questions."},{"Start":"02:20.570 ","End":"02:24.020","Text":"I\u0027m going to use the idea of energy conservation."},{"Start":"02:24.020 ","End":"02:28.355","Text":"I have my potential energy due to the dipole."},{"Start":"02:28.355 ","End":"02:30.690","Text":"I already know what that is,"},{"Start":"02:31.330 ","End":"02:38.970","Text":"and potential energy and the electric field is always a conservative quantity,"},{"Start":"02:38.970 ","End":"02:43.430","Text":"so that means that my potential energy from my dipole field"},{"Start":"02:43.430 ","End":"02:49.330","Text":"is conservative and therefore I can use it in order to solve this question."},{"Start":"02:49.330 ","End":"02:53.769","Text":"Now let\u0027s write out our energy equation,"},{"Start":"02:53.769 ","End":"02:56.450","Text":"my initial energy, which is at the start."},{"Start":"02:56.450 ","End":"02:58.460","Text":"First I have my kinetic energy,"},{"Start":"02:58.460 ","End":"03:01.609","Text":"which is 1/2 multiplied by the mass of the electron,"},{"Start":"03:01.609 ","End":"03:03.379","Text":"multiplied by its velocity,"},{"Start":"03:03.379 ","End":"03:09.270","Text":"which is v^2 plus the potential energy there."},{"Start":"03:09.270 ","End":"03:11.570","Text":"The potential energy, just to remind you,"},{"Start":"03:11.570 ","End":"03:13.610","Text":"u is equal to q,"},{"Start":"03:13.610 ","End":"03:15.590","Text":"the charge that\u0027s moving,"},{"Start":"03:15.590 ","End":"03:18.975","Text":"multiplied by the potential."},{"Start":"03:18.975 ","End":"03:25.010","Text":"The charge of the electron is just equal to negative e or e"},{"Start":"03:25.010 ","End":"03:31.955","Text":"minus and then we\u0027re multiplying it by the potential at this point."},{"Start":"03:31.955 ","End":"03:38.800","Text":"Let\u0027s call this point r_a."},{"Start":"03:38.800 ","End":"03:42.343","Text":"I\u0027ll write it up here,"},{"Start":"03:42.343 ","End":"03:45.080","Text":"this is the equation for the potential of a dipole,"},{"Start":"03:45.080 ","End":"03:55.730","Text":"and this is equal to k of p vector dot-product with r vector divided by r^3."},{"Start":"03:55.730 ","End":"04:00.300","Text":"Therefore, at our point a over here,"},{"Start":"04:00.300 ","End":"04:02.685","Text":"this will be equal to k,"},{"Start":"04:02.685 ","End":"04:06.120","Text":"our p vector is p in the x-direction."},{"Start":"04:06.120 ","End":"04:12.640","Text":"I can write here p dot-product with a,"},{"Start":"04:12.640 ","End":"04:15.285","Text":"the position of this point over here,"},{"Start":"04:15.285 ","End":"04:17.550","Text":"a in the x-direction."},{"Start":"04:17.550 ","End":"04:24.050","Text":"I\u0027ll have kpa and then I\u0027ll have x-hat dot-product with x-hat which is equal to 1,"},{"Start":"04:24.050 ","End":"04:26.885","Text":"divided by r^3,"},{"Start":"04:26.885 ","End":"04:32.495","Text":"which is from the origin until a, so it\u0027s a^3."},{"Start":"04:32.495 ","End":"04:42.750","Text":"Therefore, I can write this out as 1/2mv^2 plus e minus multiplied by"},{"Start":"04:42.750 ","End":"04:47.750","Text":"kp and then we can cancel out this a over here"},{"Start":"04:47.750 ","End":"04:53.600","Text":", divided by a^2."},{"Start":"04:53.600 ","End":"04:55.145","Text":"From conservation of energy,"},{"Start":"04:55.145 ","End":"04:59.690","Text":"I know this is equal to the energy right at the end, my final energy."},{"Start":"04:59.690 ","End":"05:05.070","Text":"I\u0027m being told that my electron comes to a stop at b, 0, 0."},{"Start":"05:05.070 ","End":"05:08.120","Text":"First of all it comes to a stop, so at this point b,"},{"Start":"05:08.120 ","End":"05:09.780","Text":"it has a velocity equal to 0,"},{"Start":"05:09.780 ","End":"05:11.600","Text":"so there\u0027s no kinetic energy,"},{"Start":"05:11.600 ","End":"05:13.410","Text":"but there is potential energy."},{"Start":"05:13.410 ","End":"05:15.980","Text":"The potential energy is the same as what we have here,"},{"Start":"05:15.980 ","End":"05:19.355","Text":"just with b instead of a."},{"Start":"05:19.355 ","End":"05:24.545","Text":"It\u0027s simply going to be the charge of our electron which is a minus charge"},{"Start":"05:24.545 ","End":"05:31.440","Text":"multiplied by kp, divided by b^2."},{"Start":"05:32.120 ","End":"05:37.970","Text":"My question was to calculate the magnitude of p. All I"},{"Start":"05:37.970 ","End":"05:43.220","Text":"have to do is I have to isolate out my p. After some algebra,"},{"Start":"05:43.220 ","End":"05:46.220","Text":"you can check this on paper, you can pause the video,"},{"Start":"05:46.220 ","End":"05:48.600","Text":"p is equal to 1/2mv^2ek,"},{"Start":"05:52.210 ","End":"06:02.910","Text":"and then multiplied by a^2 b^2 divided by b^2 minus a^2."},{"Start":"06:04.340 ","End":"06:07.530","Text":"This is the answer to question Number 1,"},{"Start":"06:07.530 ","End":"06:11.020","Text":"now let\u0027s move on to question Number 2."},{"Start":"06:11.180 ","End":"06:14.360","Text":"Question 2 is to now again,"},{"Start":"06:14.360 ","End":"06:17.120","Text":"calculate the magnitude of p,"},{"Start":"06:17.120 ","End":"06:20.870","Text":"such that if an electron was located at a,"},{"Start":"06:20.870 ","End":"06:23.960","Text":"negative root 2a, 0."},{"Start":"06:23.960 ","End":"06:29.705","Text":"Now the electron is located at different place with a velocity of 0, 0, v,"},{"Start":"06:29.705 ","End":"06:32.911","Text":"now it has a velocity in the z-direction,"},{"Start":"06:32.911 ","End":"06:36.120","Text":"it would have circular motion."},{"Start":"06:36.120 ","End":"06:42.835","Text":"What we\u0027ll get is that our electron is located approximately over here."},{"Start":"06:42.835 ","End":"06:51.345","Text":"We have that this distance over here in the x-direction is a distance of a,"},{"Start":"06:51.345 ","End":"07:00.450","Text":"and this distance in the y-direction is equal to negative root 2a."},{"Start":"07:02.360 ","End":"07:06.630","Text":"Now as for the z-direction,"},{"Start":"07:06.630 ","End":"07:11.855","Text":"the z-direction is going to be pointing towards us,"},{"Start":"07:11.855 ","End":"07:18.230","Text":"so it\u0027s going to be an axis coming out of the screen and in towards your eye."},{"Start":"07:18.230 ","End":"07:22.040","Text":"Because if we do x to y,"},{"Start":"07:22.040 ","End":"07:25.958","Text":"then we see that the z-axis is coming out towards us,"},{"Start":"07:25.958 ","End":"07:31.760","Text":"so that will be the positive z-direction out towards us and then we can"},{"Start":"07:31.760 ","End":"07:37.982","Text":"see that the velocity also of the electron is in the z,"},{"Start":"07:37.982 ","End":"07:42.990","Text":"so it\u0027s coming out towards us."},{"Start":"07:42.990 ","End":"07:48.547","Text":"If we\u0027re being told that it will have circular motion,"},{"Start":"07:48.547 ","End":"07:54.205","Text":"so that means that we\u0027re going to have some force acting over here."},{"Start":"07:54.205 ","End":"07:58.645","Text":"Let\u0027s calculate the electric force."},{"Start":"07:58.645 ","End":"08:00.910","Text":"What we can see is because we have a dipole,"},{"Start":"08:00.910 ","End":"08:05.649","Text":"so we have to calculate the electric field of a dipole."},{"Start":"08:05.649 ","End":"08:11.170","Text":"The electric field of a dipole is equal to k multiplied by"},{"Start":"08:11.170 ","End":"08:20.560","Text":"3(p.r-hat) and the r-hat direction minus our p vector."},{"Start":"08:20.560 ","End":"08:25.610","Text":"Then all of this is divided by r^3."},{"Start":"08:26.010 ","End":"08:29.560","Text":"I have p, my vector over here."},{"Start":"08:29.560 ","End":"08:31.000","Text":"I know it\u0027s in my x-direction,"},{"Start":"08:31.000 ","End":"08:33.070","Text":"but I don\u0027t know the magnitude of it."},{"Start":"08:33.070 ","End":"08:36.340","Text":"Then I have my r vector."},{"Start":"08:36.340 ","End":"08:39.520","Text":"My r vector is this,"},{"Start":"08:39.520 ","End":"08:40.960","Text":"so we\u0027re giving it over here."},{"Start":"08:40.960 ","End":"08:43.074","Text":"So a in the x-direction,"},{"Start":"08:43.074 ","End":"08:45.760","Text":"negative 2a in the y,"},{"Start":"08:45.760 ","End":"08:47.620","Text":"and 0 in the z."},{"Start":"08:47.620 ","End":"08:54.160","Text":"The magnitude of my r vector is simply equal to the square root of"},{"Start":"08:54.160 ","End":"09:02.080","Text":"a^2 plus the square root of negative 2a^2 plus 0^2,"},{"Start":"09:02.080 ","End":"09:04.225","Text":"which is obviously 0."},{"Start":"09:04.225 ","End":"09:13.495","Text":"This is equal to the square root of a^2 plus 2a^2."},{"Start":"09:13.495 ","End":"09:15.640","Text":"Or we can cancel this out,"},{"Start":"09:15.640 ","End":"09:18.980","Text":"so we\u0027ll get the square root of 3a."},{"Start":"09:21.570 ","End":"09:24.145","Text":"Now we\u0027ve simplified this."},{"Start":"09:24.145 ","End":"09:26.500","Text":"So now we want the r-hat vector,"},{"Start":"09:26.500 ","End":"09:33.730","Text":"so r-hat vector is simply equal to the r vector divided by its magnitude."},{"Start":"09:33.730 ","End":"09:37.060","Text":"We have the magnitude,"},{"Start":"09:37.060 ","End":"09:42.160","Text":"so it\u0027s 1 divided by root 3 a multiplied by"},{"Start":"09:42.160 ","End":"09:48.190","Text":"r vector which has a negative root 2a, 0."},{"Start":"09:48.190 ","End":"09:53.770","Text":"Now I\u0027m just going to simplify this by canceling out my a\u0027s."},{"Start":"09:53.770 ","End":"09:58.660","Text":"I\u0027m going to keep my 1 divided by root 3 as a common factor outside."},{"Start":"09:58.660 ","End":"10:00.940","Text":"Then we have this a divided by this a,"},{"Start":"10:00.940 ","End":"10:03.220","Text":"so we have 1 in the x-direction."},{"Start":"10:03.220 ","End":"10:04.720","Text":"This a divided by this a,"},{"Start":"10:04.720 ","End":"10:11.110","Text":"so we have negative root 2 in the y-direction and 0 here in the z-direction."},{"Start":"10:11.760 ","End":"10:17.680","Text":"Now we want to plug in all of this into our equation for the electric field,"},{"Start":"10:17.680 ","End":"10:21.685","Text":"so e is equal to k multiplied by 3,"},{"Start":"10:21.685 ","End":"10:23.800","Text":"and then we have p.r."},{"Start":"10:23.800 ","End":"10:27.910","Text":"We know that our p is in the x-direction."},{"Start":"10:27.910 ","End":"10:33.700","Text":"Our r-hat, we can see we have in the x and y-direction."},{"Start":"10:33.700 ","End":"10:37.036","Text":"When we take the dot-product,"},{"Start":"10:37.036 ","End":"10:40.150","Text":"we have 0,1,0 multiplied over here,"},{"Start":"10:40.150 ","End":"10:43.645","Text":"so we\u0027re going to lose our y and z components,"},{"Start":"10:43.645 ","End":"10:48.385","Text":"and we\u0027re just going to be left with our x component, so p.r."},{"Start":"10:48.385 ","End":"10:52.630","Text":"We\u0027re going to be left with 3 multiplied by this,"},{"Start":"10:52.630 ","End":"10:56.150","Text":"so 1 divided by root 3."},{"Start":"10:56.160 ","End":"10:59.320","Text":"Then p multiplied by 1,"},{"Start":"10:59.320 ","End":"11:05.575","Text":"which is p. We lose the other components because they\u0027re multiplied by 0 over here."},{"Start":"11:05.575 ","End":"11:09.790","Text":"Then, so that was our p.r-hat."},{"Start":"11:09.790 ","End":"11:12.459","Text":"All of this is in the r-hat direction,"},{"Start":"11:12.459 ","End":"11:17.335","Text":"so the r-hat direction is 1 divided by root 3,"},{"Start":"11:17.335 ","End":"11:23.575","Text":"and then we have 1 negative root 2, 0."},{"Start":"11:23.575 ","End":"11:27.565","Text":"That\u0027s all of that. Then we have negative p vector,"},{"Start":"11:27.565 ","End":"11:33.175","Text":"so that\u0027s p, 0, 0 over here."},{"Start":"11:33.175 ","End":"11:39.595","Text":"Then all of this is divided by r^3."},{"Start":"11:39.595 ","End":"11:42.310","Text":"We found what our r is equal to,"},{"Start":"11:42.310 ","End":"11:43.855","Text":"so we just have to cube it."},{"Start":"11:43.855 ","End":"11:47.185","Text":"What we\u0027re going to get is that i^3 is equal to 3,"},{"Start":"11:47.185 ","End":"11:53.155","Text":"root 3 multiplied by a^3."},{"Start":"11:53.155 ","End":"11:57.280","Text":"Now let\u0027s scroll down a little bit to give us some more space."},{"Start":"11:57.280 ","End":"11:58.840","Text":"Let\u0027s see what we have here."},{"Start":"11:58.840 ","End":"12:04.010","Text":"Here we have 1 divided by root 3 multiplied by 1 divided by root 3."},{"Start":"12:04.010 ","End":"12:06.690","Text":"Together that\u0027s 1 divided by 3."},{"Start":"12:06.690 ","End":"12:08.550","Text":"Then we have 3 over here,"},{"Start":"12:08.550 ","End":"12:10.665","Text":"so 3 divided by 3 is 1."},{"Start":"12:10.665 ","End":"12:13.800","Text":"We can cancel out all of this."},{"Start":"12:13.800 ","End":"12:17.505","Text":"This 3 over here is a mistake."},{"Start":"12:17.505 ","End":"12:19.560","Text":"Backs and wrote that in."},{"Start":"12:19.560 ","End":"12:22.720","Text":"We cancel it out with the 3."},{"Start":"12:22.890 ","End":"12:27.430","Text":"Now we can see that we\u0027re just left with p over here."},{"Start":"12:27.430 ","End":"12:31.510","Text":"Then we have p and then 1 in the x-direction,"},{"Start":"12:31.510 ","End":"12:32.790","Text":"negative 2 in the y-direction,"},{"Start":"12:32.790 ","End":"12:35.185","Text":"and 0 in the z-direction."},{"Start":"12:35.185 ","End":"12:38.350","Text":"Now we\u0027re going to do vector subtraction."},{"Start":"12:38.350 ","End":"12:40.030","Text":"In the x-direction,"},{"Start":"12:40.030 ","End":"12:42.190","Text":"we have p multiplied by 1,"},{"Start":"12:42.190 ","End":"12:44.095","Text":"which is p minus."},{"Start":"12:44.095 ","End":"12:46.240","Text":"Over here, we have p as well."},{"Start":"12:46.240 ","End":"12:49.360","Text":"We\u0027re going to have 0 in the x-direction."},{"Start":"12:49.360 ","End":"12:52.520","Text":"Let\u0027s write this out."},{"Start":"12:52.520 ","End":"13:00.375","Text":"First of all, we\u0027re going to have k divided by over here 3 root 3 multiplied by a^3."},{"Start":"13:00.375 ","End":"13:02.370","Text":"These are the constants."},{"Start":"13:02.370 ","End":"13:03.900","Text":"Then we\u0027re going to have,"},{"Start":"13:03.900 ","End":"13:08.215","Text":"so p minus p is 0 in the x-direction."},{"Start":"13:08.215 ","End":"13:13.225","Text":"Then we have negative root 2p minus 0."},{"Start":"13:13.225 ","End":"13:16.990","Text":"We\u0027ll have negative root 2p."},{"Start":"13:16.990 ","End":"13:20.710","Text":"Then we have 0 minus 0,"},{"Start":"13:20.710 ","End":"13:23.330","Text":"so that\u0027s just 0."},{"Start":"13:24.510 ","End":"13:28.780","Text":"This is our electric field and we can see that it only has"},{"Start":"13:28.780 ","End":"13:33.010","Text":"a component at this point over here in the y-direction."},{"Start":"13:33.010 ","End":"13:37.465","Text":"Then we can draw in our electric field over here."},{"Start":"13:37.465 ","End":"13:40.450","Text":"It\u0027s going in the negative y-direction."},{"Start":"13:40.450 ","End":"13:46.760","Text":"This is the direction of our E-field at this point over here."},{"Start":"13:48.030 ","End":"13:53.980","Text":"Now we remember that our equation for force is"},{"Start":"13:53.980 ","End":"14:00.115","Text":"that it is equal to the charge multiplied by the electric field."},{"Start":"14:00.115 ","End":"14:05.710","Text":"Our charge is that of an electron which is a negative charge."},{"Start":"14:05.710 ","End":"14:11.330","Text":"It\u0027s a negative charge multiplied by our electric field."},{"Start":"14:11.670 ","End":"14:18.415","Text":"What we can see is that the force is in the opposite direction to our electric field."},{"Start":"14:18.415 ","End":"14:22.375","Text":"Our electric field was in the negative y-direction."},{"Start":"14:22.375 ","End":"14:24.190","Text":"Now we\u0027re in the opposite direction,"},{"Start":"14:24.190 ","End":"14:30.080","Text":"so our force is going to be in the positive y-direction."},{"Start":"14:32.310 ","End":"14:38.140","Text":"Now what we can see is that we have velocity going in the z-direction,"},{"Start":"14:38.140 ","End":"14:41.365","Text":"so out of the screen and towards u and a force,"},{"Start":"14:41.365 ","End":"14:45.370","Text":"over here, in this positive y-direction."},{"Start":"14:45.370 ","End":"14:51.130","Text":"Therefore, we can see that we\u0027re going to have this circular motion."},{"Start":"14:51.130 ","End":"14:55.495","Text":"If we draw the trajectory,"},{"Start":"14:55.495 ","End":"14:59.620","Text":"again, imagine that this is properly drawn."},{"Start":"14:59.620 ","End":"15:06.295","Text":"What we can see is that we\u0027re going to have this circular motion on the yz-plane."},{"Start":"15:06.295 ","End":"15:11.455","Text":"Our electron is going to be moving like so."},{"Start":"15:11.455 ","End":"15:15.985","Text":"What we can see is that a force is always going to be constant"},{"Start":"15:15.985 ","End":"15:22.075","Text":"and into the center of this circle,"},{"Start":"15:22.075 ","End":"15:25.130","Text":"the circular motion trajectory."},{"Start":"15:25.980 ","End":"15:29.140","Text":"Or rather this is going like this."},{"Start":"15:29.140 ","End":"15:32.580","Text":"It\u0027s always going to look something like this."},{"Start":"15:32.610 ","End":"15:39.830","Text":"Of course, it\u0027s important to note that this radius away from the origin is constant."},{"Start":"15:39.830 ","End":"15:43.620","Text":"That\u0027s also why we are going to be doing circular motion."},{"Start":"15:43.620 ","End":"15:48.740","Text":"Now all we have to do is we have to find the equation for"},{"Start":"15:48.740 ","End":"15:54.105","Text":"circular motion and then just substitute in this equation for the force."},{"Start":"15:54.105 ","End":"16:01.280","Text":"In circular motion, we have that the sum of all of the forces is equal"},{"Start":"16:01.280 ","End":"16:08.175","Text":"to mv^2 divided by r. Let\u0027s see."},{"Start":"16:08.175 ","End":"16:11.615","Text":"The sum of all forces is just this electrical force."},{"Start":"16:11.615 ","End":"16:13.520","Text":"We have the charge,"},{"Start":"16:13.520 ","End":"16:15.440","Text":"so we\u0027re just going to take the size of the charge."},{"Start":"16:15.440 ","End":"16:16.790","Text":"We already know the direction."},{"Start":"16:16.790 ","End":"16:20.150","Text":"That\u0027s the size of the charge of the electron, or the magnitude,"},{"Start":"16:20.150 ","End":"16:30.015","Text":"multiplied by e. We saw that e is equal to k divided by 3 root 3a^3."},{"Start":"16:30.015 ","End":"16:32.660","Text":"Of course, we have to multiply this by p."},{"Start":"16:32.660 ","End":"16:36.650","Text":"We already know the directions that\u0027s why I\u0027m not putting that in."},{"Start":"16:36.650 ","End":"16:40.210","Text":"I\u0027m trying to find the magnitude of p, I\u0027m reminding you."},{"Start":"16:40.210 ","End":"16:42.340","Text":"The only thing I need to know is r,"},{"Start":"16:42.340 ","End":"16:48.800","Text":"so r as we know is the radius of the circle for the circular motion."},{"Start":"16:48.800 ","End":"16:56.600","Text":"R if I draw it in over here in pink will be this radius over here."},{"Start":"16:56.760 ","End":"17:03.185","Text":"As we can see, our circular motion is on the yz-plane."},{"Start":"17:03.185 ","End":"17:05.915","Text":"That means that the radius of this,"},{"Start":"17:05.915 ","End":"17:09.690","Text":"we can just see is just this value over here."},{"Start":"17:10.260 ","End":"17:16.135","Text":"Our radius is simply equal to root 2a."},{"Start":"17:16.135 ","End":"17:21.500","Text":"Here it\u0027s a negative because we\u0027re specifically here going in the negative y-direction,"},{"Start":"17:21.500 ","End":"17:23.930","Text":"but here it will be just root 2a."},{"Start":"17:23.930 ","End":"17:27.295","Text":"The magnitude of r is a root 2a."},{"Start":"17:27.295 ","End":"17:34.585","Text":"Then all we have to do is we have to isolate out our p. Let\u0027s do that."},{"Start":"17:34.585 ","End":"17:42.800","Text":"We\u0027ll have that p is equal to mv^2 divided by root 2a."},{"Start":"17:42.810 ","End":"17:51.410","Text":"Then it\u0027s going to be multiplied by 3 root 3a^3."},{"Start":"17:51.750 ","End":"17:59.360","Text":"Then divided by the charge of an electron multiplied by k. Then,"},{"Start":"17:59.360 ","End":"18:03.155","Text":"of course, this a can cancel out with this a."},{"Start":"18:03.155 ","End":"18:05.855","Text":"Then if we scroll down a little bit more,"},{"Start":"18:05.855 ","End":"18:11.365","Text":"we\u0027ll get that our answer to Question 2 is simply mv^2,"},{"Start":"18:11.365 ","End":"18:23.080","Text":"or rather 3 root 3 mv^2 multiplied by a^2 divided by root 2 e,"},{"Start":"18:23.080 ","End":"18:27.500","Text":"k. This is the answer to Question 2,"},{"Start":"18:27.500 ","End":"18:30.480","Text":"and this is the end of our lesson."}],"ID":21376},{"Watched":false,"Name":"Dipole Moment Of A System","Duration":"15m 39s","ChapterTopicVideoID":21441,"CourseChapterTopicPlaylistID":99476,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.040","Text":"Hello. In this lesson,"},{"Start":"00:02.040 ","End":"00:06.290","Text":"we\u0027re going to speak about how to find the dipole moment of a system."},{"Start":"00:06.290 ","End":"00:09.990","Text":"So that means of a case when we have a bit more of"},{"Start":"00:09.990 ","End":"00:12.360","Text":"a complicated question because instead of just"},{"Start":"00:12.360 ","End":"00:15.255","Text":"having 1 positive charge and 1 negative charge,"},{"Start":"00:15.255 ","End":"00:19.290","Text":"we\u0027re going to have a system of positive and negative charges."},{"Start":"00:19.290 ","End":"00:23.324","Text":"A dipole moment is symbolized"},{"Start":"00:23.324 ","End":"00:27.750","Text":"by the letter P and it has 3 components in the x-direction,"},{"Start":"00:27.750 ","End":"00:30.420","Text":"in the y-direction and in the z-direction."},{"Start":"00:30.420 ","End":"00:36.570","Text":"Let\u0027s give as an example a dipole moments in the x-direction is going"},{"Start":"00:36.570 ","End":"00:43.450","Text":"to be equal to the sum of x_i q_i."},{"Start":"00:43.450 ","End":"00:49.495","Text":"This calculation is similar to what we saw in finding the center of mass."},{"Start":"00:49.495 ","End":"00:55.834","Text":"Here\u0027s an example so we have our axes and we have a few charges and our system."},{"Start":"00:55.834 ","End":"00:59.570","Text":"We have the charge and also their position so minus q"},{"Start":"00:59.570 ","End":"01:03.850","Text":"minus q plus q plus q and their positions in space."},{"Start":"01:03.850 ","End":"01:10.220","Text":"Let\u0027s work out the dipole moment in the different various directions."},{"Start":"01:10.220 ","End":"01:15.635","Text":"Let\u0027s start with our dipole moment in the x-direction."},{"Start":"01:15.635 ","End":"01:19.625","Text":"We know it\u0027s the sum of the x_i multiply it by q_i."},{"Start":"01:19.625 ","End":"01:23.255","Text":"Let\u0027s start with this charge over here."},{"Start":"01:23.255 ","End":"01:25.475","Text":"We\u0027re going to have its x position,"},{"Start":"01:25.475 ","End":"01:30.535","Text":"which is 0 multiplied by its charge so that\u0027s going to be 0."},{"Start":"01:30.535 ","End":"01:33.690","Text":"Let\u0027s go on to this charge."},{"Start":"01:33.690 ","End":"01:36.560","Text":"We\u0027re going to have its x position,"},{"Start":"01:36.560 ","End":"01:40.835","Text":"which is 0 multiplied by the charge so that\u0027s against 0."},{"Start":"01:40.835 ","End":"01:44.960","Text":"Then let\u0027s go to this charge over here so it\u0027s x exposition,"},{"Start":"01:44.960 ","End":"01:48.965","Text":"which is 1 multiplied by its charge of positive q,"},{"Start":"01:48.965 ","End":"01:51.845","Text":"so q and then here, again,"},{"Start":"01:51.845 ","End":"01:58.860","Text":"the x position is 1 multiplied by the charge positive q so again plus q."},{"Start":"01:58.860 ","End":"02:03.485","Text":"We get that our dipole moments in the x-direction is equal to 2q."},{"Start":"02:03.485 ","End":"02:09.560","Text":"Now the dipole moment in the y-direction is going to be our position in the y-direction."},{"Start":"02:09.560 ","End":"02:14.130","Text":"Negative 1 multiplied by our charge, negative q."},{"Start":"02:14.130 ","End":"02:16.595","Text":"Negative times a negative is a positive."},{"Start":"02:16.595 ","End":"02:21.509","Text":"We have q, then we\u0027ll have our y position,"},{"Start":"02:21.509 ","End":"02:26.655","Text":"so 1 multiplied by negative q so that\u0027s going to be negative q."},{"Start":"02:26.655 ","End":"02:33.085","Text":"Then our y-position here is 0 multiplied by positive q, so that\u0027s 0."},{"Start":"02:33.085 ","End":"02:36.410","Text":"Then here we\u0027re going to have our y-position,"},{"Start":"02:36.410 ","End":"02:41.735","Text":"which is 2 multiplied by positive q so that\u0027s plus 2q."},{"Start":"02:41.735 ","End":"02:47.210","Text":"We can see that these 2 cancel out and we can see that our dipole moment in"},{"Start":"02:47.210 ","End":"02:52.550","Text":"the y-direction is also equal to 2q, so that\u0027s great."},{"Start":"02:52.550 ","End":"02:53.870","Text":"We had a system of"},{"Start":"02:53.870 ","End":"02:59.045","Text":"4 discrete charges and we worked out the dipole moments in each direction."},{"Start":"02:59.045 ","End":"03:03.455","Text":"Now what happens if we have a system of continuous charges?"},{"Start":"03:03.455 ","End":"03:07.040","Text":"Instead of spaces between each charge"},{"Start":"03:07.040 ","End":"03:09.050","Text":"there\u0027s going to be so many charges in"},{"Start":"03:09.050 ","End":"03:12.305","Text":"such a small space that we say that it\u0027s continuous."},{"Start":"03:12.305 ","End":"03:18.425","Text":"Then as with anything that moves from a discrete equation to a continuous equation,"},{"Start":"03:18.425 ","End":"03:23.135","Text":"our Sigma over here becomes an integral on x,"},{"Start":"03:23.135 ","End":"03:26.350","Text":"so the position dq."},{"Start":"03:26.350 ","End":"03:30.260","Text":"Then of course, the dipole moment in the y-direction was simply be"},{"Start":"03:30.260 ","End":"03:34.680","Text":"an integral on ydq but it\u0027s the same idea."},{"Start":"03:34.850 ","End":"03:42.385","Text":"Now let\u0027s solve an example that deals with this continuous charge system."},{"Start":"03:42.385 ","End":"03:45.245","Text":"Here we have a spherical shell,"},{"Start":"03:45.245 ","End":"03:47.465","Text":"it has area but it doesn\u0027t have volume,"},{"Start":"03:47.465 ","End":"03:52.505","Text":"so shell and we\u0027re told that it has charge density of Sigma,"},{"Start":"03:52.505 ","End":"03:57.528","Text":"which is given by the equation of Sigma_naught multiplied by cosine Phi."},{"Start":"03:57.528 ","End":"04:01.650","Text":"Our angle so Phi is the angle from the z-axis."},{"Start":"04:01.650 ","End":"04:04.645","Text":"This angle over here is going to be Phi."},{"Start":"04:04.645 ","End":"04:09.560","Text":"What we can see from our charge density is that when we\u0027re right at the top,"},{"Start":"04:09.560 ","End":"04:17.985","Text":"we have the largest value for Sigma for charge density as we hit Pi over 2,"},{"Start":"04:17.985 ","End":"04:19.610","Text":"so this point over here,"},{"Start":"04:19.610 ","End":"04:22.910","Text":"we can see that our charge density is equal to"},{"Start":"04:22.910 ","End":"04:28.895","Text":"0 and as we go from Pi over 2 until we get to Pi,"},{"Start":"04:28.895 ","End":"04:32.855","Text":"we can see that our charged density becomes a negative,"},{"Start":"04:32.855 ","End":"04:36.110","Text":"but it becomes a maximum in the negative direction."},{"Start":"04:36.110 ","End":"04:40.790","Text":"Then as we go to here,"},{"Start":"04:40.790 ","End":"04:46.100","Text":"so this is 3-quarters Pi so we can see that again,"},{"Start":"04:46.100 ","End":"04:52.570","Text":"our Sigma is going to be equal to 0 and then we go back up and hit a maximum."},{"Start":"04:52.570 ","End":"04:55.280","Text":"Now what we\u0027re going to do is we\u0027re going to work out"},{"Start":"04:55.280 ","End":"04:59.385","Text":"the dipole moment of this spherical shell."},{"Start":"04:59.385 ","End":"05:06.410","Text":"We\u0027re trying to find our dipole moment and let\u0027s start each axes separately."},{"Start":"05:06.410 ","End":"05:10.925","Text":"First let\u0027s find our dipole moment in the x-direction."},{"Start":"05:10.925 ","End":"05:14.525","Text":"As we can see from our equation,"},{"Start":"05:14.525 ","End":"05:18.870","Text":"this is going to be an integral, an xdq."},{"Start":"05:18.870 ","End":"05:21.615","Text":"Let\u0027s take a look at what a dq is."},{"Start":"05:21.615 ","End":"05:23.820","Text":"Our dq, as we know,"},{"Start":"05:23.820 ","End":"05:29.135","Text":"is going to be Sigma and because we\u0027re dealing with area and also Sigma"},{"Start":"05:29.135 ","End":"05:35.315","Text":"is charge per unit area so we have to integrate along area ds."},{"Start":"05:35.315 ","End":"05:39.500","Text":"Now because we have a spherical shell so we\u0027re going to need to work in"},{"Start":"05:39.500 ","End":"05:43.760","Text":"spherical coordinates so that means that our ds is"},{"Start":"05:43.760 ","End":"05:53.470","Text":"going to be equal to r^2 sine Phi d Theta d Phi."},{"Start":"05:53.470 ","End":"05:56.035","Text":"Because we\u0027re dealing with a spherical shell,"},{"Start":"05:56.035 ","End":"06:00.730","Text":"that means our r is actually a constant, it\u0027s not a variable,"},{"Start":"06:00.730 ","End":"06:02.275","Text":"it\u0027s just going to be R,"},{"Start":"06:02.275 ","End":"06:04.630","Text":"so we can switch this over to"},{"Start":"06:04.630 ","End":"06:11.305","Text":"R. Now let\u0027s go back to our dipole moment in the x-direction."},{"Start":"06:11.305 ","End":"06:14.840","Text":"It\u0027s going to be the integral on xdq."},{"Start":"06:14.840 ","End":"06:16.980","Text":"Let\u0027s have in dq,"},{"Start":"06:16.980 ","End":"06:20.950","Text":"so it\u0027s going to be Sigma so we can also substitute in our value for"},{"Start":"06:20.950 ","End":"06:25.775","Text":"sigma so that\u0027s Sigma_naught multiplied by cosine of Phi."},{"Start":"06:25.775 ","End":"06:28.800","Text":"That\u0027s was Sigma and then multiplied by"},{"Start":"06:28.800 ","End":"06:37.410","Text":"r^2 sine Phi d Theta d Phi and now because we"},{"Start":"06:37.410 ","End":"06:41.345","Text":"have 2 variables that we\u0027re integrating along them because we\u0027re doing"},{"Start":"06:41.345 ","End":"06:46.650","Text":"an integration over area so they have to be 2 integration signs."},{"Start":"06:46.650 ","End":"06:53.150","Text":"Our Theta integration is going to be around a full circle in the xy-plane so that\u0027s"},{"Start":"06:53.150 ","End":"06:59.535","Text":"going to be from 0 until 2Pi and then an r Phi integral,"},{"Start":"06:59.535 ","End":"07:05.430","Text":"we\u0027re going from Phi is equal to 0 and then up until 1/2."},{"Start":"07:05.430 ","End":"07:10.320","Text":"We\u0027re going from 0 until Pi, 1/2 a circle."},{"Start":"07:10.320 ","End":"07:13.745","Text":"Now before we can integrate,"},{"Start":"07:13.745 ","End":"07:19.894","Text":"we know that our x is also very well and because we\u0027re using spherical coordinates,"},{"Start":"07:19.894 ","End":"07:24.120","Text":"we have to convert our x also into spherical coordinates."},{"Start":"07:24.120 ","End":"07:29.345","Text":"We know that our x value is equal to and spherical coordinates r,"},{"Start":"07:29.345 ","End":"07:32.540","Text":"the radius where here it\u0027s specifically R because we\u0027re dealing"},{"Start":"07:32.540 ","End":"07:36.334","Text":"with a constant radius but if it wasn\u0027t,"},{"Start":"07:36.334 ","End":"07:38.735","Text":"if we were integrating along volume,"},{"Start":"07:38.735 ","End":"07:48.560","Text":"so we would put a small r over here so it\u0027s going to be equal to R sine Phi cosine Theta."},{"Start":"07:48.560 ","End":"07:52.836","Text":"This is x in spherical coordinates."},{"Start":"07:52.836 ","End":"07:58.420","Text":"Now let\u0027s rewrite our integration and we can take out all of our constants."},{"Start":"07:58.420 ","End":"08:03.310","Text":"We see that our constants are equal Sigma_naughts R^3."},{"Start":"08:03.310 ","End":"08:08.875","Text":"Because we have this R^2 multiplied this R when we plug in our x and then we have"},{"Start":"08:08.875 ","End":"08:15.325","Text":"our integration from 0 to Pi and our other one from 0-2Pi."},{"Start":"08:15.325 ","End":"08:18.280","Text":"Then we have cosine of"},{"Start":"08:18.280 ","End":"08:22.780","Text":"Phi sine^2 Phi and"},{"Start":"08:22.780 ","End":"08:29.140","Text":"then cosine of Theta d Theta d Phi."},{"Start":"08:29.140 ","End":"08:32.290","Text":"This looks pretty complicated."},{"Start":"08:32.290 ","End":"08:35.605","Text":"We have cosine Phi is sine^2 Phi cosine Theta."},{"Start":"08:35.605 ","End":"08:37.690","Text":"How do we even begin?"},{"Start":"08:37.690 ","End":"08:40.615","Text":"A tip for the grades,"},{"Start":"08:40.615 ","End":"08:43.600","Text":"let\u0027s see how we do this. Let\u0027s use blue."},{"Start":"08:43.600 ","End":"08:48.490","Text":"Every time you see that we have to integrate along cosine of"},{"Start":"08:48.490 ","End":"08:54.070","Text":"Theta or cosine Phi from 0 until 2Pi."},{"Start":"08:54.070 ","End":"08:59.080","Text":"Every time we do a full period when dealing with cosine of something,"},{"Start":"08:59.080 ","End":"09:03.520","Text":"cosine Alpha from 0 until 2Pi,"},{"Start":"09:03.520 ","End":"09:06.415","Text":"where our angle is a variable."},{"Start":"09:06.415 ","End":"09:09.670","Text":"That it\u0027s not cosine 60."},{"Start":"09:09.670 ","End":"09:13.255","Text":"When our angles are variable and we have to integrate"},{"Start":"09:13.255 ","End":"09:17.365","Text":"along our cosine variable from 0 until 2Pi,"},{"Start":"09:17.365 ","End":"09:21.560","Text":"this is always going to be equal to 0."},{"Start":"09:21.870 ","End":"09:30.760","Text":"Let\u0027s put when integrating from 0 and 2Pi."},{"Start":"09:30.760 ","End":"09:36.265","Text":"Then anything multiplied by 0 is going to be equal to 0"},{"Start":"09:36.265 ","End":"09:42.505","Text":"so we got our dipole moment in the x-direction is equal to 0."},{"Start":"09:42.505 ","End":"09:46.300","Text":"We didn\u0027t even have to do all of this ugly integration."},{"Start":"09:46.300 ","End":"09:50.710","Text":"Remember that tip. That\u0027s great."},{"Start":"09:50.710 ","End":"09:52.915","Text":"That\u0027s our dipole moment in the x-direction."},{"Start":"09:52.915 ","End":"09:56.090","Text":"Now what about in the y-direction?"},{"Start":"09:56.460 ","End":"09:58.900","Text":"Instead of rewriting everything,"},{"Start":"09:58.900 ","End":"10:01.840","Text":"I\u0027m just going to write in a green pen, if we had y."},{"Start":"10:01.840 ","End":"10:03.670","Text":"In the y-direction,"},{"Start":"10:03.670 ","End":"10:08.305","Text":"our dipole moment in the y-direction will be the integral on ydq."},{"Start":"10:08.305 ","End":"10:10.540","Text":"Then our dq is going to be equal to"},{"Start":"10:10.540 ","End":"10:16.250","Text":"the exact same thing and then our integral instead of our x here will have a y."},{"Start":"10:16.350 ","End":"10:19.330","Text":"Then the rest will be the same."},{"Start":"10:19.330 ","End":"10:24.280","Text":"Then that means that we\u0027re going to have to write in our y in spherical coordinates."},{"Start":"10:24.280 ","End":"10:26.605","Text":"Our y in spherical coordinates,"},{"Start":"10:26.605 ","End":"10:29.860","Text":"instead of R sine Phi cosine Theta,"},{"Start":"10:29.860 ","End":"10:34.525","Text":"it\u0027s going to be R sine Phi sine Theta."},{"Start":"10:34.525 ","End":"10:36.670","Text":"That\u0027s the only difference."},{"Start":"10:36.670 ","End":"10:39.145","Text":"Then when we rewrite everything here,"},{"Start":"10:39.145 ","End":"10:40.870","Text":"all of this is going to be the same."},{"Start":"10:40.870 ","End":"10:43.825","Text":"Aside from here, instead of cosine Theta d Theta,"},{"Start":"10:43.825 ","End":"10:47.560","Text":"we\u0027ll have sine Theta d Theta."},{"Start":"10:47.560 ","End":"10:52.315","Text":"Now let\u0027s see what\u0027s happening here."},{"Start":"10:52.315 ","End":"10:58.270","Text":"We\u0027re again left with this gross integral with all these angles and cosine and sine."},{"Start":"10:58.270 ","End":"11:02.560","Text":"However, also with our sine Theta,"},{"Start":"11:02.560 ","End":"11:04.870","Text":"if we\u0027re integrating on our sine Theta,"},{"Start":"11:04.870 ","End":"11:08.034","Text":"where Theta is a variable,"},{"Start":"11:08.034 ","End":"11:14.680","Text":"d Theta and also our sine Theta as being integrated for a whole period."},{"Start":"11:14.680 ","End":"11:17.230","Text":"As in from 0 until 2Pi,"},{"Start":"11:17.230 ","End":"11:20.125","Text":"then this also becomes 0."},{"Start":"11:20.125 ","End":"11:22.210","Text":"We can write this as well."},{"Start":"11:22.210 ","End":"11:27.310","Text":"Sine Theta from 0 until 2Pi integration"},{"Start":"11:27.310 ","End":"11:34.105","Text":"becomes equal to 0 and then 0 times anything is equal to 0."},{"Start":"11:34.105 ","End":"11:36.250","Text":"We can see therefore,"},{"Start":"11:36.250 ","End":"11:42.295","Text":"that our dipole moment in the y-direction is also going to be equal to 0."},{"Start":"11:42.295 ","End":"11:47.365","Text":"I\u0027ve written this nifty little law up at the top over here."},{"Start":"11:47.365 ","End":"11:49.780","Text":"Remember that because it will save you"},{"Start":"11:49.780 ","End":"11:51.880","Text":"loads of times instead of trying to figure this out and"},{"Start":"11:51.880 ","End":"11:57.680","Text":"then the most frustrating thing is working through all of this and getting 0 in the end."},{"Start":"11:57.720 ","End":"12:01.180","Text":"The only components of our dipole moment,"},{"Start":"12:01.180 ","End":"12:03.325","Text":"which isn\u0027t going to cancel out,"},{"Start":"12:03.325 ","End":"12:08.410","Text":"is going to be a dipole moment in the z-direction."},{"Start":"12:08.410 ","End":"12:11.065","Text":"Let\u0027s calculate it."},{"Start":"12:11.065 ","End":"12:14.800","Text":"A lot of this is going to be the same as our p_x,"},{"Start":"12:14.800 ","End":"12:17.275","Text":"but let\u0027s just write it quickly anyway."},{"Start":"12:17.275 ","End":"12:26.890","Text":"Our p_z is an integral on zdq where our dq is equal to the same thing as it is over here."},{"Start":"12:26.890 ","End":"12:35.635","Text":"That\u0027s Sigma R^2 sine Phi d Theta d Phi,"},{"Start":"12:35.635 ","End":"12:39.250","Text":"where our Sigma is equal to this over here."},{"Start":"12:39.250 ","End":"12:42.100","Text":"Then we can carry on with our p_z."},{"Start":"12:42.100 ","End":"12:47.770","Text":"Our p_z is going to be equal to our double integral just like from here,"},{"Start":"12:47.770 ","End":"12:52.945","Text":"from 0 to Pi and 0-2Pi."},{"Start":"12:52.945 ","End":"12:59.020","Text":"Then we\u0027re going to have z Sigma_naught cosine of"},{"Start":"12:59.020 ","End":"13:06.550","Text":"PhiR^2 sine Phi d Theta d Phi."},{"Start":"13:06.550 ","End":"13:09.625","Text":"Then we have to plug in our z."},{"Start":"13:09.625 ","End":"13:17.470","Text":"Our z in spherical coordinates is equal to R cosine of Phi."},{"Start":"13:17.470 ","End":"13:23.890","Text":"Then we can plug everything back in and take out our constants."},{"Start":"13:23.890 ","End":"13:33.520","Text":"We\u0027ll get Sigma_naught R^3 because R^2 multiplied by R. Then R double integral from 0 to"},{"Start":"13:33.520 ","End":"13:38.530","Text":"Pi and 0 till 2Pi and then we have that it\u0027s going"},{"Start":"13:38.530 ","End":"13:44.110","Text":"to be equal to cosine^2 Phi because we have a cosine Phi here and a cosine Phi here."},{"Start":"13:44.110 ","End":"13:52.100","Text":"Then sine Phi d Theta d Phi."},{"Start":"13:53.340 ","End":"14:00.670","Text":"Then this integral is going to be equal to Sigma_naught R^3"},{"Start":"14:00.670 ","End":"14:08.680","Text":"multiplied by 2Pi from our integral along Theta because there\u0027s no Theta variable."},{"Start":"14:08.680 ","End":"14:18.020","Text":"It\u0027s just going to be 2Pi and then multiplied by 4/3 from our integral along Phi."},{"Start":"14:19.170 ","End":"14:25.510","Text":"We can see that our dipole moment has 0 in the x components,"},{"Start":"14:25.510 ","End":"14:30.309","Text":"0 in the y components and then this over here in the z components."},{"Start":"14:30.309 ","End":"14:32.125","Text":"Let\u0027s just write this here."},{"Start":"14:32.125 ","End":"14:34.690","Text":"Our dipole moments therefore"},{"Start":"14:34.690 ","End":"14:38.605","Text":"is a vector quantity first of all and it\u0027s going to be equal to 0,"},{"Start":"14:38.605 ","End":"14:40.930","Text":"0 and then our p_z."},{"Start":"14:40.930 ","End":"14:45.175","Text":"Let\u0027s just write p_z, this over here."},{"Start":"14:45.175 ","End":"14:49.615","Text":"Then if I want to find out my potential in space,"},{"Start":"14:49.615 ","End":"14:51.970","Text":"so we can use our equation."},{"Start":"14:51.970 ","End":"15:00.505","Text":"That was equal to k dipole moment dot product with our r vector divided by r^3."},{"Start":"15:00.505 ","End":"15:07.540","Text":"If I want to find my electric field far away from the dipole, so it\u0027s k,"},{"Start":"15:07.540 ","End":"15:13.360","Text":"3 multiplied by our dipole moment dot product"},{"Start":"15:13.360 ","End":"15:21.534","Text":"r-hat in r-hat direction minus our dipole moment."},{"Start":"15:21.534 ","End":"15:26.065","Text":"This is also divided by r^3."},{"Start":"15:26.065 ","End":"15:30.895","Text":"Remember that here it\u0027s in the r-hat direction in both of these."},{"Start":"15:30.895 ","End":"15:34.550","Text":"Now we have to do is just plug this right in."},{"Start":"15:35.340 ","End":"15:38.810","Text":"That\u0027s the end of this lesson."}],"ID":22327},{"Watched":false,"Name":"Torque Experienced By Dipole","Duration":"10m 19s","ChapterTopicVideoID":21442,"CourseChapterTopicPlaylistID":99476,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.055","Text":"Hello. In this lesson,"},{"Start":"00:02.055 ","End":"00:05.580","Text":"we\u0027re going to be speaking about the moment or the torque"},{"Start":"00:05.580 ","End":"00:11.340","Text":"experienced by a dipole when it\u0027s placed in a uniform electric field."},{"Start":"00:11.340 ","End":"00:18.045","Text":"Up until now, we\u0027ve spoken about how an electric dipole affects its surroundings."},{"Start":"00:18.045 ","End":"00:21.210","Text":"So what the potential will be at certain points in space,"},{"Start":"00:21.210 ","End":"00:24.300","Text":"what the electric field will be far away from the dipole,"},{"Start":"00:24.300 ","End":"00:25.950","Text":"and so far and so forth."},{"Start":"00:25.950 ","End":"00:33.600","Text":"But now, let\u0027s see what affects the surroundings will have on the dipole itself."},{"Start":"00:33.600 ","End":"00:37.485","Text":"First, a brief reminder of what is an electric dipole."},{"Start":"00:37.485 ","End":"00:41.520","Text":"An electric dipole is a pair of"},{"Start":"00:41.520 ","End":"00:46.155","Text":"charges where 1 charge is a minus and 1 charge is positive,"},{"Start":"00:46.155 ","End":"00:53.550","Text":"and the dipole moment is given by the letter P. P is a vector quantity,"},{"Start":"00:53.550 ","End":"00:56.985","Text":"and it\u0027s given by the equation q,"},{"Start":"00:56.985 ","End":"01:01.350","Text":"the charge multiplied by d vector,"},{"Start":"01:01.350 ","End":"01:06.635","Text":"where the magnitude of this vector is the distance between the 2 charges,"},{"Start":"01:06.635 ","End":"01:12.185","Text":"and the direction of the vector is from the negative charge until the positive charge."},{"Start":"01:12.185 ","End":"01:14.810","Text":"Now, we assume that d,"},{"Start":"01:14.810 ","End":"01:19.430","Text":"the distance between the 2 dipoles doesn\u0027t change."},{"Start":"01:19.460 ","End":"01:22.680","Text":"The 2 charges in the dipole doesn\u0027t change."},{"Start":"01:22.680 ","End":"01:25.100","Text":"A negative charge and"},{"Start":"01:25.100 ","End":"01:30.180","Text":"a positive charge will never move closer or further away from one another."},{"Start":"01:30.440 ","End":"01:35.770","Text":"Now let\u0027s see what happens when we place this dipole in an electric field."},{"Start":"01:35.770 ","End":"01:39.470","Text":"Because we have a negative charge and a positive charge,"},{"Start":"01:39.470 ","End":"01:44.285","Text":"our charges are going to feel a force according to the equation"},{"Start":"01:44.285 ","End":"01:49.835","Text":"of f is equal to q multiplied by E, the electric field."},{"Start":"01:49.835 ","End":"01:54.770","Text":"Remember that this force isn\u0027t always in the direction of our electric field."},{"Start":"01:54.770 ","End":"01:57.079","Text":"Because if our charge is negative,"},{"Start":"01:57.079 ","End":"02:02.430","Text":"then the force is going to be in the opposite direction to our electric field."},{"Start":"02:03.410 ","End":"02:08.980","Text":"As we can see, our positive charge will experience a force upwards,"},{"Start":"02:08.980 ","End":"02:11.140","Text":"in this example here specifically,"},{"Start":"02:11.140 ","End":"02:16.460","Text":"and our negative charge will experience a force downwards."},{"Start":"02:16.760 ","End":"02:20.905","Text":"We can see that the forces are equal and opposite,"},{"Start":"02:20.905 ","End":"02:27.670","Text":"so the sum of the forces on our dipole is going to be equal to 0."},{"Start":"02:27.670 ","End":"02:30.590","Text":"Our dipole won\u0027t move."},{"Start":"02:31.820 ","End":"02:35.815","Text":"We can say that either our dipole won\u0027t move or"},{"Start":"02:35.815 ","End":"02:39.535","Text":"the center of mass of our dipole won\u0027t move."},{"Start":"02:39.535 ","End":"02:42.505","Text":"In this case, nothing changes."},{"Start":"02:42.505 ","End":"02:50.100","Text":"But what happens if our dipole is at some angle to our electric field?"},{"Start":"02:50.410 ","End":"02:56.545","Text":"Here, we can see that our electric dipole isn\u0027t parallel to our electric field."},{"Start":"02:56.545 ","End":"03:04.025","Text":"Now, if we draw on the forces acting on the electric dipole,"},{"Start":"03:04.025 ","End":"03:06.260","Text":"just like we did before,"},{"Start":"03:06.260 ","End":"03:08.600","Text":"so we can see that our force on"},{"Start":"03:08.600 ","End":"03:13.894","Text":"our positive charge is still going to be in the upwards direction,"},{"Start":"03:13.894 ","End":"03:15.879","Text":"in the direction of the electric field,"},{"Start":"03:15.879 ","End":"03:19.430","Text":"and the force on our negative electron is also"},{"Start":"03:19.430 ","End":"03:23.605","Text":"still going to be in the opposite direction to our electric field."},{"Start":"03:23.605 ","End":"03:25.740","Text":"The forces are always parallel."},{"Start":"03:25.740 ","End":"03:31.730","Text":"We can see that our sum of all of the forces on"},{"Start":"03:31.730 ","End":"03:38.735","Text":"our dipole is still going to be equal to 0 because these forces are equal and opposite."},{"Start":"03:38.735 ","End":"03:43.115","Text":"The sum of the forces will still be equal to 0, they cancel out."},{"Start":"03:43.115 ","End":"03:49.070","Text":"However, the torque experienced by the dipole or the sum of the torques,"},{"Start":"03:49.070 ","End":"03:53.170","Text":"the sum of the moments will not be equal to 0."},{"Start":"03:53.170 ","End":"03:57.440","Text":"The torque that will be experienced by this dipole,"},{"Start":"03:57.440 ","End":"04:00.200","Text":"Tau is equal to p,"},{"Start":"04:00.200 ","End":"04:01.565","Text":"the dipole moment,"},{"Start":"04:01.565 ","End":"04:05.940","Text":"cross E are external electric field."},{"Start":"04:06.890 ","End":"04:09.775","Text":"This is the equation we wanted to get to."},{"Start":"04:09.775 ","End":"04:12.935","Text":"This is what you have to remember and write in your books."},{"Start":"04:12.935 ","End":"04:17.200","Text":"We can see that our dipole is going to want to"},{"Start":"04:17.200 ","End":"04:23.460","Text":"rotate around in order to be parallel to our electric field."},{"Start":"04:23.460 ","End":"04:28.000","Text":"We can see that the center of mass of the dipole isn\u0027t going to move,"},{"Start":"04:28.000 ","End":"04:29.695","Text":"it\u0027s not going to shift around."},{"Start":"04:29.695 ","End":"04:32.230","Text":"However, it is going to rotate."},{"Start":"04:32.230 ","End":"04:35.507","Text":"Our center of mass is going to become the axis of rotation,"},{"Start":"04:35.507 ","End":"04:40.900","Text":"where our positive and negative masses rotate around in order to"},{"Start":"04:40.900 ","End":"04:44.440","Text":"formulate or in order to make"},{"Start":"04:44.440 ","End":"04:49.460","Text":"our dipole be in a position that it\u0027s parallel to our electric field."},{"Start":"04:49.790 ","End":"04:55.520","Text":"If we look at the example that we have over here and we want to find the torque."},{"Start":"04:55.520 ","End":"05:00.095","Text":"it\u0027s going to be equal to P cross E. If I do that,"},{"Start":"05:00.095 ","End":"05:05.195","Text":"cross product, we\u0027ll see that our torque is coming out of the page."},{"Start":"05:05.195 ","End":"05:07.940","Text":"Let\u0027s symbolize that with a dot."},{"Start":"05:07.940 ","End":"05:11.560","Text":"Our torque is coming out of the page towards us,"},{"Start":"05:11.560 ","End":"05:15.500","Text":"and that means that the torque"},{"Start":"05:15.500 ","End":"05:19.594","Text":"or the moment experienced by the dipole is going in this direction,"},{"Start":"05:19.594 ","End":"05:22.070","Text":"in the anticlockwise direction."},{"Start":"05:22.070 ","End":"05:25.130","Text":"What we can see is that if this red arrow is"},{"Start":"05:25.130 ","End":"05:28.880","Text":"our electric field and the black arrow is our dipole moment."},{"Start":"05:28.880 ","End":"05:32.030","Text":"Our dipole moment is always going to want to"},{"Start":"05:32.030 ","End":"05:37.530","Text":"rotate so that it\u0027s in the direction of the electric field."},{"Start":"05:37.530 ","End":"05:44.665","Text":"It\u0027s always going to rotate such that it eventually looks like so."},{"Start":"05:44.665 ","End":"05:47.975","Text":"This is the equation that we wanted to get to."},{"Start":"05:47.975 ","End":"05:51.365","Text":"Now, I\u0027m going to show how we derive this equation."},{"Start":"05:51.365 ","End":"05:53.460","Text":"If that doesn\u0027t interest you or you don\u0027t need it,"},{"Start":"05:53.460 ","End":"05:57.745","Text":"you\u0027re welcome to end this video and move on to the next."},{"Start":"05:57.745 ","End":"06:01.670","Text":"We\u0027re looking again at this diagram where we have an electric dipole which has"},{"Start":"06:01.670 ","End":"06:05.840","Text":"some kind of angle to our electric field."},{"Start":"06:05.840 ","End":"06:09.710","Text":"What we\u0027re going to do is we\u0027re going to define an axis,"},{"Start":"06:09.710 ","End":"06:14.839","Text":"which its origin is at the center of the dipole,"},{"Start":"06:14.839 ","End":"06:18.815","Text":"halfway between the 2 charges."},{"Start":"06:18.815 ","End":"06:24.155","Text":"We\u0027re going to define vector r,"},{"Start":"06:24.155 ","End":"06:27.530","Text":"this green line, which goes from the origin of"},{"Start":"06:27.530 ","End":"06:32.225","Text":"our axis and points towards our positive charge."},{"Start":"06:32.225 ","End":"06:35.210","Text":"We\u0027re going to call that vector r plus."},{"Start":"06:35.210 ","End":"06:38.179","Text":"Then what we\u0027re going to do is we\u0027re going to define"},{"Start":"06:38.179 ","End":"06:43.955","Text":"a similar vector going in this direction from the origin to the negative charge."},{"Start":"06:43.955 ","End":"06:48.630","Text":"We\u0027re going to say that that vector is called r minus."},{"Start":"06:49.310 ","End":"06:55.760","Text":"The torque experienced by a body which is rotating is"},{"Start":"06:55.760 ","End":"07:01.920","Text":"equal to r vector cross our force."},{"Start":"07:01.920 ","End":"07:04.785","Text":"This is from mechanics course."},{"Start":"07:04.785 ","End":"07:09.110","Text":"Let\u0027s work out the torque on our positive charge."},{"Start":"07:09.110 ","End":"07:12.140","Text":"Let\u0027s call that Tau plus."},{"Start":"07:12.140 ","End":"07:15.275","Text":"That is going to be equal to our r vector,"},{"Start":"07:15.275 ","End":"07:23.325","Text":"which is r plus cross the force experienced by our positive charge, so F plus."},{"Start":"07:23.325 ","End":"07:29.720","Text":"Now notice that our r plus vector is exactly half of our d vector because it"},{"Start":"07:29.720 ","End":"07:36.210","Text":"starts from the center of mass and it points in the same direction as our d vector."},{"Start":"07:36.210 ","End":"07:39.945","Text":"It\u0027s going towards the positive charge."},{"Start":"07:39.945 ","End":"07:50.865","Text":"In that case, I can say that this is equal to half d vector cross our F plus."},{"Start":"07:50.865 ","End":"07:52.625","Text":"This is our F plus,"},{"Start":"07:52.625 ","End":"07:55.820","Text":"and then I can substitute in what our F plus is equal to,"},{"Start":"07:55.820 ","End":"08:02.040","Text":"so I have half d vector cross and then our F plus is equal to qE."},{"Start":"08:03.410 ","End":"08:09.235","Text":"Really, F plus is in the direction of our E vector."},{"Start":"08:09.235 ","End":"08:12.050","Text":"Great, that\u0027s our Tau plus."},{"Start":"08:12.050 ","End":"08:15.245","Text":"Now let\u0027s see what we get for a negative charge."},{"Start":"08:15.245 ","End":"08:19.370","Text":"Tau minus is going to be equal to our r minus"},{"Start":"08:19.370 ","End":"08:23.510","Text":"vector cross the force on our negative charge,"},{"Start":"08:23.510 ","End":"08:29.945","Text":"which is going to be equal to our r minus vector is exactly half d,"},{"Start":"08:29.945 ","End":"08:32.000","Text":"but in the opposite direction to"},{"Start":"08:32.000 ","End":"08:36.185","Text":"our d vector because it\u0027s going towards the negative charge."},{"Start":"08:36.185 ","End":"08:42.364","Text":"It\u0027s in the opposite direction so we\u0027ll have negative half d vector,"},{"Start":"08:42.364 ","End":"08:48.435","Text":"and then cross product with our force, now F minus."},{"Start":"08:48.435 ","End":"08:50.610","Text":"Let\u0027s take a look."},{"Start":"08:50.610 ","End":"08:54.450","Text":"It\u0027s going in this direction."},{"Start":"08:54.450 ","End":"08:58.060","Text":"That\u0027s our F minus."},{"Start":"08:59.080 ","End":"09:05.599","Text":"We can see that that is equal to a negative charge,"},{"Start":"09:05.599 ","End":"09:12.145","Text":"so negative q multiplied by our electric field."},{"Start":"09:12.145 ","End":"09:16.910","Text":"Then we can see that this negative and this negative cancel out,"},{"Start":"09:16.910 ","End":"09:23.645","Text":"meaning that we will be left with half d cross qE,"},{"Start":"09:23.645 ","End":"09:26.845","Text":"just like our Tau plus."},{"Start":"09:26.845 ","End":"09:31.210","Text":"Therefore, the sum of all torques,"},{"Start":"09:31.210 ","End":"09:36.830","Text":"sum of Tau, is going to be equal to our Tau plus,"},{"Start":"09:36.830 ","End":"09:39.844","Text":"plus our Tau minus,"},{"Start":"09:39.844 ","End":"09:44.690","Text":"which is simply going to be equal to 2 Tau plus,"},{"Start":"09:44.690 ","End":"09:50.040","Text":"which is going to be equal to d cross qE,"},{"Start":"09:50.040 ","End":"09:54.195","Text":"so d cross q,"},{"Start":"09:54.195 ","End":"10:04.275","Text":"dq is equal to our P vector cross our E vector over here."},{"Start":"10:04.275 ","End":"10:09.285","Text":"We can have a d vector multiplied by q is our P,"},{"Start":"10:09.285 ","End":"10:13.730","Text":"and then we have it cross product with our E. That\u0027s what we get over here."},{"Start":"10:13.730 ","End":"10:17.845","Text":"This is what we got for our equation like so."},{"Start":"10:17.845 ","End":"10:20.860","Text":"That\u0027s the end of our lesson."}],"ID":22328},{"Watched":false,"Name":"Multipole Expansion","Duration":"40m 5s","ChapterTopicVideoID":21443,"CourseChapterTopicPlaylistID":99476,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.529","Text":"Hello, in this lesson,"},{"Start":"00:01.529 ","End":"00:04.905","Text":"we\u0027re going to be dealing with the multiple expansion."},{"Start":"00:04.905 ","End":"00:10.030","Text":"We\u0027re going to see what it means and we\u0027re also going to derive it."},{"Start":"00:11.330 ","End":"00:18.284","Text":"Let\u0027s say that we have some kind of general system of charges."},{"Start":"00:18.284 ","End":"00:22.770","Text":"Let\u0027s say that this has volume."},{"Start":"00:22.770 ","End":"00:30.030","Text":"But of course, we can also have a system that just has area or just has length."},{"Start":"00:30.030 ","End":"00:36.960","Text":"Here, we\u0027re dealing with volume and we\u0027re taking a general shape."},{"Start":"00:36.960 ","End":"00:41.410","Text":"This is correct for everything."},{"Start":"00:42.380 ","End":"00:46.620","Text":"This has charge density per unit volume Rho,"},{"Start":"00:46.620 ","End":"00:51.220","Text":"and it\u0027s dependent on the position."},{"Start":"00:51.710 ","End":"00:58.100","Text":"Now let\u0027s say we want to find the potential at some random point. Let\u0027s say over here."},{"Start":"00:58.100 ","End":"01:02.840","Text":"Of course, this potential is also dependent on its position r,"},{"Start":"01:02.840 ","End":"01:06.769","Text":"and just to differentiate between the position of the point where we\u0027re finding"},{"Start":"01:06.769 ","End":"01:12.724","Text":"the potential and the charge density per unit volume,"},{"Start":"01:12.724 ","End":"01:16.114","Text":"depending on where we are within this material."},{"Start":"01:16.114 ","End":"01:21.240","Text":"Let\u0027s call this r vector, r tag vector."},{"Start":"01:23.180 ","End":"01:27.980","Text":"R vector is where we are calculating the potential."},{"Start":"01:27.980 ","End":"01:35.490","Text":"This is potential and r tag vector is the position within the charged material."},{"Start":"01:36.260 ","End":"01:41.540","Text":"Now let\u0027s imagine that we have some axes."},{"Start":"01:41.540 ","End":"01:42.920","Text":"Here is the x direction,"},{"Start":"01:42.920 ","End":"01:44.930","Text":"here is the y direction."},{"Start":"01:44.930 ","End":"01:50.234","Text":"Then from the origin to some point within the material."},{"Start":"01:50.234 ","End":"01:53.589","Text":"This is our r tag vector."},{"Start":"01:53.589 ","End":"02:01.105","Text":"From the origin up until the point where we\u0027re calculating our potential is our r vector."},{"Start":"02:01.105 ","End":"02:05.599","Text":"What we\u0027ll see when we do the multiple expansion is"},{"Start":"02:05.599 ","End":"02:09.730","Text":"that we will get that the potential at this point,"},{"Start":"02:09.730 ","End":"02:11.210","Text":"r over here,"},{"Start":"02:11.210 ","End":"02:13.909","Text":"this general point is going to be something."},{"Start":"02:13.909 ","End":"02:17.180","Text":"We\u0027ll see what this something is divided by r,"},{"Start":"02:17.180 ","End":"02:20.480","Text":"where of course r is the magnitude of r vector."},{"Start":"02:20.480 ","End":"02:24.005","Text":"If we take the magnitude of r vector will get this r,"},{"Start":"02:24.005 ","End":"02:29.260","Text":"and then we\u0027ll have plus another something divided by r^2"},{"Start":"02:29.260 ","End":"02:36.005","Text":"plus another something divided by r^3 and so on and so forth."},{"Start":"02:36.005 ","End":"02:39.475","Text":"What we\u0027ll have over here is,"},{"Start":"02:39.475 ","End":"02:46.984","Text":"over here we\u0027ll have k multiplied by the total charge of this material."},{"Start":"02:46.984 ","End":"02:50.000","Text":"Then, here we have the potential."},{"Start":"02:50.000 ","End":"02:52.294","Text":"This is the potential for a point charge."},{"Start":"02:52.294 ","End":"02:54.664","Text":"Then we have the potential for a dipole,"},{"Start":"02:54.664 ","End":"03:00.390","Text":"which is k multiplied by the dipole moment.r hat."},{"Start":"03:00.390 ","End":"03:08.249","Text":"Then here we\u0027ll have the potential for a quadrupole. Let\u0027s write it in."},{"Start":"03:08.300 ","End":"03:16.950","Text":"Here, we\u0027ll have, so for a quadrupole and here we\u0027ll have a for an octopole and so on."},{"Start":"03:17.800 ","End":"03:24.600","Text":"This is going to be an infinite series."},{"Start":"03:24.600 ","End":"03:27.750","Text":"It\u0027s just going to go on until infinity,"},{"Start":"03:27.750 ","End":"03:32.115","Text":"r divided by r^5, r^6, and so on."},{"Start":"03:32.115 ","End":"03:39.169","Text":"In a case where we have that r is much larger than r tag,"},{"Start":"03:39.169 ","End":"03:44.944","Text":"so r the distance where we\u0027re trying to calculate the potential"},{"Start":"03:44.944 ","End":"03:51.200","Text":"is much bigger than the distance between all of the particles in this charged material,"},{"Start":"03:51.200 ","End":"03:53.820","Text":"so that\u0027s what that means."},{"Start":"03:55.310 ","End":"04:01.190","Text":"In this case, when the point where potential has been"},{"Start":"04:01.190 ","End":"04:03.740","Text":"calculated as much farther than"},{"Start":"04:03.740 ","End":"04:08.059","Text":"the distance between the particles and the charged material."},{"Start":"04:08.059 ","End":"04:10.729","Text":"When we have that r is much bigger than r tag,"},{"Start":"04:10.729 ","End":"04:16.360","Text":"then each element is much smaller than its predecessor."},{"Start":"04:16.360 ","End":"04:25.535","Text":"This divided by r^2 is going to be much smaller than this divided by r and"},{"Start":"04:25.535 ","End":"04:30.110","Text":"this octopole divided by r^4 will be much smaller"},{"Start":"04:30.110 ","End":"04:35.550","Text":"than this quadrupole divided by r^3 and in turn much smaller than this and this."},{"Start":"04:35.550 ","End":"04:39.379","Text":"That means that if we\u0027re measuring the potential when we\u0027re very"},{"Start":"04:39.379 ","End":"04:42.859","Text":"far away when we\u0027re taking approximations,"},{"Start":"04:42.859 ","End":"04:46.355","Text":"for this large value,"},{"Start":"04:46.355 ","End":"04:51.410","Text":"then what we will take is the largest element in the series,"},{"Start":"04:51.410 ","End":"04:55.590","Text":"which is a non-0 value."},{"Start":"04:56.330 ","End":"04:59.479","Text":"To calculate the potential when using"},{"Start":"04:59.479 ","End":"05:02.934","Text":"this approximation where r is much larger than r tag,"},{"Start":"05:02.934 ","End":"05:06.610","Text":"we take the largest non-0 element."},{"Start":"05:06.610 ","End":"05:09.890","Text":"What we\u0027ve seen before is that sometimes we\u0027ve been dealing"},{"Start":"05:09.890 ","End":"05:13.025","Text":"with trying to find the potential when we have a dipole."},{"Start":"05:13.025 ","End":"05:17.234","Text":"That means that we have 1 positive and 1 negative charge."},{"Start":"05:17.234 ","End":"05:23.145","Text":"So they have the same magnitude of charge but with opposite signs."},{"Start":"05:23.145 ","End":"05:27.010","Text":"In that case, we\u0027ve seen that the total charge is equal to 0."},{"Start":"05:27.010 ","End":"05:29.060","Text":"In that case, this first element,"},{"Start":"05:29.060 ","End":"05:32.330","Text":"or this first-order potential approximation,"},{"Start":"05:32.330 ","End":"05:37.380","Text":"is going to be equal to 0 because the total charge is going to be 0."},{"Start":"05:37.380 ","End":"05:40.159","Text":"But of course, we know that there are still charges,"},{"Start":"05:40.159 ","End":"05:43.170","Text":"which means that there\u0027s still going to be potential."},{"Start":"05:43.780 ","End":"05:46.040","Text":"This was our zeroth-order."},{"Start":"05:46.040 ","End":"05:49.915","Text":"We moved over to our first-order."},{"Start":"05:49.915 ","End":"05:53.899","Text":"That is the second element in this series."},{"Start":"05:53.899 ","End":"05:57.150","Text":"This is the potential for a dipole."},{"Start":"05:57.400 ","End":"06:01.399","Text":"Then we saw that when we use this calculation,"},{"Start":"06:01.399 ","End":"06:02.704","Text":"we find the dipole moment,"},{"Start":"06:02.704 ","End":"06:06.920","Text":"and then we can in fact find the potential very far away."},{"Start":"06:06.920 ","End":"06:10.250","Text":"What happens if we have 2 dipoles together?"},{"Start":"06:10.250 ","End":"06:13.520","Text":"We have 2 positive charges and 2 negative charges."},{"Start":"06:13.520 ","End":"06:16.045","Text":"That means we have a quadrupole."},{"Start":"06:16.045 ","End":"06:20.645","Text":"If we try to find the potential of a dipole when we have a quadrupole,"},{"Start":"06:20.645 ","End":"06:24.170","Text":"we\u0027re going to get that this is also equal to 0"},{"Start":"06:24.170 ","End":"06:28.520","Text":"in which case we have to go to the next element along in the series,"},{"Start":"06:28.520 ","End":"06:31.370","Text":"which is the potential for a quadrupole."},{"Start":"06:31.370 ","End":"06:34.384","Text":"That means that whatever our system,"},{"Start":"06:34.384 ","End":"06:40.125","Text":"we always take the potential where we have the largest non-0 elements."},{"Start":"06:40.125 ","End":"06:41.660","Text":"Depending on our system,"},{"Start":"06:41.660 ","End":"06:47.190","Text":"we\u0027re going to choose either the zeroth order up until the nth order."},{"Start":"06:47.330 ","End":"06:53.029","Text":"Of course, before our equation for a quadrupole and octopole and so on,"},{"Start":"06:53.029 ","End":"06:56.735","Text":"we\u0027re going to have k\u0027s multiplying everything out."},{"Start":"06:56.735 ","End":"07:03.770","Text":"Now we\u0027ve understood why we have this multipole expansion and now let\u0027s see what it is."},{"Start":"07:03.770 ","End":"07:06.260","Text":"What\u0027s the equation for the quadrupole and octopole,"},{"Start":"07:06.260 ","End":"07:11.350","Text":"and etc. Let\u0027s begin."},{"Start":"07:11.350 ","End":"07:16.290","Text":"Let\u0027s take this little point over here."},{"Start":"07:16.290 ","End":"07:24.094","Text":"We have this volume over here and it has some kind of charge dq tag."},{"Start":"07:24.094 ","End":"07:35.040","Text":"What we want to do is we want to know what potential it induces over here at this point."},{"Start":"07:36.110 ","End":"07:39.770","Text":"We\u0027ll draw a vector from this piece"},{"Start":"07:39.770 ","End":"07:43.279","Text":"over here up until where we\u0027re measuring the potential."},{"Start":"07:43.279 ","End":"07:51.150","Text":"This vector is equal to our r vector minus our r tag vector."},{"Start":"07:51.150 ","End":"07:58.099","Text":"Because if we add on this r tag vectors in the negative direction,"},{"Start":"07:58.099 ","End":"08:02.910","Text":"so we have r vector minus the r tag vector."},{"Start":"08:04.160 ","End":"08:08.450","Text":"Now if we want to calculate the potential from this point,"},{"Start":"08:08.450 ","End":"08:11.015","Text":"so let\u0027s scroll down a little bit."},{"Start":"08:11.015 ","End":"08:13.205","Text":"We\u0027ll have d Phi,"},{"Start":"08:13.205 ","End":"08:18.510","Text":"so the potential is going to be equal to 4 point charge,"},{"Start":"08:18.510 ","End":"08:21.499","Text":"so k multiplied by the total charge,"},{"Start":"08:21.499 ","End":"08:23.554","Text":"which is dq tag,"},{"Start":"08:23.554 ","End":"08:25.894","Text":"the total charge of this point is dq tag,"},{"Start":"08:25.894 ","End":"08:33.155","Text":"divided by r. The vector over here is r minus r tag."},{"Start":"08:33.155 ","End":"08:38.910","Text":"We take the magnitude of r minus r tag."},{"Start":"08:41.460 ","End":"08:48.130","Text":"Now, if I wanted the total potential at this point over here,"},{"Start":"08:48.130 ","End":"08:54.175","Text":"due to all of the particles in this shape,"},{"Start":"08:54.175 ","End":"09:00.160","Text":"so I want the total potential at this point"},{"Start":"09:00.160 ","End":"09:08.470","Text":"r. It will simply be equal to the integral along d Phi."},{"Start":"09:08.470 ","End":"09:11.350","Text":"We\u0027ll be integrating along this."},{"Start":"09:11.350 ","End":"09:19.940","Text":"Our integrating constants will be integrating along q and along r tag."},{"Start":"09:20.280 ","End":"09:25.195","Text":"What\u0027s important to note is that our r vector over here,"},{"Start":"09:25.195 ","End":"09:27.744","Text":"this is a constant."},{"Start":"09:27.744 ","End":"09:29.454","Text":"It isn\u0027t changing,"},{"Start":"09:29.454 ","End":"09:31.255","Text":"and we can see that from the diagram."},{"Start":"09:31.255 ","End":"09:34.269","Text":"This is our r vector in green and it\u0027s not changing."},{"Start":"09:34.269 ","End":"09:37.705","Text":"We are always calculating the potential at this point."},{"Start":"09:37.705 ","End":"09:40.539","Text":"This r vector is just from the origin until this point."},{"Start":"09:40.539 ","End":"09:42.414","Text":"So there\u0027s no reason for it to change."},{"Start":"09:42.414 ","End":"09:45.039","Text":"But our r tag is constantly"},{"Start":"09:45.039 ","End":"09:49.960","Text":"changing because we\u0027re summing up all of the charges in the shape."},{"Start":"09:49.960 ","End":"09:52.638","Text":"So r tag is constantly changing,"},{"Start":"09:52.638 ","End":"09:56.559","Text":"and so therefore our dq is also changing."},{"Start":"09:56.559 ","End":"10:01.550","Text":"That\u0027s what we\u0027re integrating along."},{"Start":"10:02.190 ","End":"10:05.710","Text":"We\u0027ll see that inside the dq tag,"},{"Start":"10:05.710 ","End":"10:07.930","Text":"we have our dr tag hiding."},{"Start":"10:07.930 ","End":"10:10.090","Text":"We\u0027ll get there in a second."},{"Start":"10:10.090 ","End":"10:13.284","Text":"The k is a constant so we can take it out,"},{"Start":"10:13.284 ","End":"10:15.446","Text":"and then we\u0027re integrating along dq."},{"Start":"10:15.446 ","End":"10:17.870","Text":"What is dq?"},{"Start":"10:17.940 ","End":"10:23.650","Text":"Dq tag is our charge density, which is Rho,"},{"Start":"10:23.650 ","End":"10:26.320","Text":"which is dependent on r tag,"},{"Start":"10:26.320 ","End":"10:31.660","Text":"as we saw, multiplied by its volume."},{"Start":"10:31.660 ","End":"10:34.360","Text":"A small element of volume,"},{"Start":"10:34.360 ","End":"10:36.295","Text":"so dv tag,"},{"Start":"10:36.295 ","End":"10:40.539","Text":"and this is of course still divided by the magnitude of the radius,"},{"Start":"10:40.539 ","End":"10:44.600","Text":"so that\u0027s r minus r tag."},{"Start":"10:44.640 ","End":"10:50.090","Text":"Now what I want to do is I want to deal with this size over here."},{"Start":"10:50.880 ","End":"10:56.058","Text":"Let\u0027s write this out over here."},{"Start":"10:56.058 ","End":"11:04.780","Text":"Finding the magnitude of this vector minus this vector is the same as doing"},{"Start":"11:04.780 ","End":"11:09.309","Text":"this multiplication of r vector minus r tag vector"},{"Start":"11:09.309 ","End":"11:15.385","Text":"multiplied by r vector minus r tag vector."},{"Start":"11:15.385 ","End":"11:20.539","Text":"These two mean the exact same thing."},{"Start":"11:20.730 ","End":"11:23.230","Text":"When we open up the brackets,"},{"Start":"11:23.230 ","End":"11:26.680","Text":"when we have r vector multiplied by r vector,"},{"Start":"11:26.680 ","End":"11:29.155","Text":"so we\u0027re taking the dot product."},{"Start":"11:29.155 ","End":"11:32.590","Text":"What we\u0027re just going to get is r^2."},{"Start":"11:32.590 ","End":"11:36.710","Text":"It\u0027s just the magnitude of r vector squared."},{"Start":"11:37.020 ","End":"11:42.805","Text":"I\u0027m reminding you that r vector is simply rr hat."},{"Start":"11:42.805 ","End":"11:46.330","Text":"When we have r vector multiplied by r vector,"},{"Start":"11:46.330 ","End":"11:49.765","Text":"what we have is r multiplied by r,"},{"Start":"11:49.765 ","End":"11:54.010","Text":"so r^2 multiplied by r hat.r hat,"},{"Start":"11:54.010 ","End":"11:56.350","Text":"which is simply equal to r^2."},{"Start":"11:56.350 ","End":"11:58.660","Text":"This is equal to 1."},{"Start":"11:58.660 ","End":"12:02.950","Text":"Now I\u0027m going to rub this out to not confuse,"},{"Start":"12:02.950 ","End":"12:11.748","Text":"but we have r^2 and then we have negative 2r vector.r tag vector,"},{"Start":"12:11.748 ","End":"12:18.830","Text":"and then we have plus r tag squared for the same reason that we got r^2."},{"Start":"12:19.260 ","End":"12:21.699","Text":"Here we have a magnitude,"},{"Start":"12:21.699 ","End":"12:23.365","Text":"here we have a magnitude,"},{"Start":"12:23.365 ","End":"12:28.090","Text":"and here we have the dot product or scalar multiplication."},{"Start":"12:28.090 ","End":"12:33.280","Text":"Let\u0027s just write dot product or scalar multiplication."},{"Start":"12:33.280 ","End":"12:36.519","Text":"Whenever we have a dot product or scalar multiplication,"},{"Start":"12:36.519 ","End":"12:40.974","Text":"we\u0027re going to get some magnitude,"},{"Start":"12:40.974 ","End":"12:44.665","Text":"so this value also isn\u0027t a vector."},{"Start":"12:44.665 ","End":"12:47.154","Text":"It\u0027s just a magnitude."},{"Start":"12:47.154 ","End":"12:50.724","Text":"I don\u0027t want to write this like this."},{"Start":"12:50.724 ","End":"12:54.220","Text":"What I\u0027m going to do is I\u0027m going to write that r"},{"Start":"12:54.220 ","End":"13:01.779","Text":"vector.r tag vector is simply equal to our equation for the dot product,"},{"Start":"13:01.779 ","End":"13:04.780","Text":"which is the magnitude of r vector, which is r,"},{"Start":"13:04.780 ","End":"13:07.869","Text":"multiplied by the magnitude of our r tag vector,"},{"Start":"13:07.869 ","End":"13:13.140","Text":"which is r tag, multiplied by cosine of the angle between the 2."},{"Start":"13:13.140 ","End":"13:17.470","Text":"I\u0027m going to call the angle between the 2, Theta tag."},{"Start":"13:18.150 ","End":"13:23.380","Text":"Theta tag is the angle between r and r tag."},{"Start":"13:23.380 ","End":"13:27.100","Text":"If we come to our diagram over here,"},{"Start":"13:27.100 ","End":"13:31.596","Text":"this is our angle Theta tag,"},{"Start":"13:31.596 ","End":"13:37.159","Text":"between our r vector and our r tag vector like so."},{"Start":"13:38.160 ","End":"13:41.829","Text":"I\u0027ve labeled Theta tag as Theta tag and not"},{"Start":"13:41.829 ","End":"13:46.585","Text":"Theta because my Theta tag is dependent on r tag."},{"Start":"13:46.585 ","End":"13:50.720","Text":"As my r tag vector moves around,"},{"Start":"13:50.910 ","End":"13:54.340","Text":"therefore my angle Theta tag will change."},{"Start":"13:54.340 ","End":"13:57.303","Text":"First, my r vector will be here and then it will be here,"},{"Start":"13:57.303 ","End":"13:58.689","Text":"and my Theta tag will change,"},{"Start":"13:58.689 ","End":"14:04.315","Text":"and of course, my r vector is constantly constant, it\u0027s not changing."},{"Start":"14:04.315 ","End":"14:07.600","Text":"My Theta tag is not dependent on my r vector,"},{"Start":"14:07.600 ","End":"14:10.180","Text":"but it is dependent on my r tag vector,"},{"Start":"14:10.180 ","End":"14:13.880","Text":"so that\u0027s why there\u0027s the tag here to remind you that."},{"Start":"14:14.400 ","End":"14:19.165","Text":"Now let\u0027s go back over here."},{"Start":"14:19.165 ","End":"14:22.569","Text":"Now I\u0027m going to write this like so."},{"Start":"14:22.569 ","End":"14:25.840","Text":"I have r^2 minus 2,"},{"Start":"14:25.840 ","End":"14:35.750","Text":"and then I have r.r tag multiplied by cosine of Theta tag plus r tag squared."},{"Start":"14:35.820 ","End":"14:42.820","Text":"Now, something very important that I forgot is to put in the square root signs."},{"Start":"14:42.820 ","End":"14:50.800","Text":"The magnitude of r minus r tag vector is the square root of this squared."},{"Start":"14:50.800 ","End":"15:00.700","Text":"Every way I have to add in this square root sign, like so."},{"Start":"15:00.700 ","End":"15:04.975","Text":"Now I\u0027m going to carry this on over here,"},{"Start":"15:04.975 ","End":"15:07.585","Text":"and what I\u0027m going to do is I\u0027m going to take my r"},{"Start":"15:07.585 ","End":"15:12.519","Text":"squared out as a common factor or a common multiple."},{"Start":"15:12.519 ","End":"15:15.445","Text":"I\u0027ll have r^2,"},{"Start":"15:15.445 ","End":"15:18.639","Text":"I\u0027ll put in the square root in a second multiplied by,"},{"Start":"15:18.639 ","End":"15:26.379","Text":"so r^2 divided by r^2 is 1 minus 2r tag cosine of"},{"Start":"15:26.379 ","End":"15:35.394","Text":"Theta tag divided by r plus r tag divided by r^2,"},{"Start":"15:35.394 ","End":"15:39.175","Text":"and then the square root of all of this."},{"Start":"15:39.175 ","End":"15:44.109","Text":"If you multiply out this bracket over here,"},{"Start":"15:44.109 ","End":"15:47.480","Text":"you\u0027ll get the exact same thing that we had over here."},{"Start":"15:48.690 ","End":"15:52.945","Text":"Now I\u0027m going to take this r^2 out of the square root."},{"Start":"15:52.945 ","End":"16:00.054","Text":"I\u0027ll just have r multiplied by the square root of 1 minus 2r tag cosine of Theta tag"},{"Start":"16:00.054 ","End":"16:06.978","Text":"divided by r plus r tagged divided by r^2,"},{"Start":"16:06.978 ","End":"16:09.740","Text":"and the square root of all of this."},{"Start":"16:10.470 ","End":"16:15.625","Text":"Now what I\u0027m going to do, instead of writing out all of this over and over again,"},{"Start":"16:15.625 ","End":"16:17.619","Text":"I\u0027m going to take this,"},{"Start":"16:17.619 ","End":"16:19.134","Text":"let\u0027s mark it in blue,"},{"Start":"16:19.134 ","End":"16:22.383","Text":"including the minus sign over here."},{"Start":"16:22.383 ","End":"16:27.110","Text":"All of this, I\u0027m going to call Epsilon."},{"Start":"16:27.180 ","End":"16:29.695","Text":"Let\u0027s write this out."},{"Start":"16:29.695 ","End":"16:38.080","Text":"Epsilon is equal to r tag divided by r^2 minus 2r tag,"},{"Start":"16:38.080 ","End":"16:44.270","Text":"cosine of Theta tag divided by r. This is Epsilon."},{"Start":"16:45.120 ","End":"16:48.910","Text":"What we\u0027ll have is that this is equal to,"},{"Start":"16:48.910 ","End":"16:51.894","Text":"therefore, taking this into account,"},{"Start":"16:51.894 ","End":"16:59.510","Text":"this is equal to r multiplied by the square root of 1 plus Epsilon."},{"Start":"17:00.660 ","End":"17:04.900","Text":"Then instead of writing the square root like this,"},{"Start":"17:04.900 ","End":"17:08.680","Text":"I\u0027m going to write it like so."},{"Start":"17:08.680 ","End":"17:11.365","Text":"This is the exact same thing,"},{"Start":"17:11.365 ","End":"17:13.525","Text":"1 plus Epsilon,"},{"Start":"17:13.525 ","End":"17:15.550","Text":"and then we take the square root of that,"},{"Start":"17:15.550 ","End":"17:19.460","Text":"which I can just write as this to the power of a half."},{"Start":"17:19.620 ","End":"17:23.109","Text":"Now let\u0027s go back to our potential."},{"Start":"17:23.109 ","End":"17:25.539","Text":"Let\u0027s just scroll down a bit."},{"Start":"17:25.539 ","End":"17:29.245","Text":"We\u0027re back to our potential over here."},{"Start":"17:29.245 ","End":"17:36.130","Text":"We have that our potential as a function of r is equal to k multiplied by the"},{"Start":"17:36.130 ","End":"17:43.104","Text":"integral of 1 divided by the magnitude of r minus r tag vector,"},{"Start":"17:43.104 ","End":"17:46.075","Text":"multiplied by Rho,"},{"Start":"17:46.075 ","End":"17:50.935","Text":"which is a function of r tag dv tag."},{"Start":"17:50.935 ","End":"17:54.445","Text":"I\u0027ve just written this out and now what I want to do"},{"Start":"17:54.445 ","End":"18:00.350","Text":"is I want to work on this over here, this fraction."},{"Start":"18:00.840 ","End":"18:03.535","Text":"Let\u0027s work on this."},{"Start":"18:03.535 ","End":"18:10.912","Text":"We have that 1 divided by the magnitude of r vector minus r tag vector,"},{"Start":"18:10.912 ","End":"18:13.794","Text":"so r minus r tag vector is this,"},{"Start":"18:13.794 ","End":"18:16.809","Text":"which we saw is equal to this over here."},{"Start":"18:16.809 ","End":"18:20.380","Text":"That\u0027s like saying that it\u0027s equal to 1 divided by r,"},{"Start":"18:20.380 ","End":"18:24.970","Text":"1 plus Epsilon to the power of 1/2."},{"Start":"18:24.970 ","End":"18:28.930","Text":"This I can write as 1 divided by r multiplied by"},{"Start":"18:28.930 ","End":"18:34.165","Text":"1 plus Epsilon to the power of negative 1.5."},{"Start":"18:34.165 ","End":"18:37.450","Text":"Up until here, very simple."},{"Start":"18:37.450 ","End":"18:46.750","Text":"Now what I\u0027m going to do is I\u0027m going to take this and use Taylor series approximation,"},{"Start":"18:46.750 ","End":"18:48.849","Text":"in order to approximate it."},{"Start":"18:48.849 ","End":"18:50.589","Text":"We\u0027re going to be using Taylor,"},{"Start":"18:50.589 ","End":"18:58.160","Text":"where Epsilon is my variable and that\u0027s what we\u0027re going to do now."},{"Start":"18:59.070 ","End":"19:06.205","Text":"Right now we\u0027re doing the Taylor expansion about Epsilon is equal to 0."},{"Start":"19:06.205 ","End":"19:09.685","Text":"I\u0027m assuming that you know how to do the Taylor expansion."},{"Start":"19:09.685 ","End":"19:17.335","Text":"You have to take the derivatives and divide it by 2 factorial and 3 factorial, etc."},{"Start":"19:17.335 ","End":"19:20.094","Text":"If you don\u0027t know it, please look it up."},{"Start":"19:20.094 ","End":"19:22.135","Text":"I\u0027m just going to write it out."},{"Start":"19:22.135 ","End":"19:26.484","Text":"We\u0027re going to have 1 divided by r multiplied by the Taylor expansion of this,"},{"Start":"19:26.484 ","End":"19:47.544","Text":"which is simply 1 minus 1/2 Epsilon plus"},{"Start":"19:47.544 ","End":"19:49.105","Text":"3/8 Epsilon squared,"},{"Start":"19:49.105 ","End":"19:55.150","Text":"minus 5/16 Epsilon cubed,"},{"Start":"19:55.150 ","End":"19:58.640","Text":"and so on and so forth."},{"Start":"20:00.060 ","End":"20:05.540","Text":"Now let\u0027s substitute in our Epsilon."},{"Start":"20:06.180 ","End":"20:09.819","Text":"Let\u0027s scroll down over here,"},{"Start":"20:09.819 ","End":"20:14.200","Text":"and here, we\u0027re subbing in Epsilon."},{"Start":"20:14.200 ","End":"20:16.795","Text":"Let\u0027s just rewrite our Epsilon,"},{"Start":"20:16.795 ","End":"20:21.579","Text":"because we\u0027ll see that when we have to square and cube Epsilon, it\u0027s easier like this."},{"Start":"20:21.579 ","End":"20:25.029","Text":"We\u0027re going to take r tag divided by r as a common factor."},{"Start":"20:25.029 ","End":"20:27.925","Text":"We have r tag divided by r over here,"},{"Start":"20:27.925 ","End":"20:36.595","Text":"and then it\u0027s multiplied by r tag divided by r minus 2 cosine of Theta tag."},{"Start":"20:36.595 ","End":"20:41.545","Text":"This is just Epsilon written out a bit easier."},{"Start":"20:41.545 ","End":"20:44.214","Text":"Now we\u0027re substituting this in over here."},{"Start":"20:44.214 ","End":"20:51.770","Text":"We\u0027ll have 1 divided by r multiplied by 1 minus a 1/2 of Epsilon."},{"Start":"20:53.040 ","End":"20:58.600","Text":"We\u0027ll have multiplied by r tag divided by r"},{"Start":"20:58.600 ","End":"21:04.074","Text":"multiplied by r tag divided by r minus 2 cosine of Theta tag."},{"Start":"21:04.074 ","End":"21:10.679","Text":"Then we have plus 3/8 of Epsilon squared."},{"Start":"21:10.679 ","End":"21:14.190","Text":"Here we have r tag divided by r squared,"},{"Start":"21:14.190 ","End":"21:18.089","Text":"so here you can see it\u0027s easier multiplied by r tag divided by r"},{"Start":"21:18.089 ","End":"21:23.280","Text":"minus 2 cosine of Theta tag squared,"},{"Start":"21:23.280 ","End":"21:27.420","Text":"and then minus 5/16 of Epsilon cubed."},{"Start":"21:27.420 ","End":"21:32.549","Text":"We have r tag divided by r cubed multiplied by"},{"Start":"21:32.549 ","End":"21:39.450","Text":"r tag divided by r minus 2 cosine of Theta tag cubed,"},{"Start":"21:39.450 ","End":"21:42.730","Text":"and then so on and so forth."},{"Start":"21:43.950 ","End":"21:48.115","Text":"Now what we\u0027ll notice is that when we open up these brackets,"},{"Start":"21:48.115 ","End":"21:51.010","Text":"we\u0027re going to get all mixed elements."},{"Start":"21:51.010 ","End":"21:57.235","Text":"Because here r tag divided by r multiplied by this will give us squared,"},{"Start":"21:57.235 ","End":"21:59.185","Text":"r tag divided by r squared."},{"Start":"21:59.185 ","End":"22:05.499","Text":"Then here it will give us r tag divided by r multiplied by some value."},{"Start":"22:05.499 ","End":"22:12.010","Text":"Here we have r tag divided by r^2 multiplied by this so then we\u0027ll have it cubed."},{"Start":"22:12.010 ","End":"22:16.945","Text":"Of course, this is also to the power of 2 and so on."},{"Start":"22:16.945 ","End":"22:19.585","Text":"We have lots of mixed elements."},{"Start":"22:19.585 ","End":"22:21.520","Text":"Let\u0027s just rewrite it."},{"Start":"22:21.520 ","End":"22:26.095","Text":"I\u0027ve already taken the liberty of opening up the brackets myself."},{"Start":"22:26.095 ","End":"22:27.610","Text":"If you want to do it yourself,"},{"Start":"22:27.610 ","End":"22:29.665","Text":"obviously pause the video and go for it."},{"Start":"22:29.665 ","End":"22:31.959","Text":"We have 1 plus,"},{"Start":"22:31.959 ","End":"22:39.145","Text":"and then we have r tag divided by r multiplied by cosine of Theta tag."},{"Start":"22:39.145 ","End":"22:42.925","Text":"This is to the power of 1."},{"Start":"22:42.925 ","End":"22:50.155","Text":"Then we have plus r tag divided by r to the power of 2 multiplied by"},{"Start":"22:50.155 ","End":"22:58.330","Text":"3 cosine squared Theta tag minus 1 divided by 2."},{"Start":"22:58.330 ","End":"23:02.035","Text":"Then after that we have r tag divided by"},{"Start":"23:02.035 ","End":"23:09.835","Text":"r cubed multiplied by 5 cosine cubed,"},{"Start":"23:09.835 ","End":"23:14.980","Text":"Theta tag minus 3 cosine Theta tag,"},{"Start":"23:14.980 ","End":"23:18.249","Text":"divided by 2 and of course,"},{"Start":"23:18.249 ","End":"23:19.644","Text":"so on and so forth."},{"Start":"23:19.644 ","End":"23:21.980","Text":"I won\u0027t complete this."},{"Start":"23:23.100 ","End":"23:26.349","Text":"Now what I want you to notice,"},{"Start":"23:26.349 ","End":"23:27.715","Text":"let\u0027s do this in blue,"},{"Start":"23:27.715 ","End":"23:30.475","Text":"is that aside from our r tag divided by r,"},{"Start":"23:30.475 ","End":"23:36.759","Text":"we also have all of these cosines over here, functions with cosine."},{"Start":"23:36.759 ","End":"23:39.010","Text":"We see all of this."},{"Start":"23:39.010 ","End":"23:42.774","Text":"What these 3 are,"},{"Start":"23:42.774 ","End":"23:46.531","Text":"it\u0027s very interesting,"},{"Start":"23:46.531 ","End":"23:49.345","Text":"is this is the Legendre polynomial."},{"Start":"23:49.345 ","End":"23:52.190","Text":"Legendre."},{"Start":"23:54.780 ","End":"24:01.105","Text":"The Legendre polynomial can be written as Pn and then in brackets, cosine of theta."},{"Start":"24:01.105 ","End":"24:02.529","Text":"If you don\u0027t know about it,"},{"Start":"24:02.529 ","End":"24:04.659","Text":"it\u0027s useful to maybe look it up,"},{"Start":"24:04.659 ","End":"24:06.309","Text":"but you don\u0027t necessarily have to know it."},{"Start":"24:06.309 ","End":"24:09.984","Text":"But it\u0027s very interesting that we get these elements over here"},{"Start":"24:09.984 ","End":"24:14.990","Text":"which completely match this Legendre polynomial."},{"Start":"24:15.060 ","End":"24:19.525","Text":"Then knowing that, let\u0027s scroll down."},{"Start":"24:19.525 ","End":"24:23.420","Text":"What do we can see is that we just have a series."},{"Start":"24:23.610 ","End":"24:30.580","Text":"What we can say therefore is that 1 divided by the magnitude of I minus I"},{"Start":"24:30.580 ","End":"24:37.195","Text":"tag vector can be simply written as 1 divided by r from over here,"},{"Start":"24:37.195 ","End":"24:47.410","Text":"multiplied by the sum of r tag divided by r to the power of n. Okay,"},{"Start":"24:47.410 ","End":"24:56.800","Text":"so we\u0027re summing along n multiplied by the Legendre polynomial Pn cosine of theta."},{"Start":"24:56.800 ","End":"25:00.500","Text":"Then specifically this is of theta tag,"},{"Start":"25:01.560 ","End":"25:08.660","Text":"and of course, we\u0027re summing from n is equal to zero up until infinity."},{"Start":"25:08.900 ","End":"25:12.315","Text":"This function is very useful to know."},{"Start":"25:12.315 ","End":"25:20.740","Text":"Let\u0027s say this over here would be P2 of cosine of theta in the Legendre polynomial,"},{"Start":"25:20.740 ","End":"25:23.755","Text":"this would be P1, this would be P0,"},{"Start":"25:23.755 ","End":"25:28.129","Text":"and this would be P3, and so on and so forth."},{"Start":"25:28.590 ","End":"25:31.794","Text":"So now we have this."},{"Start":"25:31.794 ","End":"25:35.050","Text":"Let\u0027s go back to our potential over here,"},{"Start":"25:35.050 ","End":"25:36.654","Text":"remember what it looks like?"},{"Start":"25:36.654 ","End":"25:41.479","Text":"Okay, so now we\u0027re just going to write this out again."},{"Start":"25:42.210 ","End":"25:44.844","Text":"Let\u0027s write it in black."},{"Start":"25:44.844 ","End":"25:50.185","Text":"Again, we have that our potential as a function of r is equal to,"},{"Start":"25:50.185 ","End":"25:53.170","Text":"and now we\u0027re going to integrate along each 1 of these things."},{"Start":"25:53.170 ","End":"25:56.725","Text":"We have r and the integral of 1 divided by r,"},{"Start":"25:56.725 ","End":"25:58.945","Text":"and then we\u0027re taking the first element."},{"Start":"25:58.945 ","End":"26:03.715","Text":"We have 1 divided by r multiplied by 1 and then we have it rho dv."},{"Start":"26:03.715 ","End":"26:08.620","Text":"Rho, which is as a function of r tag dv tag,"},{"Start":"26:08.620 ","End":"26:11.740","Text":"and then plus the integral on the second."},{"Start":"26:11.740 ","End":"26:13.989","Text":"Again we have k which is a constant,"},{"Start":"26:13.989 ","End":"26:21.009","Text":"and then we have the integral on 1 divided by r multiplied by r tag"},{"Start":"26:21.009 ","End":"26:29.529","Text":"divided by r multiplied by cosine of theta tag d,"},{"Start":"26:29.529 ","End":"26:35.949","Text":"not yet d, and then rho as a function of r tag,"},{"Start":"26:35.949 ","End":"26:38.590","Text":"and then dv tag."},{"Start":"26:38.590 ","End":"26:41.440","Text":"Then on the third element,"},{"Start":"26:41.440 ","End":"26:45.100","Text":"plus k, the constant and then the integral,"},{"Start":"26:45.100 ","End":"26:52.044","Text":"so then this time let\u0027s just write it as r tag squared divided by r cubed, okay?"},{"Start":"26:52.044 ","End":"26:55.930","Text":"Because here we have I squared but then divided by another r,"},{"Start":"26:55.930 ","End":"26:57.865","Text":"so r cubed, okay?"},{"Start":"26:57.865 ","End":"27:04.104","Text":"Then this is multiplied by 3. cosine squared of theta tag minus 1"},{"Start":"27:04.104 ","End":"27:12.519","Text":"divided by 2 multiplied by rho as a function of r tag dv tag."},{"Start":"27:12.519 ","End":"27:16.450","Text":"Then let\u0027s just do our last 1."},{"Start":"27:16.450 ","End":"27:18.985","Text":"We have plus k and then the integral."},{"Start":"27:18.985 ","End":"27:26.290","Text":"Here we can write r tag cubed divided by r cubed multiplied by 1 divided by r, so forth."},{"Start":"27:26.290 ","End":"27:28.465","Text":"Okay, remember to always multiply it by this,"},{"Start":"27:28.465 ","End":"27:37.810","Text":"and then multiplied by 5 cosine cubed of theta tag minus 3 cosine theta tag,"},{"Start":"27:37.810 ","End":"27:40.525","Text":"all of this divided by 2."},{"Start":"27:40.525 ","End":"27:45.805","Text":"Then row as a function of r tag, dv tag,"},{"Start":"27:45.805 ","End":"27:49.059","Text":"and of course, so on and so forth."},{"Start":"27:49.059 ","End":"27:54.280","Text":"Now I want you to remember that this r,"},{"Start":"27:54.280 ","End":"27:57.909","Text":"okay, all of these over here are constants."},{"Start":"27:57.909 ","End":"28:01.569","Text":"Let\u0027s just go back to the diagram."},{"Start":"28:01.569 ","End":"28:04.074","Text":"Look how much we\u0027ve completed so far."},{"Start":"28:04.074 ","End":"28:07.644","Text":"I\u0027m reminding you r is the magnitude of this r vector,"},{"Start":"28:07.644 ","End":"28:10.030","Text":"which as we said, it\u0027s constant where calculating"},{"Start":"28:10.030 ","End":"28:12.595","Text":"the potential at a specific point that we chose."},{"Start":"28:12.595 ","End":"28:16.045","Text":"This r vector is from the origin to the specific point,"},{"Start":"28:16.045 ","End":"28:18.535","Text":"so it\u0027s constant, it\u0027s not changing."},{"Start":"28:18.535 ","End":"28:20.860","Text":"Let\u0027s come back,"},{"Start":"28:20.860 ","End":"28:22.689","Text":"and in that case,"},{"Start":"28:22.689 ","End":"28:27.129","Text":"we can take out our r from this integral sign,"},{"Start":"28:27.129 ","End":"28:36.489","Text":"so we have k divided by r integral of rho of r tag vector dv tag plus k"},{"Start":"28:36.489 ","End":"28:41.410","Text":"divided by this time r squared integral of r tag"},{"Start":"28:41.410 ","End":"28:46.975","Text":"multiplied by cosine of the theta tag,"},{"Start":"28:46.975 ","End":"28:56.030","Text":"rho r tag vector dv tag plus k divided by r cubed"},{"Start":"28:56.030 ","End":"29:06.360","Text":"integral of r tag squared multiplied by 3 cosine squared theta tag minus 1 divided"},{"Start":"29:06.360 ","End":"29:12.689","Text":"by 2 rho r tag dv tag"},{"Start":"29:12.689 ","End":"29:17.215","Text":"plus k divided by r to the power 4."},{"Start":"29:17.215 ","End":"29:25.735","Text":"Then the integral of r tag cubed multiplied by 5 cosine cubed theta tag minus 3,"},{"Start":"29:25.735 ","End":"29:30.385","Text":"cosine of theta tag divided by 2."},{"Start":"29:30.385 ","End":"29:34.255","Text":"Then row as a function of r tag,"},{"Start":"29:34.255 ","End":"29:38.034","Text":"dv tag, and so on."},{"Start":"29:38.034 ","End":"29:46.765","Text":"Now we can note that here we have our first element."},{"Start":"29:46.765 ","End":"29:54.879","Text":"Here we have our second element and our third element and our fourth element."},{"Start":"29:54.879 ","End":"29:56.500","Text":"I have 1, 2,"},{"Start":"29:56.500 ","End":"29:58.690","Text":"3, 4 elements."},{"Start":"29:58.690 ","End":"30:03.535","Text":"Just like if we go all the way back up to here,"},{"Start":"30:03.535 ","End":"30:04.975","Text":"we had our first element,"},{"Start":"30:04.975 ","End":"30:08.020","Text":"second element, third element, fourth element."},{"Start":"30:08.020 ","End":"30:11.260","Text":"Notice our first is divided by r,"},{"Start":"30:11.260 ","End":"30:12.849","Text":"then divide it by r squared,"},{"Start":"30:12.849 ","End":"30:17.695","Text":"then the third divided by r cubed and the fourth divided by r to the power of 4,"},{"Start":"30:17.695 ","End":"30:19.959","Text":"which is what we have here."},{"Start":"30:19.959 ","End":"30:21.700","Text":"First element divided by r,"},{"Start":"30:21.700 ","End":"30:23.890","Text":"second element divided by r^2,"},{"Start":"30:23.890 ","End":"30:25.885","Text":"the third element divided by r^3,"},{"Start":"30:25.885 ","End":"30:28.974","Text":"and fourth element divided by r 4."},{"Start":"30:28.974 ","End":"30:38.425","Text":"Our first element over here. Let\u0027s take a look at this."},{"Start":"30:38.425 ","End":"30:41.740","Text":"Integrating along rho dv."},{"Start":"30:41.740 ","End":"30:44.215","Text":"Yes, it\u0027s rho r tag dv tag,"},{"Start":"30:44.215 ","End":"30:47.365","Text":"but we\u0027re integrating along rho dv."},{"Start":"30:47.365 ","End":"30:50.365","Text":"If we go back here,"},{"Start":"30:50.365 ","End":"30:55.720","Text":"so we can see that we\u0027re integrating along all of the charges."},{"Start":"30:55.720 ","End":"31:00.760","Text":"Okay, so we\u0027re taking the charge density and integrating along the volume."},{"Start":"31:00.760 ","End":"31:04.435","Text":"What we\u0027ll get is the total charge,"},{"Start":"31:04.435 ","End":"31:06.099","Text":"which is what we have here,"},{"Start":"31:06.099 ","End":"31:09.789","Text":"k multiplied by the total charge,"},{"Start":"31:09.789 ","End":"31:16.030","Text":"which is true if this is just has some kind of total charge Q."},{"Start":"31:16.030 ","End":"31:18.910","Text":"The potential that will measure over here will be"},{"Start":"31:18.910 ","End":"31:22.468","Text":"the same potential that will measure for a point charge."},{"Start":"31:22.468 ","End":"31:25.165","Text":"That\u0027s where this comes from."},{"Start":"31:25.165 ","End":"31:32.120","Text":"If we integrate along rho dv we\u0027ll just get this total charge qt."},{"Start":"31:32.760 ","End":"31:35.290","Text":"Then we\u0027ll get,"},{"Start":"31:35.290 ","End":"31:38.935","Text":"so all of this in blue is just the total charge."},{"Start":"31:38.935 ","End":"31:41.425","Text":"Then we get k,"},{"Start":"31:41.425 ","End":"31:43.270","Text":"Q divided by r,"},{"Start":"31:43.270 ","End":"31:46.434","Text":"which is the potential of a point charge."},{"Start":"31:46.434 ","End":"31:49.750","Text":"Now let\u0027s look at the second 1."},{"Start":"31:49.750 ","End":"31:52.810","Text":"Let\u0027s label this 1."},{"Start":"31:52.810 ","End":"31:56.680","Text":"It was simpler, we just got Q of t. Now let\u0027s look at 2,"},{"Start":"31:56.680 ","End":"32:02.450","Text":"2 is slightly more complicated, so we\u0027ll write it out."},{"Start":"32:02.490 ","End":"32:06.354","Text":"In 2, we have the integral on r tag,"},{"Start":"32:06.354 ","End":"32:13.735","Text":"cosine of theta tag multiplied by rho of r tag dv tag."},{"Start":"32:13.735 ","End":"32:16.555","Text":"R tag cosine of theta tag,"},{"Start":"32:16.555 ","End":"32:19.045","Text":"we can just write this out."},{"Start":"32:19.045 ","End":"32:27.339","Text":"If you remember, r vector dot product with r tag vector,"},{"Start":"32:27.339 ","End":"32:29.230","Text":"so we got r tag,"},{"Start":"32:29.230 ","End":"32:33.009","Text":"cosine of theta tag and then also"},{"Start":"32:33.009 ","End":"32:37.600","Text":"multiplied by r. What we have here is that just divided by"},{"Start":"32:37.600 ","End":"32:46.660","Text":"r. We have r vector dot r tag of vector divided by r. I\u0027ll just write this,"},{"Start":"32:46.660 ","End":"32:51.039","Text":"r multiplied by r tag,"},{"Start":"32:51.039 ","End":"32:56.379","Text":"multiplied by cosine of theta tag is this, okay?"},{"Start":"32:56.379 ","End":"32:58.825","Text":"This top product over here."},{"Start":"32:58.825 ","End":"33:00.130","Text":"And then we divide it by r,"},{"Start":"33:00.130 ","End":"33:04.854","Text":"and then we\u0027re left with r\u0027cos Theta tag which is exactly what we have over here."},{"Start":"33:04.854 ","End":"33:08.380","Text":"You see why we get that and then this is multiplied by"},{"Start":"33:08.380 ","End":"33:13.069","Text":"rho is a function of I tag, dv tag."},{"Start":"33:13.350 ","End":"33:17.875","Text":"Then we know that r i vector is constant."},{"Start":"33:17.875 ","End":"33:19.210","Text":"We\u0027ve already spoken about that."},{"Start":"33:19.210 ","End":"33:21.189","Text":"That\u0027s the vector that points from the origin to"},{"Start":"33:21.189 ","End":"33:24.189","Text":"the point where we\u0027re calculating the potential."},{"Start":"33:24.189 ","End":"33:27.385","Text":"Of course, the magnitude of the r vector is also constant."},{"Start":"33:27.385 ","End":"33:29.530","Text":"We can take that out of the integral sign."},{"Start":"33:29.530 ","End":"33:33.610","Text":"We have r vector divided by r dot-product,"},{"Start":"33:33.610 ","End":"33:40.340","Text":"where the integral of r\u0027 vector multiplied by rho(r\u0027)dv\u0027."},{"Start":"33:44.490 ","End":"33:53.780","Text":"We\u0027ve already seen that this is simply the equation for p vector."},{"Start":"33:54.030 ","End":"33:57.925","Text":"We\u0027ve already seen this completely."},{"Start":"33:57.925 ","End":"34:01.869","Text":"In that case, we can just write this as that this is"},{"Start":"34:01.869 ","End":"34:09.819","Text":"our p vector dot r vector divided by r. As we know,"},{"Start":"34:09.819 ","End":"34:12.190","Text":"r vector divided by its magnitude,"},{"Start":"34:12.190 ","End":"34:18.420","Text":"so r vector divided by I is the same as writing p.r hat."},{"Start":"34:18.420 ","End":"34:19.950","Text":"All of this is r-hat."},{"Start":"34:19.950 ","End":"34:23.169","Text":"R vector divided by r is r hat."},{"Start":"34:24.030 ","End":"34:29.980","Text":"This is the second element and of course,"},{"Start":"34:29.980 ","End":"34:32.935","Text":"multiplied by k divided by r^2."},{"Start":"34:32.935 ","End":"34:36.850","Text":"P.r multiplied by k divided by r^2."},{"Start":"34:36.850 ","End":"34:41.304","Text":"Then if we go up lo and behold,"},{"Start":"34:41.304 ","End":"34:44.905","Text":"k divided by r^2 multiplied by p vector."},{"Start":"34:44.905 ","End":"34:48.560","Text":"r hat for the second element."},{"Start":"34:49.410 ","End":"34:54.820","Text":"Just a note on this dipole moment,"},{"Start":"34:54.820 ","End":"34:57.219","Text":"here when we\u0027re doing an integral,"},{"Start":"34:57.219 ","End":"34:59.454","Text":"that means that we have some continuous shape,"},{"Start":"34:59.454 ","End":"35:01.149","Text":"like what we saw above."},{"Start":"35:01.149 ","End":"35:08.244","Text":"But if we have some discrete system where we just have lots of different point charges."},{"Start":"35:08.244 ","End":"35:11.754","Text":"It\u0027s not continuous, it\u0027s a discrete system."},{"Start":"35:11.754 ","End":"35:16.600","Text":"In that case, what we can do is we can write this as a sum."},{"Start":"35:16.600 ","End":"35:26.395","Text":"We have that p= the sum on r\u0027_iq_i,"},{"Start":"35:26.395 ","End":"35:30.100","Text":"so a summing on each individual charge."},{"Start":"35:30.100 ","End":"35:32.214","Text":"Because in actual fact,"},{"Start":"35:32.214 ","End":"35:36.729","Text":"what we have over here when we integrate,"},{"Start":"35:36.729 ","End":"35:37.915","Text":"we lose the i."},{"Start":"35:37.915 ","End":"35:43.195","Text":"But what we have over here over Rho dv is we have just dq_i."},{"Start":"35:43.195 ","End":"35:48.039","Text":"Of course, we lose the i when we\u0027re integrating because i tells us"},{"Start":"35:48.039 ","End":"35:50.169","Text":"that we\u0027re dealing with something discrete and the"},{"Start":"35:50.169 ","End":"35:53.529","Text":"integral tells us that we\u0027re dealing with something continuous."},{"Start":"35:53.529 ","End":"35:55.630","Text":"We can have something like this."},{"Start":"35:55.630 ","End":"36:00.370","Text":"Let\u0027s say we\u0027re trying to find the dipole moment along the x-axis,"},{"Start":"36:00.370 ","End":"36:08.620","Text":"so we can write p_x is equal to the sum and then ri\u0027 tag simply becomes x_i."},{"Start":"36:08.620 ","End":"36:13.300","Text":"All the particles along the x-axis multiplied by q_i."},{"Start":"36:13.300 ","End":"36:17.320","Text":"Okay. This is just another note for"},{"Start":"36:17.320 ","End":"36:21.740","Text":"this over here if we\u0027re dealing with a discrete system."},{"Start":"36:21.990 ","End":"36:30.530","Text":"Of course P_y, the dipole moment along the y-axis would be the sum on y_iq_i."},{"Start":"36:32.790 ","End":"36:43.060","Text":"Afterwards, we have our third element over here on our fourth element in blue over here."},{"Start":"36:43.060 ","End":"36:46.225","Text":"The third element represents the quadrupole,"},{"Start":"36:46.225 ","End":"36:48.475","Text":"I\u0027m not going to calculate it."},{"Start":"36:48.475 ","End":"36:53.275","Text":"But all you have to do is this integral and you\u0027ll get your value for the quadrupole."},{"Start":"36:53.275 ","End":"36:55.840","Text":"For number 4, in blue,"},{"Start":"36:55.840 ","End":"37:00.475","Text":"we\u0027ll get the element representing the octopole."},{"Start":"37:00.475 ","End":"37:03.550","Text":"Again, just do this integration."},{"Start":"37:03.550 ","End":"37:06.324","Text":"Usually, if you\u0027re calculating the potential,"},{"Start":"37:06.324 ","End":"37:10.180","Text":"you\u0027ll either get that this number 1,"},{"Start":"37:10.180 ","End":"37:13.675","Text":"Q_T is some non-zero value,"},{"Start":"37:13.675 ","End":"37:16.974","Text":"or that you\u0027re dipole is some non-zero value."},{"Start":"37:16.974 ","End":"37:21.475","Text":"These are the two values of potential that you\u0027ll most often have to deal with."},{"Start":"37:21.475 ","End":"37:24.610","Text":"If however, you have a super intense question"},{"Start":"37:24.610 ","End":"37:27.640","Text":"where you have to deal with a quadrupole or an octopole,"},{"Start":"37:27.640 ","End":"37:30.115","Text":"which is very rare, so here you know,"},{"Start":"37:30.115 ","End":"37:34.389","Text":"you have K/r^3 multiplied by this integration."},{"Start":"37:34.389 ","End":"37:42.235","Text":"Or for the octopole, k/r^4 multiplied by this integration."},{"Start":"37:42.235 ","End":"37:45.969","Text":"So that\u0027s what you do and then you find the potential."},{"Start":"37:45.969 ","End":"37:50.350","Text":"Then for instance, if you wanted to find the electric field,"},{"Start":"37:50.350 ","End":"37:53.440","Text":"you just take the negative derivative of the potential,"},{"Start":"37:53.440 ","End":"37:56.305","Text":"and then you get the electric field."},{"Start":"37:56.305 ","End":"38:00.070","Text":"Of course, when you do the negative derivative of the potential,"},{"Start":"38:00.070 ","End":"38:06.085","Text":"because the potential is as a function of r. so r vector not r\u0027 vector,"},{"Start":"38:06.085 ","End":"38:10.175","Text":"here the potential as a function of r vector."},{"Start":"38:10.175 ","End":"38:14.065","Text":"Then if you want to find the electric field,"},{"Start":"38:14.065 ","End":"38:21.620","Text":"so you take the negative derivative of the potential as a function of r vector."},{"Start":"38:24.450 ","End":"38:29.290","Text":"Because potential isn\u0027t a vector quantity,"},{"Start":"38:29.290 ","End":"38:36.830","Text":"so as a function of r not r\u0027 and then you\u0027ll get the E field."},{"Start":"38:36.900 ","End":"38:42.520","Text":"Then, of course, it\u0027s in the r direction."},{"Start":"38:42.520 ","End":"38:48.879","Text":"Okay. That is the end of this lesson."},{"Start":"38:48.879 ","End":"38:50.905","Text":"It\u0027s a little bit of a difficult lesson."},{"Start":"38:50.905 ","End":"38:53.770","Text":"But what\u0027s important to take away from this is one,"},{"Start":"38:53.770 ","End":"38:58.615","Text":"understanding what r vector and r\u0027 represent."},{"Start":"38:58.615 ","End":"39:06.010","Text":"That the potential is just an infinite series where if we\u0027re using this approximation,"},{"Start":"39:06.010 ","End":"39:11.739","Text":"so when a point where we\u0027re calculating the potential is"},{"Start":"39:11.739 ","End":"39:17.680","Text":"much further away than the distance between all the particles in this charge material,"},{"Start":"39:17.680 ","End":"39:22.150","Text":"then each element is much smaller than its predecessor."},{"Start":"39:22.150 ","End":"39:24.850","Text":"Then in order to calculate the potential,"},{"Start":"39:24.850 ","End":"39:29.875","Text":"we use this approximation and we take the largest nonzero element."},{"Start":"39:29.875 ","End":"39:34.675","Text":"If we\u0027re dealing with lots of plus and minus charges in here,"},{"Start":"39:34.675 ","End":"39:38.305","Text":"we might get that the total charge is equal to 0."},{"Start":"39:38.305 ","End":"39:40.300","Text":"However, there are charges present,"},{"Start":"39:40.300 ","End":"39:42.655","Text":"so we know that there is going to be a potential,"},{"Start":"39:42.655 ","End":"39:47.515","Text":"in which case we will have to calculate the potential for"},{"Start":"39:47.515 ","End":"39:50.409","Text":"a dipole and if that still is equal to"},{"Start":"39:50.409 ","End":"39:54.009","Text":"0 then we\u0027ll go to calculating the potential for the quadrupole,"},{"Start":"39:54.009 ","End":"39:59.784","Text":"which you saw how to do lower down and so on and so forth."},{"Start":"39:59.784 ","End":"40:03.940","Text":"That is what is important to take away from this lesson."},{"Start":"40:03.940 ","End":"40:06.680","Text":"That\u0027s the end of this lesson."}],"ID":22329},{"Watched":false,"Name":"Exercise 3","Duration":"29m 4s","ChapterTopicVideoID":21444,"CourseChapterTopicPlaylistID":99476,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.725","Text":"Hello. In this lesson,"},{"Start":"00:01.725 ","End":"00:04.370","Text":"we\u0027re going to be answering the following question."},{"Start":"00:04.370 ","End":"00:09.055","Text":"We\u0027re being told that a charge q is located at 0,0,d,"},{"Start":"00:09.055 ","End":"00:11.073","Text":"so on the z-axis,"},{"Start":"00:11.073 ","End":"00:16.242","Text":"and a charge of minus q is located at 0,0,minus d,"},{"Start":"00:16.242 ","End":"00:21.750","Text":"so also in the z-axis but at minus d. Question number 1,"},{"Start":"00:21.750 ","End":"00:23.430","Text":"which is what we\u0027ll start with,"},{"Start":"00:23.430 ","End":"00:29.620","Text":"is to calculate the exact potential at a point along the z-axis."},{"Start":"00:29.750 ","End":"00:34.615","Text":"Notice we\u0027re being asked to calculate the exact potential."},{"Start":"00:34.615 ","End":"00:37.685","Text":"We\u0027re not going to use the dipole approximation,"},{"Start":"00:37.685 ","End":"00:42.395","Text":"we\u0027re going to do the whole entire equation, the whole shebang."},{"Start":"00:42.395 ","End":"00:46.910","Text":"Let\u0027s choose this point over here along the z-axis and we\u0027re"},{"Start":"00:46.910 ","End":"00:52.410","Text":"trying to find the potential at this point."},{"Start":"00:52.670 ","End":"01:00.035","Text":"We already know that the potential for a point charge is"},{"Start":"01:00.035 ","End":"01:06.950","Text":"equal to k multiplied by q divided by r. In that case,"},{"Start":"01:06.950 ","End":"01:09.365","Text":"if we take the potential at our point z,"},{"Start":"01:09.365 ","End":"01:12.536","Text":"so we have 2 point charges."},{"Start":"01:12.536 ","End":"01:16.730","Text":"We have the potential due to this charge q,"},{"Start":"01:16.730 ","End":"01:20.600","Text":"so we have kq divided by r. r is"},{"Start":"01:20.600 ","End":"01:24.965","Text":"the distance between this point charge to where we\u0027re calculating the potential."},{"Start":"01:24.965 ","End":"01:29.310","Text":"We can see the potential at this point is at a point z along"},{"Start":"01:29.310 ","End":"01:36.965","Text":"the z-axis and here a point charge q is at a position d,"},{"Start":"01:36.965 ","End":"01:41.075","Text":"so therefore, this distance is"},{"Start":"01:41.075 ","End":"01:45.230","Text":"z minus d. Then we\u0027re going to superimpose"},{"Start":"01:45.230 ","End":"01:50.395","Text":"onto this the potential due to this point charge negative q."},{"Start":"01:50.395 ","End":"01:53.510","Text":"Then we have k multiplied by the charge,"},{"Start":"01:53.510 ","End":"01:56.690","Text":"which is negative q divided by,"},{"Start":"01:56.690 ","End":"02:00.770","Text":"so here all of this distance over here"},{"Start":"02:00.770 ","End":"02:06.500","Text":"is z and then this is another distance d away from the origin."},{"Start":"02:06.500 ","End":"02:10.430","Text":"Here we have z plus d,"},{"Start":"02:10.430 ","End":"02:13.500","Text":"this whole length over here."},{"Start":"02:14.630 ","End":"02:21.670","Text":"Now let\u0027s add this all up so we can take kq out as a common factor."},{"Start":"02:21.670 ","End":"02:26.015","Text":"Then we have 1 divided by z minus d,"},{"Start":"02:26.015 ","End":"02:33.260","Text":"minus 1 divided by z plus d. Then this is equal to,"},{"Start":"02:33.260 ","End":"02:35.915","Text":"so let\u0027s make a common denominator."},{"Start":"02:35.915 ","End":"02:45.275","Text":"We\u0027ll have kq multiplied by z plus d minus z minus d"},{"Start":"02:45.275 ","End":"02:50.280","Text":"divided by z minus d multiplied by"},{"Start":"02:50.280 ","End":"02:57.500","Text":"z plus d. If we open up all of the brackets,"},{"Start":"02:57.500 ","End":"02:59.924","Text":"we can see that the z\u0027s cancel out,"},{"Start":"02:59.924 ","End":"03:03.120","Text":"so we\u0027ll be left with 2kqd."},{"Start":"03:04.120 ","End":"03:06.650","Text":"We have d minus minus d,"},{"Start":"03:06.650 ","End":"03:09.800","Text":"so that\u0027s 2d divided by,"},{"Start":"03:09.800 ","End":"03:16.320","Text":"and here we\u0027ll have z^2 minus d^2."},{"Start":"03:17.120 ","End":"03:20.950","Text":"This is the answer to question number 1."},{"Start":"03:20.950 ","End":"03:23.305","Text":"Now let\u0027s move on to question number 2."},{"Start":"03:23.305 ","End":"03:25.015","Text":"Question number 2 is,"},{"Start":"03:25.015 ","End":"03:31.120","Text":"what is the minimal value of z such that the approximation of the potential at"},{"Start":"03:31.120 ","End":"03:38.360","Text":"that point won\u0027t deviate more than 1 percent from the actual value?"},{"Start":"03:38.430 ","End":"03:41.200","Text":"What does that mean?"},{"Start":"03:41.200 ","End":"03:43.570","Text":"We\u0027re going to find the minimum value of"},{"Start":"03:43.570 ","End":"03:47.836","Text":"z such that the approximation of the potential at that point."},{"Start":"03:47.836 ","End":"03:50.785","Text":"What we mean by the approximation of the potential"},{"Start":"03:50.785 ","End":"03:54.829","Text":"is we\u0027re going to take the potential of a dipole."},{"Start":"03:54.829 ","End":"03:58.930","Text":"We\u0027ve seen that equation before and that\u0027s an approximation."},{"Start":"03:58.930 ","End":"04:02.840","Text":"This is the real potential of what we have in front of"},{"Start":"04:02.840 ","End":"04:08.550","Text":"us and our equation for the potential of a dipole is our approximation."},{"Start":"04:09.230 ","End":"04:13.744","Text":"At that point won\u0027t deviate more than 1 percent,"},{"Start":"04:13.744 ","End":"04:17.390","Text":"so what we take is we calculate the potential of"},{"Start":"04:17.390 ","End":"04:23.390","Text":"the dipole minus the real potential that we calculated over here,"},{"Start":"04:23.390 ","End":"04:25.830","Text":"so this is real."},{"Start":"04:26.050 ","End":"04:32.945","Text":"Then we divide this all by the real potential,"},{"Start":"04:32.945 ","End":"04:35.000","Text":"so divided by this,"},{"Start":"04:35.000 ","End":"04:37.795","Text":"and then we\u0027re going to take the absolute value of this"},{"Start":"04:37.795 ","End":"04:42.929","Text":"because this deviation can be in the positive or negative direction."},{"Start":"04:42.929 ","End":"04:45.145","Text":"We want to get a positive number."},{"Start":"04:45.145 ","End":"04:48.705","Text":"This has to be smaller than 1 percent."},{"Start":"04:48.705 ","End":"04:49.965","Text":"What is 1 percent?"},{"Start":"04:49.965 ","End":"04:51.740","Text":"It\u0027s 1 divided by 100,"},{"Start":"04:51.740 ","End":"04:56.220","Text":"or in other words, 0.01."},{"Start":"04:57.320 ","End":"05:04.055","Text":"Another way of writing this is that the potential for a dipole,"},{"Start":"05:04.055 ","End":"05:06.410","Text":"according to the equation that we know,"},{"Start":"05:06.410 ","End":"05:13.590","Text":"has to be smaller than or equal to the real potential that we calculated in Question 1,"},{"Start":"05:13.590 ","End":"05:17.520","Text":"plus 1 percent of the real potential,"},{"Start":"05:17.520 ","End":"05:25.922","Text":"so 1 percent is just 0.01 multiplied by the real potential,"},{"Start":"05:25.922 ","End":"05:30.110","Text":"and/or it has to be larger or equal to"},{"Start":"05:30.110 ","End":"05:36.020","Text":"the real potential that we calculated in Question 1 minus 1 percent,"},{"Start":"05:36.020 ","End":"05:42.100","Text":"so minus 0.01 of the potential."},{"Start":"05:43.250 ","End":"05:46.820","Text":"Both of these equations are the exact same thing."},{"Start":"05:46.820 ","End":"05:48.050","Text":"If you play around with it,"},{"Start":"05:48.050 ","End":"05:50.225","Text":"so you\u0027ll get this equation."},{"Start":"05:50.225 ","End":"05:52.958","Text":"We can use either one of these."},{"Start":"05:52.958 ","End":"05:57.409","Text":"Then we\u0027ll find that if our potential for the dipole,"},{"Start":"05:57.409 ","End":"06:02.930","Text":"the value at z falls somewhere in this region,"},{"Start":"06:02.930 ","End":"06:06.380","Text":"then we\u0027re good, then we\u0027re answering the question."},{"Start":"06:06.380 ","End":"06:14.675","Text":"First of all, let\u0027s use our equation for the potential of a dipole,"},{"Start":"06:14.675 ","End":"06:16.730","Text":"and let\u0027s work out what this is."},{"Start":"06:16.730 ","End":"06:23.870","Text":"We know that the potential of a dipole is equal to kp.r hat divided by"},{"Start":"06:23.870 ","End":"06:33.660","Text":"r^2 or we can also write it as kp.r vector divided by r^3."},{"Start":"06:33.770 ","End":"06:36.868","Text":"But we\u0027re going to use this version,"},{"Start":"06:36.868 ","End":"06:39.470","Text":"so I\u0027m just going to rub this out."},{"Start":"06:39.470 ","End":"06:44.210","Text":"Now let\u0027s just calculate what p and r hat,"},{"Start":"06:44.210 ","End":"06:46.440","Text":"and r is equal to."},{"Start":"06:46.940 ","End":"06:49.700","Text":"p is the dipole moment."},{"Start":"06:49.700 ","End":"06:51.755","Text":"We saw it\u0027s equal to the charge,"},{"Start":"06:51.755 ","End":"06:58.010","Text":"which is q multiplied by the distance between the 2 charges."},{"Start":"06:58.010 ","End":"07:02.860","Text":"Here we have d in the positive direction and d in the negative direction,"},{"Start":"07:02.860 ","End":"07:04.760","Text":"so the distance is 2d."},{"Start":"07:04.760 ","End":"07:08.720","Text":"Then we know that p is also a vector,"},{"Start":"07:08.720 ","End":"07:10.910","Text":"and specifically with dipoles,"},{"Start":"07:10.910 ","End":"07:18.440","Text":"we know that the vector is pointing from the negative to the positive charge."},{"Start":"07:18.440 ","End":"07:25.670","Text":"This is the direction of p. It\u0027s always in a dipole from the negative to the positive."},{"Start":"07:25.670 ","End":"07:31.380","Text":"We can see that that is in the positive z direction, so z hat."},{"Start":"07:31.430 ","End":"07:34.845","Text":"Now we have to see what r hat is."},{"Start":"07:34.845 ","End":"07:38.570","Text":"r hat is the direction from"},{"Start":"07:38.570 ","End":"07:43.385","Text":"the origin up until the point where we\u0027re measuring the potential."},{"Start":"07:43.385 ","End":"07:46.540","Text":"Let\u0027s take a look."},{"Start":"07:46.540 ","End":"07:49.700","Text":"r hat is this over here,"},{"Start":"07:49.700 ","End":"07:53.090","Text":"so from the origin until this point over here."},{"Start":"07:53.090 ","End":"07:55.490","Text":"We can see it\u0027s going like this."},{"Start":"07:55.490 ","End":"07:57.530","Text":"In red we have our r hat."},{"Start":"07:57.530 ","End":"08:02.035","Text":"We can see that that is also pointing in the positive z direction."},{"Start":"08:02.035 ","End":"08:06.500","Text":"The next thing we need is r. r is the distance"},{"Start":"08:06.500 ","End":"08:11.705","Text":"between the center of the dipole and the point where we\u0027re measuring the potential."},{"Start":"08:11.705 ","End":"08:14.885","Text":"The center of the dipole is at the origin,"},{"Start":"08:14.885 ","End":"08:17.210","Text":"and we\u0027re measuring the potential at z,"},{"Start":"08:17.210 ","End":"08:22.200","Text":"so r is simply just z."},{"Start":"08:22.210 ","End":"08:27.980","Text":"Now let\u0027s plug everything into our equation."},{"Start":"08:27.980 ","End":"08:30.758","Text":"We have k then p.r,"},{"Start":"08:30.758 ","End":"08:35.075","Text":"so we\u0027ll have 2d multiplied by q."},{"Start":"08:35.075 ","End":"08:38.660","Text":"Then we have z hat dot-product with the z hat,"},{"Start":"08:38.660 ","End":"08:43.635","Text":"which is just 1, so we have k multiplied by q2d."},{"Start":"08:43.635 ","End":"08:45.973","Text":"We\u0027ll sort this out in a second."},{"Start":"08:45.973 ","End":"08:48.950","Text":"Then all of this is divided by r^2."},{"Start":"08:48.950 ","End":"08:52.950","Text":"r^2 is z, so z^2."},{"Start":"08:53.150 ","End":"08:59.240","Text":"Now what we want to do is we want to check for which value of z,"},{"Start":"08:59.240 ","End":"09:04.295","Text":"this over here, what we just calculated falls within this region."},{"Start":"09:04.295 ","End":"09:05.615","Text":"Now, first of all,"},{"Start":"09:05.615 ","End":"09:06.800","Text":"let\u0027s take a look."},{"Start":"09:06.800 ","End":"09:11.375","Text":"We can see that the real potential and the potential of the dipole,"},{"Start":"09:11.375 ","End":"09:18.210","Text":"we can see that they both have the same numerator, kq2d, kq2d."},{"Start":"09:18.210 ","End":"09:20.195","Text":"They have the same numerator."},{"Start":"09:20.195 ","End":"09:23.000","Text":"However, the denominator is slightly different."},{"Start":"09:23.000 ","End":"09:24.770","Text":"Both have z^2, however,"},{"Start":"09:24.770 ","End":"09:30.844","Text":"the denominator of the real potential has negative d^2."},{"Start":"09:30.844 ","End":"09:33.170","Text":"Now, d is, of course a distance,"},{"Start":"09:33.170 ","End":"09:34.445","Text":"so it\u0027s a positive value."},{"Start":"09:34.445 ","End":"09:36.485","Text":"You can\u0027t have negative distance."},{"Start":"09:36.485 ","End":"09:42.515","Text":"We have z^2 minus this positive value,"},{"Start":"09:42.515 ","End":"09:45.380","Text":"which means that the denominator over here for"},{"Start":"09:45.380 ","End":"09:50.000","Text":"the real potential is smaller and therefore,"},{"Start":"09:50.000 ","End":"09:53.705","Text":"if the denominator for the real potential is smaller,"},{"Start":"09:53.705 ","End":"09:56.225","Text":"that means that the real potential itself,"},{"Start":"09:56.225 ","End":"10:00.580","Text":"the entire fraction, is bigger than the potential for the dipole."},{"Start":"10:00.580 ","End":"10:06.080","Text":"We can say that the real potential is"},{"Start":"10:06.080 ","End":"10:12.240","Text":"bigger than our dipole approximation for the potential."},{"Start":"10:13.750 ","End":"10:19.210","Text":"This means that I can cancel one side of this equation."},{"Start":"10:19.210 ","End":"10:23.870","Text":"Here I have for any value of z that I put in,"},{"Start":"10:23.870 ","End":"10:28.670","Text":"my dipole potential will always be smaller than my real potential."},{"Start":"10:28.670 ","End":"10:33.575","Text":"If my dipole potential is always smaller than my real potential, then of course,"},{"Start":"10:33.575 ","End":"10:35.120","Text":"it\u0027s always going to be smaller than"},{"Start":"10:35.120 ","End":"10:40.060","Text":"my real potential plus this added 1 percent of real potential."},{"Start":"10:40.060 ","End":"10:43.975","Text":"Therefore, I can say that this side,"},{"Start":"10:43.975 ","End":"10:48.890","Text":"I already fall in this region just from the equation itself."},{"Start":"10:48.890 ","End":"10:52.820","Text":"Now, all I have to do is I have to check this region over here."},{"Start":"10:52.820 ","End":"10:57.470","Text":"Is for which value of a z my dipole potential is going to be"},{"Start":"10:57.470 ","End":"11:03.930","Text":"bigger or equal to my real potential minus 1 percent of my real potential."},{"Start":"11:05.060 ","End":"11:10.025","Text":"Let\u0027s just scroll down to give us a little bit more space."},{"Start":"11:10.025 ","End":"11:15.065","Text":"What we need is that the dipole potential,"},{"Start":"11:15.065 ","End":"11:18.665","Text":"which we got over here,"},{"Start":"11:18.665 ","End":"11:27.890","Text":"is equal to 2kqd divided by z^2 has to"},{"Start":"11:27.890 ","End":"11:31.925","Text":"be bigger than or equal to"},{"Start":"11:31.925 ","End":"11:38.554","Text":"the real potential minus 0.01 multiplied by the real potential."},{"Start":"11:38.554 ","End":"11:46.750","Text":"I can just write real potential and then I have 1 minus 0.01."},{"Start":"11:46.900 ","End":"11:48.920","Text":"Or in other words,"},{"Start":"11:48.920 ","End":"11:53.939","Text":"this is equal to the real potential multiplied"},{"Start":"11:53.939 ","End":"11:59.837","Text":"by 1 minus 0.01 is equal to 0.99."},{"Start":"11:59.837 ","End":"12:03.570","Text":"This is equal to,"},{"Start":"12:03.570 ","End":"12:05.535","Text":"so my real potential is"},{"Start":"12:05.535 ","End":"12:13.890","Text":"2kqd divided by z^2 minus d^2,"},{"Start":"12:13.890 ","End":"12:17.205","Text":"and this is multiplied by 0.99."},{"Start":"12:17.205 ","End":"12:23.835","Text":"Now, let\u0027s just play around with it."},{"Start":"12:23.835 ","End":"12:27.180","Text":"What we have, I\u0027m just going to take out all the middle,"},{"Start":"12:27.180 ","End":"12:33.090","Text":"but we have the 2kqd divided by z^2 has to be bigger or equal to here."},{"Start":"12:33.090 ","End":"12:38.010","Text":"I\u0027m going to rub out the middle. Now, let\u0027s play around."},{"Start":"12:38.010 ","End":"12:39.450","Text":"I can see first of all,"},{"Start":"12:39.450 ","End":"12:40.890","Text":"the numerator is the same,"},{"Start":"12:40.890 ","End":"12:44.355","Text":"so I can divide both sides by 2kqd,"},{"Start":"12:44.355 ","End":"12:46.755","Text":"and that will cancel out."},{"Start":"12:46.755 ","End":"12:57.090","Text":"Then what I can do is multiply this side by z^2 or multiply both sides by z^2 and by"},{"Start":"12:57.090 ","End":"13:07.410","Text":"z^2 minus d. Then I\u0027ll get z^2 multiplied by 0.99."},{"Start":"13:07.410 ","End":"13:09.765","Text":"I\u0027m just doing cross multiplication."},{"Start":"13:09.765 ","End":"13:13.760","Text":"This has to be smaller or equal"},{"Start":"13:13.760 ","End":"13:21.170","Text":"to z^2 minus d^2."},{"Start":"13:21.170 ","End":"13:23.450","Text":"It looks as if I flipped over this side,"},{"Start":"13:23.450 ","End":"13:24.875","Text":"but I in fact haven\u0027t."},{"Start":"13:24.875 ","End":"13:28.555","Text":"All I\u0027ve done is done that"},{"Start":"13:28.555 ","End":"13:32.385","Text":"z^2 minus d^2 has moved onto"},{"Start":"13:32.385 ","End":"13:36.135","Text":"this side because I multiplied both sides and then it\u0027s bigger or equal to,"},{"Start":"13:36.135 ","End":"13:39.840","Text":"and then I multiply both sides by z^2."},{"Start":"13:39.840 ","End":"13:42.270","Text":"Then this z^2 will move over here,"},{"Start":"13:42.270 ","End":"13:46.590","Text":"z^2 multiplied by 0.99."},{"Start":"13:46.590 ","End":"13:48.705","Text":"The equation is just flipped over."},{"Start":"13:48.705 ","End":"13:51.254","Text":"It looks like the sign has changed direction,"},{"Start":"13:51.254 ","End":"13:53.610","Text":"but in fact hasn\u0027t."},{"Start":"13:53.610 ","End":"14:04.005","Text":"Then what I can do is I can take z^2 over to this side and make it a common factor."},{"Start":"14:04.005 ","End":"14:07.665","Text":"Then I\u0027ll have z^2 and then I have here,"},{"Start":"14:07.665 ","End":"14:10.815","Text":"it\u0027s multiplied by 0.99."},{"Start":"14:10.815 ","End":"14:14.735","Text":"Then here I\u0027m subtracting z^2 from both sides."},{"Start":"14:14.735 ","End":"14:16.430","Text":"Then I have negative 1,"},{"Start":"14:16.430 ","End":"14:21.575","Text":"and this is smaller or equal to negative d^2."},{"Start":"14:21.575 ","End":"14:27.085","Text":"Then I can just multiply both sides by negative 1."},{"Start":"14:27.085 ","End":"14:31.300","Text":"What I\u0027ll get is I\u0027ll have z^2,"},{"Start":"14:31.300 ","End":"14:34.870","Text":"and then here I\u0027ll have 1 minus 0.99."},{"Start":"14:35.480 ","End":"14:41.610","Text":"Then this will be bigger or equal to d^2."},{"Start":"14:44.270 ","End":"14:47.280","Text":"Then I can just simplify this."},{"Start":"14:47.280 ","End":"14:53.970","Text":"So I\u0027ll have the z^2 and then 1 minus 0.99 is just 0.01."},{"Start":"14:53.970 ","End":"15:00.730","Text":"So z^2 minus 0.01 is bigger or equal to d^2."},{"Start":"15:02.420 ","End":"15:10.680","Text":"Then what we\u0027re going to do is we\u0027re just going to take the square root of both sides."},{"Start":"15:10.680 ","End":"15:21.600","Text":"We\u0027re going to get that z and then the square root of 0.01 is just 0.1."},{"Start":"15:21.600 ","End":"15:27.480","Text":"So z multiplied by 0.1 is bigger"},{"Start":"15:27.480 ","End":"15:33.780","Text":"or equal to d. We took the square root of both sides."},{"Start":"15:33.780 ","End":"15:36.810","Text":"The square root of 0.01 is 0.1."},{"Start":"15:36.810 ","End":"15:39.930","Text":"Then we just have to isolate out our z."},{"Start":"15:39.930 ","End":"15:47.475","Text":"So z has to be greater or equal to,"},{"Start":"15:47.475 ","End":"15:51.285","Text":"so 0.1 is just 1 divided by 10."},{"Start":"15:51.285 ","End":"15:55.090","Text":"That means if we multiply both sides by 10,"},{"Start":"15:55.670 ","End":"15:59.680","Text":"we\u0027ll get rid of the 0.1 over here."},{"Start":"16:00.740 ","End":"16:06.574","Text":"Therefore, if I need z to be greater or equal to 10d,"},{"Start":"16:06.574 ","End":"16:08.010","Text":"then I can say, okay,"},{"Start":"16:08.010 ","End":"16:11.970","Text":"I\u0027m reminding you the question was to find the minimum value of"},{"Start":"16:11.970 ","End":"16:17.340","Text":"z such that we don\u0027t get more than a 1 percent deviation."},{"Start":"16:17.340 ","End":"16:20.115","Text":"Then I will get that my z_min,"},{"Start":"16:20.115 ","End":"16:25.600","Text":"my minimum z value has to be equal to 10d."},{"Start":"16:25.970 ","End":"16:29.055","Text":"This is the answer to question number 2."},{"Start":"16:29.055 ","End":"16:32.470","Text":"Now let\u0027s move on to question number 3."},{"Start":"16:34.250 ","End":"16:41.210","Text":"Question number 3 is the same thing just with the electric field."},{"Start":"16:41.210 ","End":"16:46.850","Text":"What is the minimal value of z such that the approximation of the electric field,"},{"Start":"16:46.850 ","End":"16:50.330","Text":"the electric field for a dipole at this point,"},{"Start":"16:50.330 ","End":"16:55.830","Text":"won\u0027t deviate more than 1 percent from the actual value of the electric field?"},{"Start":"16:57.020 ","End":"17:00.240","Text":"The first thing that we have to do is we have to"},{"Start":"17:00.240 ","End":"17:03.180","Text":"find the electric field for the real value."},{"Start":"17:03.180 ","End":"17:06.870","Text":"The E field for a point charge, as we know,"},{"Start":"17:06.870 ","End":"17:12.310","Text":"is equal to kq divided by r^2,"},{"Start":"17:12.710 ","End":"17:16.650","Text":"and of course it\u0027s in the radial direction."},{"Start":"17:16.650 ","End":"17:18.960","Text":"We have 2 point charges,"},{"Start":"17:18.960 ","End":"17:21.240","Text":"so we\u0027re going to superimpose them."},{"Start":"17:21.240 ","End":"17:27.585","Text":"The E_real is going to be equal to the electric field from this charge q."},{"Start":"17:27.585 ","End":"17:31.215","Text":"That\u0027s kq divided by r^2,"},{"Start":"17:31.215 ","End":"17:38.700","Text":"where r^2 we already saw was z minus d^2 and of course,"},{"Start":"17:38.700 ","End":"17:40.245","Text":"in the radial direction."},{"Start":"17:40.245 ","End":"17:42.135","Text":"Then we\u0027re adding on k,"},{"Start":"17:42.135 ","End":"17:47.715","Text":"and then this charge is minus q divided by r^2,"},{"Start":"17:47.715 ","End":"17:51.220","Text":"where here we have z plus d^2."},{"Start":"17:53.120 ","End":"17:57.420","Text":"This is also in the radial direction."},{"Start":"17:57.420 ","End":"18:04.350","Text":"Now, because we are specifically trying to measure the electric field now over here,"},{"Start":"18:04.350 ","End":"18:06.615","Text":"along the z axis."},{"Start":"18:06.615 ","End":"18:09.315","Text":"Specifically in this example,"},{"Start":"18:09.315 ","End":"18:14.385","Text":"we\u0027re measuring this in the z direction."},{"Start":"18:14.385 ","End":"18:16.500","Text":"Because we\u0027re trying to measure over here,"},{"Start":"18:16.500 ","End":"18:19.680","Text":"which means that it\u0027s going up here in the z direction."},{"Start":"18:19.680 ","End":"18:25.649","Text":"Now we can take out kqz hat as common factors."},{"Start":"18:25.649 ","End":"18:31.320","Text":"Then we have 1 divided by z minus d^2,"},{"Start":"18:31.320 ","End":"18:34.125","Text":"minus, because of this minus q over here,"},{"Start":"18:34.125 ","End":"18:39.375","Text":"1 divided by z plus d^2."},{"Start":"18:39.375 ","End":"18:43.260","Text":"Then we can do this common denominator again."},{"Start":"18:43.260 ","End":"18:46.620","Text":"We have kqz hat,"},{"Start":"18:46.620 ","End":"18:50.385","Text":"and then we\u0027re going to have up here."},{"Start":"18:50.385 ","End":"18:56.280","Text":"We\u0027ll have z^2 plus 2dz."},{"Start":"18:56.280 ","End":"18:58.305","Text":"Let\u0027s just scroll over here,"},{"Start":"18:58.305 ","End":"19:03.975","Text":"plus d^2, so that will be over here."},{"Start":"19:03.975 ","End":"19:13.230","Text":"Then we have minus z^2 minus 2dz plus d^2,"},{"Start":"19:13.230 ","End":"19:17.610","Text":"and then all of this is divided by z"},{"Start":"19:17.610 ","End":"19:24.600","Text":"minus d^2, z plus d^2."},{"Start":"19:24.600 ","End":"19:28.575","Text":"Here we have positive z,"},{"Start":"19:28.575 ","End":"19:30.210","Text":"and then here we have"},{"Start":"19:30.210 ","End":"19:35.265","Text":"positive z^2 and then here we have negative z^2 so we can cross this out."},{"Start":"19:35.265 ","End":"19:41.250","Text":"Then here we have positive d^2 and then here we have minus d^2."},{"Start":"19:41.250 ","End":"19:46.635","Text":"We can cross out our d^2 so then we have 2dz-(-2dz)."},{"Start":"19:46.635 ","End":"19:51.857","Text":"Then we\u0027ll have kq multiplied by"},{"Start":"19:51.857 ","End":"19:58.665","Text":"2dz+2dz is 4dz in the z direction divided by,"},{"Start":"19:58.665 ","End":"20:08.880","Text":"so this denominator over here can be written also as z-d,"},{"Start":"20:08.880 ","End":"20:14.550","Text":"z+d, and then all of this squared."},{"Start":"20:14.550 ","End":"20:25.215","Text":"It\u0027s the same thing. Then this is simply equal to z^2-d^2 and all of this squared."},{"Start":"20:25.215 ","End":"20:34.090","Text":"Then we can write our denominator as just (z^2-d^2)^2."},{"Start":"20:34.850 ","End":"20:38.850","Text":"This is the real E field,"},{"Start":"20:38.850 ","End":"20:41.640","Text":"and this is what we\u0027re going to need at the end,"},{"Start":"20:41.640 ","End":"20:45.585","Text":"so let\u0027s just box this over here so that we don\u0027t lose it,"},{"Start":"20:45.585 ","End":"20:50.130","Text":"so that it\u0027s easier to find and now let\u0027s just scroll down."},{"Start":"20:50.130 ","End":"20:54.750","Text":"Now let\u0027s work out the E field for a dipole."},{"Start":"20:54.750 ","End":"20:57.450","Text":"This is using approximations,"},{"Start":"20:57.450 ","End":"21:05.610","Text":"so it is equal to k multiplied by 3p.r hat in"},{"Start":"21:05.610 ","End":"21:08.865","Text":"the r hat direction minus"},{"Start":"21:08.865 ","End":"21:15.630","Text":"p divided by r^3."},{"Start":"21:15.630 ","End":"21:22.425","Text":"Just a reminder before we calculated that our p vectors 2qd in the z direction,"},{"Start":"21:22.425 ","End":"21:28.020","Text":"that our r hat is just the same as z hat and then r is equal to z."},{"Start":"21:28.020 ","End":"21:30.015","Text":"Remember we use the diagram."},{"Start":"21:30.015 ","End":"21:34.320","Text":"Then what we have is k multiplied by 3 times,"},{"Start":"21:34.320 ","End":"21:37.050","Text":"so we have p vector,"},{"Start":"21:37.050 ","End":"21:42.105","Text":"which is 2qd in the z direction,"},{"Start":"21:42.105 ","End":"21:46.530","Text":"z hat multiplied by or dot product with r hat,"},{"Start":"21:46.530 ","End":"21:53.925","Text":"which is z hat so this becomes equal to 1 and this is in the r hat direction,"},{"Start":"21:53.925 ","End":"21:58.770","Text":"which is just the z hat direction,"},{"Start":"21:58.770 ","End":"22:01.185","Text":"minus our p vector,"},{"Start":"22:01.185 ","End":"22:07.590","Text":"which is minus 2qd in the z direction."},{"Start":"22:07.880 ","End":"22:11.595","Text":"Then all of this is divided by r^3,"},{"Start":"22:11.595 ","End":"22:13.950","Text":"which is just z^3,"},{"Start":"22:13.950 ","End":"22:16.380","Text":"so then we can write this."},{"Start":"22:16.380 ","End":"22:25.260","Text":"Let\u0027s scroll over here as k and then we have 3 times 2qd times 1 in the z direction,"},{"Start":"22:25.260 ","End":"22:26.850","Text":"so we have 6."},{"Start":"22:26.850 ","End":"22:36.540","Text":"We have k multiplied by 6qd and then the z direction comes from here,"},{"Start":"22:36.540 ","End":"22:46.665","Text":"minus 2qd in the z direction divided by z^3."},{"Start":"22:46.665 ","End":"22:56.820","Text":"Then we can just carry this on over here so therefore we will get that our E dipole."},{"Start":"22:56.820 ","End":"23:00.645","Text":"With approximation, we just have 6 minus 2 is 4,"},{"Start":"23:00.645 ","End":"23:11.280","Text":"so we have 4kqd divided by z^3 in the z direction."},{"Start":"23:11.280 ","End":"23:13.380","Text":"This is also something that we need,"},{"Start":"23:13.380 ","End":"23:16.215","Text":"so we\u0027ll box it over here so that\u0027s easy to find."},{"Start":"23:16.215 ","End":"23:17.715","Text":"There\u0027s just a mistake here."},{"Start":"23:17.715 ","End":"23:22.830","Text":"This is meant to be minus d^2."},{"Start":"23:22.830 ","End":"23:28.605","Text":"Now we\u0027re going to do the same trick as we did for question number 2."},{"Start":"23:28.605 ","End":"23:37.965","Text":"We know that the E field for the dipole has to be smaller or equal to,"},{"Start":"23:37.965 ","End":"23:48.060","Text":"then the real E field plus 1 percent of it so plus 0.01 multiplied"},{"Start":"23:48.060 ","End":"23:55.695","Text":"by E real and the dipole field has to be bigger or equal to"},{"Start":"23:55.695 ","End":"24:00.570","Text":"the real E-field minus"},{"Start":"24:00.570 ","End":"24:07.080","Text":"0.01 multiplied by the real E field."},{"Start":"24:07.080 ","End":"24:09.120","Text":"In question number 2,"},{"Start":"24:09.120 ","End":"24:13.470","Text":"we saw that the dipole potential,"},{"Start":"24:13.470 ","End":"24:19.710","Text":"is always, for any value of z is going to be smaller than the real potential."},{"Start":"24:19.710 ","End":"24:25.110","Text":"In that case, we can already know that the dipole field is also therefore"},{"Start":"24:25.110 ","End":"24:30.140","Text":"going to be smaller than the real field and if it\u0027s smaller than a real field,"},{"Start":"24:30.140 ","End":"24:34.400","Text":"then of course it\u0027s smaller than the real field plus some value."},{"Start":"24:34.400 ","End":"24:38.960","Text":"This side of the equation is good and now we\u0027re going to try and find"},{"Start":"24:38.960 ","End":"24:44.370","Text":"the value for z such that this side of the equation also occurs."},{"Start":"24:44.630 ","End":"24:47.940","Text":"Let\u0027s scroll down so we\u0027re doing this."},{"Start":"24:47.940 ","End":"24:52.791","Text":"Here we have a common multiple of E real."},{"Start":"24:52.791 ","End":"24:55.920","Text":"If we put E real outside of the brackets,"},{"Start":"24:55.920 ","End":"25:00.225","Text":"we\u0027ll have E real multiplied by 1 minus 0.001,"},{"Start":"25:00.225 ","End":"25:02.970","Text":"so what we\u0027ll get is E real,"},{"Start":"25:02.970 ","End":"25:08.850","Text":"which is 4kqdz divided"},{"Start":"25:08.850 ","End":"25:17.475","Text":"by (z^2-d^2)^2 multiplied by 1 minus 0.01,"},{"Start":"25:17.475 ","End":"25:23.040","Text":"which is multiplied by 0.99."},{"Start":"25:23.040 ","End":"25:26.250","Text":"I\u0027ve just rearranged this so that it makes more sense and"},{"Start":"25:26.250 ","End":"25:29.729","Text":"obviously I\u0027m not going to put in this z hat over here the direction,"},{"Start":"25:29.729 ","End":"25:33.105","Text":"because we know that the dipole field and the real field are,"},{"Start":"25:33.105 ","End":"25:35.790","Text":"of course, pointing in the same direction."},{"Start":"25:35.790 ","End":"25:40.590","Text":"This has to be smaller or equal to my dipole field,"},{"Start":"25:40.590 ","End":"25:49.110","Text":"which has 4kqd divided by z^3."},{"Start":"25:49.110 ","End":"25:54.727","Text":"First of all, I can see 4kqd, 4kqd."},{"Start":"25:54.727 ","End":"25:57.400","Text":"I\u0027ve divided both sides by that."},{"Start":"25:57.470 ","End":"26:07.800","Text":"Now what I\u0027m going to do is I\u0027m going to multiply both sides by z^3 and by (z^2-d^2)^2."},{"Start":"26:07.800 ","End":"26:17.055","Text":"What I\u0027ll get is that z^3 multiplied by z is z^4, multiplied by 0.99."},{"Start":"26:17.055 ","End":"26:20.400","Text":"This is smaller or equal to so here,"},{"Start":"26:20.400 ","End":"26:27.015","Text":"that will just be (z^2-d^2)^2."},{"Start":"26:27.015 ","End":"26:31.485","Text":"Now what we\u0027re going to do is we\u0027re going to square root both sides,"},{"Start":"26:31.485 ","End":"26:35.475","Text":"so z^4, square root of that is z^2."},{"Start":"26:35.475 ","End":"26:40.020","Text":"The square root of 0.99 can plug it into a calculator,"},{"Start":"26:40.020 ","End":"26:45.340","Text":"but it\u0027s approximately equal to 0.995,"},{"Start":"26:45.410 ","End":"26:51.150","Text":"and this has to be smaller or equal to the square root of this."},{"Start":"26:51.150 ","End":"26:59.250","Text":"That\u0027s just z^2-d^2 without the brackets because we square rooted of this."},{"Start":"26:59.250 ","End":"27:02.550","Text":"Now we can move this z over to"},{"Start":"27:02.550 ","End":"27:06.840","Text":"the other side or move this d to the other side and this z,"},{"Start":"27:06.840 ","End":"27:17.310","Text":"so what we\u0027ll get is that d^2 has to be smaller or equal to z^2 and then in brackets,"},{"Start":"27:17.310 ","End":"27:23.560","Text":"so we have 1 minus 0.995."},{"Start":"27:23.570 ","End":"27:29.710","Text":"Of course, 1 minus 0.995 is just equal to 0.005,"},{"Start":"27:30.650 ","End":"27:34.575","Text":"so therefore, we\u0027re going to get,"},{"Start":"27:34.575 ","End":"27:37.320","Text":"if we then square root, again both sides,"},{"Start":"27:37.320 ","End":"27:45.675","Text":"we\u0027ll get that d has to be smaller or equal to z multiplied by the square root of 0.005,"},{"Start":"27:45.675 ","End":"27:52.425","Text":"multiplied by approximately 0.07."},{"Start":"27:52.425 ","End":"27:58.275","Text":"The square root of 0.005 is approximately 0.07."},{"Start":"27:58.275 ","End":"28:03.975","Text":"Then we\u0027re going to multiply both sides by 0.07,"},{"Start":"28:03.975 ","End":"28:07.140","Text":"so what we\u0027re going to get,"},{"Start":"28:07.140 ","End":"28:15.165","Text":"so 0.07 is just equal to 7 divided by 100."},{"Start":"28:15.165 ","End":"28:19.500","Text":"We\u0027ll multiply both sides by a 100 and divide by 7."},{"Start":"28:19.500 ","End":"28:22.785","Text":"Therefore will get that"},{"Start":"28:22.785 ","End":"28:31.385","Text":"100d divided by 7 is smaller or equal to z."},{"Start":"28:31.385 ","End":"28:40.400","Text":"Then, therefore, our z min is going to be approximately equal to,"},{"Start":"28:40.400 ","End":"28:45.930","Text":"so 100 divided by 7 is approximately 14.3,"},{"Start":"28:47.240 ","End":"28:52.455","Text":"so z min is going to be approximately equal to"},{"Start":"28:52.455 ","End":"29:02.235","Text":"14.3 multiplied by d. That is the final answer for question number 3,"},{"Start":"29:02.235 ","End":"29:05.080","Text":"and that is the end of our lesson."}],"ID":22330},{"Watched":false,"Name":"Energy of Dipole in Electric Field","Duration":"7m 18s","ChapterTopicVideoID":21445,"CourseChapterTopicPlaylistID":99476,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.385","Text":"Hello. In this lesson,"},{"Start":"00:02.385 ","End":"00:08.025","Text":"we\u0027re going to be speaking about the energy of a dipole in an electric field."},{"Start":"00:08.025 ","End":"00:12.255","Text":"Let\u0027s imagine that we have an electric field,"},{"Start":"00:12.255 ","End":"00:17.250","Text":"and let\u0027s say that it\u0027s pointing in this direction."},{"Start":"00:17.250 ","End":"00:21.510","Text":"Then we have a dipole which is pointing,"},{"Start":"00:21.510 ","End":"00:24.680","Text":"let\u0027s say in this direction."},{"Start":"00:24.680 ","End":"00:28.830","Text":"The dipole moment is this p and this angle"},{"Start":"00:28.830 ","End":"00:33.810","Text":"between the dipole moment and the electric field,"},{"Start":"00:33.810 ","End":"00:36.660","Text":"let\u0027s say is Theta."},{"Start":"00:36.660 ","End":"00:44.330","Text":"We know that the electric field is going to produce a moment of force,"},{"Start":"00:44.330 ","End":"00:47.635","Text":"which is equal to Tau,"},{"Start":"00:47.635 ","End":"00:51.485","Text":"and that is equal to p, the dipole moment,"},{"Start":"00:51.485 ","End":"01:00.050","Text":"cross multiplied by E. This is a moment of force or the torque."},{"Start":"01:00.050 ","End":"01:03.160","Text":"Due to this torque,"},{"Start":"01:03.160 ","End":"01:08.300","Text":"our dipole is going to start rotating in this direction so"},{"Start":"01:08.300 ","End":"01:13.430","Text":"that it will line up with the direction of the E field."},{"Start":"01:13.430 ","End":"01:16.115","Text":"Our dipole moment is going to start rotating,"},{"Start":"01:16.115 ","End":"01:18.064","Text":"which means that it\u0027s moving,"},{"Start":"01:18.064 ","End":"01:22.410","Text":"and if it\u0027s moving that means it has kinetic energy."},{"Start":"01:22.640 ","End":"01:28.080","Text":"What we can see is that I have a potential for kinetic energy,"},{"Start":"01:28.080 ","End":"01:35.845","Text":"if I place or release a dipole such that it is at an angle to the electric field."},{"Start":"01:35.845 ","End":"01:40.250","Text":"Let\u0027s say that I had this electric field"},{"Start":"01:40.250 ","End":"01:45.140","Text":"again and I released the dipole from this angle over here."},{"Start":"01:45.140 ","End":"01:51.365","Text":"This is our p. We can see that this angle Theta is a very, very large angle."},{"Start":"01:51.365 ","End":"01:55.940","Text":"That means that my potential for kinetic energy is going to be higher."},{"Start":"01:55.940 ","End":"02:00.350","Text":"Because my dipole will have to move more in order"},{"Start":"02:00.350 ","End":"02:05.635","Text":"to become parallel to the electric field."},{"Start":"02:05.635 ","End":"02:10.355","Text":"That means that there\u0027ll be more potential for kinetic energy."},{"Start":"02:10.355 ","End":"02:15.985","Text":"Of course, we know that if our electric field is like so,"},{"Start":"02:15.985 ","End":"02:21.320","Text":"and our dipole moment p is so,"},{"Start":"02:21.320 ","End":"02:24.875","Text":"then we\u0027ll get that the torque over here is equal to 0."},{"Start":"02:24.875 ","End":"02:28.775","Text":"This is a stable position."},{"Start":"02:28.775 ","End":"02:32.090","Text":"What we\u0027re going to do now is we\u0027re going to look at"},{"Start":"02:32.090 ","End":"02:36.860","Text":"an equation that is the equation for the potential energy,"},{"Start":"02:36.860 ","End":"02:43.335","Text":"which will tell us how much kinetic energy our dipole moment can"},{"Start":"02:43.335 ","End":"02:50.820","Text":"attain depending on the angle that it is at relative to the electric field."},{"Start":"02:50.960 ","End":"02:59.930","Text":"The potential energy of a dipole p located and then electric field E is equal to"},{"Start":"02:59.930 ","End":"03:03.830","Text":"negative p dot E. We have"},{"Start":"03:03.830 ","End":"03:09.570","Text":"the negative dot-product between the dipole moment and the electric field."},{"Start":"03:10.460 ","End":"03:14.735","Text":"This is something to remember and also this equation over here."},{"Start":"03:14.735 ","End":"03:19.585","Text":"But this is the potential energy of a dipole in an electric field."},{"Start":"03:19.585 ","End":"03:25.460","Text":"Of course, we have the dot-product over here okay so we can also write this as negative"},{"Start":"03:25.460 ","End":"03:31.490","Text":"multiplied by the size of our dipole moment,"},{"Start":"03:31.490 ","End":"03:37.490","Text":"which is p, multiplied by the magnitude of our electric field,"},{"Start":"03:37.490 ","End":"03:43.810","Text":"which is just E, and then multiplied by cosine of the angle between the 2."},{"Start":"03:43.810 ","End":"03:48.890","Text":"This is the magnitude of the dipole moment and the magnitude of the electric field."},{"Start":"03:48.890 ","End":"03:55.350","Text":"In these cases, we can see that if Theta is equal to 0,"},{"Start":"03:55.580 ","End":"04:03.765","Text":"then that means that our U our potential energy will equal to negative PE."},{"Start":"04:03.765 ","End":"04:06.910","Text":"Because cosine of 0 is equal to 1."},{"Start":"04:06.910 ","End":"04:11.210","Text":"If Theta is equal to 90 degrees,"},{"Start":"04:11.210 ","End":"04:18.350","Text":"then our energy is going to be equal to 0 because cosine of 90 degrees is equal to 0."},{"Start":"04:18.350 ","End":"04:23.795","Text":"If our Theta is equal to 180 degrees,"},{"Start":"04:23.795 ","End":"04:29.790","Text":"then our potential energy U is going to be equal to PE."},{"Start":"04:31.790 ","End":"04:34.890","Text":"Let\u0027s take a look."},{"Start":"04:34.890 ","End":"04:40.760","Text":"We can see that by definition over here when Theta is equal to 0."},{"Start":"04:40.760 ","End":"04:45.670","Text":"That means that our electric field looks like so,"},{"Start":"04:45.670 ","End":"04:51.615","Text":"and that our dipole moment looks like so."},{"Start":"04:51.615 ","End":"04:57.410","Text":"We can see that this is the minimum energy that we will have."},{"Start":"04:57.410 ","End":"05:01.880","Text":"We can see that the minimum energy is defined as negative PE."},{"Start":"05:01.880 ","End":"05:05.240","Text":"It\u0027s some negative value and"},{"Start":"05:05.240 ","End":"05:09.140","Text":"this is the least amount of potential energy that we can have."},{"Start":"05:09.140 ","End":"05:15.920","Text":"That will help us to explain why over here if Theta is equal to 90 so that will mean that"},{"Start":"05:15.920 ","End":"05:22.800","Text":"our electric field and a dipole moment are perpendicular to 1 another."},{"Start":"05:22.800 ","End":"05:26.900","Text":"It\u0027s clear that our dipole moment is going to want to move in"},{"Start":"05:26.900 ","End":"05:31.475","Text":"order to line up with our electric field."},{"Start":"05:31.475 ","End":"05:33.930","Text":"It\u0027s going to want to move like this."},{"Start":"05:33.930 ","End":"05:37.880","Text":"Even though our potential energy is written at 0,"},{"Start":"05:37.880 ","End":"05:40.190","Text":"we know that that isn\u0027t the minimum because"},{"Start":"05:40.190 ","End":"05:42.950","Text":"our minimum potential energy is a negative value,"},{"Start":"05:42.950 ","End":"05:44.795","Text":"okay so below the 0."},{"Start":"05:44.795 ","End":"05:50.270","Text":"The fact that this is 0 it\u0027s just calibrated that at 90 degrees our energy is equal to 0."},{"Start":"05:50.270 ","End":"05:53.600","Text":"But we know that our dipole moment is still going to"},{"Start":"05:53.600 ","End":"05:57.905","Text":"have kinetic energy up until it\u0027s lined up like so."},{"Start":"05:57.905 ","End":"06:02.365","Text":"The potential energy is defined as"},{"Start":"06:02.365 ","End":"06:08.620","Text":"this stable position of equilibrium at this negative value."},{"Start":"06:08.620 ","End":"06:11.990","Text":"That\u0027s why this is 0 and it still makes sense."},{"Start":"06:11.990 ","End":"06:18.745","Text":"What we can see here at Theta is equal to 180 so we have our E field like so."},{"Start":"06:18.745 ","End":"06:22.970","Text":"We have our dipole moment like so."},{"Start":"06:22.970 ","End":"06:24.860","Text":"We can see that we\u0027re going to need"},{"Start":"06:24.860 ","End":"06:31.145","Text":"the maximum amount of energy or we\u0027ll have the maximum amount of kinetic energy,"},{"Start":"06:31.145 ","End":"06:38.040","Text":"because our dipole moment has to move in maximum angle of 180 degrees."},{"Start":"06:38.040 ","End":"06:42.210","Text":"That\u0027s why the energy is PE,"},{"Start":"06:42.210 ","End":"06:43.910","Text":"the maximum amount of energy,"},{"Start":"06:43.910 ","End":"06:46.685","Text":"which is twice this U min."},{"Start":"06:46.685 ","End":"06:49.825","Text":"This is our U max."},{"Start":"06:49.825 ","End":"06:51.740","Text":"That\u0027s the end of the lesson."},{"Start":"06:51.740 ","End":"06:54.939","Text":"What\u0027s important to remember is this equation,"},{"Start":"06:54.939 ","End":"06:58.025","Text":"and to remember that just because U is equal to 0,"},{"Start":"06:58.025 ","End":"07:00.215","Text":"doesn\u0027t mean that there\u0027s no energy."},{"Start":"07:00.215 ","End":"07:05.555","Text":"Because we can see that our minimum energy where we do have 0 kinetic energy,"},{"Start":"07:05.555 ","End":"07:08.865","Text":"is that this negative value for U."},{"Start":"07:08.865 ","End":"07:12.320","Text":"It\u0027s calibrated a bit differently to what we\u0027re used to."},{"Start":"07:12.320 ","End":"07:14.630","Text":"All right, so that is the end of the lesson."},{"Start":"07:14.630 ","End":"07:15.695","Text":"In the next lesson,"},{"Start":"07:15.695 ","End":"07:18.870","Text":"we\u0027re going to be proving this equation."}],"ID":22331},{"Watched":false,"Name":"Energy Equation Proof","Duration":"9m 22s","ChapterTopicVideoID":21446,"CourseChapterTopicPlaylistID":99476,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.025","Text":"Hello. In this lesson,"},{"Start":"00:02.025 ","End":"00:05.550","Text":"we\u0027re going to be looking at the proof and the explanation,"},{"Start":"00:05.550 ","End":"00:10.000","Text":"for our energy equation that we saw in the previous lesson."},{"Start":"00:10.000 ","End":"00:13.550","Text":"We saw the equation in the previous lesson that said that"},{"Start":"00:13.550 ","End":"00:16.970","Text":"the potential for kinetic energy U is"},{"Start":"00:16.970 ","End":"00:23.280","Text":"equal to negative p dot E. Where p is the dipole moment,"},{"Start":"00:23.280 ","End":"00:25.500","Text":"and E is the electric field."},{"Start":"00:25.500 ","End":"00:31.635","Text":"Let\u0027s say that we have our electric field in some direction. Let\u0027s say this."},{"Start":"00:31.635 ","End":"00:37.640","Text":"We have our dipole moment in another random direction, let\u0027s say like so."},{"Start":"00:37.640 ","End":"00:40.220","Text":"What we\u0027re going to do, is we\u0027re going to"},{"Start":"00:40.220 ","End":"00:43.775","Text":"choose that the direction that the electric field is in."},{"Start":"00:43.775 ","End":"00:47.290","Text":"This direction is going to be our x-axis,"},{"Start":"00:47.290 ","End":"00:52.620","Text":"and of course perpendicular to it is the y-axis,"},{"Start":"00:52.620 ","End":"00:57.240","Text":"and perpendicular to both of those to the x,"},{"Start":"00:57.240 ","End":"01:00.180","Text":"y axis is the z-axis."},{"Start":"01:00.180 ","End":"01:05.950","Text":"Our dipole moment p is always going to be on the x, y plane."},{"Start":"01:06.170 ","End":"01:14.920","Text":"Then we\u0027ll define this angle between p and E as being Theta."},{"Start":"01:15.530 ","End":"01:22.845","Text":"Usually energy U is defined as the negative integral of F,"},{"Start":"01:22.845 ","End":"01:26.430","Text":"the force along dr."},{"Start":"01:26.430 ","End":"01:34.365","Text":"What do we have here is that our dipole moment is rotating in order to line up."},{"Start":"01:34.365 ","End":"01:37.455","Text":"It isn\u0027t exactly a force in a straight line,"},{"Start":"01:37.455 ","End":"01:40.290","Text":"but it\u0027s rather a rotation,"},{"Start":"01:40.290 ","End":"01:41.745","Text":"so as we know,"},{"Start":"01:41.745 ","End":"01:46.050","Text":"we have this moment of force or torque."},{"Start":"01:46.050 ","End":"01:51.465","Text":"In that case, we can say that similarly to this force acting in a straight line,"},{"Start":"01:51.465 ","End":"01:54.540","Text":"we have our torque which is causing this energy."},{"Start":"01:54.540 ","End":"01:57.825","Text":"We\u0027ll have that U is equal to the negative integral,"},{"Start":"01:57.825 ","End":"01:59.760","Text":"and then instead of an enforce,"},{"Start":"01:59.760 ","End":"02:02.790","Text":"we\u0027re integrating along the torque,"},{"Start":"02:02.790 ","End":"02:05.895","Text":"dot and instead of dr,"},{"Start":"02:05.895 ","End":"02:11.080","Text":"so we\u0027re going to be along d Theta."},{"Start":"02:12.440 ","End":"02:16.770","Text":"Let\u0027s begin, so we have the negative integral."},{"Start":"02:16.770 ","End":"02:24.960","Text":"We know that torque is equal to p cross multiplied by E and then dot,"},{"Start":"02:24.960 ","End":"02:29.050","Text":"so dot product with rd Theta."},{"Start":"02:29.450 ","End":"02:33.585","Text":"We know that our p cross E,"},{"Start":"02:33.585 ","End":"02:38.070","Text":"we can rewrite this as the magnitude of p,"},{"Start":"02:38.070 ","End":"02:41.085","Text":"multiplied by the magnitude of E,"},{"Start":"02:41.085 ","End":"02:44.280","Text":"multiplied by sin of the angle between the 2,"},{"Start":"02:44.280 ","End":"02:49.840","Text":"so that\u0027s Theta dot d Theta."},{"Start":"02:51.440 ","End":"02:55.005","Text":"Before I forget this p cross E,"},{"Start":"02:55.005 ","End":"02:59.040","Text":"are going to have a direction over here."},{"Start":"02:59.040 ","End":"03:01.815","Text":"If you remember the right-hand rule,"},{"Start":"03:01.815 ","End":"03:05.925","Text":"you take your fingers curling around from p,"},{"Start":"03:05.925 ","End":"03:13.065","Text":"and p is rotating into E. It\u0027s trying to line up with our E field."},{"Start":"03:13.065 ","End":"03:17.190","Text":"If you curl your fingers p like it\u0027s curling into E,"},{"Start":"03:17.190 ","End":"03:21.285","Text":"your thumb will be pointing in the negative z direction."},{"Start":"03:21.285 ","End":"03:29.925","Text":"Then what will have is that all of this is in the negative z direction,"},{"Start":"03:29.925 ","End":"03:37.240","Text":"so this all belongs to this p cross E and then dot d Theta."},{"Start":"03:38.330 ","End":"03:42.075","Text":"Let\u0027s carry this on the next line,"},{"Start":"03:42.075 ","End":"03:46.220","Text":"so far we have the negative integral of the magnitude of p"},{"Start":"03:46.220 ","End":"03:50.810","Text":"multiplied by the magnitude of E multiplied by sin(Theta),"},{"Start":"03:50.810 ","End":"03:54.935","Text":"and all of this is in the negative z direction."},{"Start":"03:54.935 ","End":"03:58.205","Text":"What is this d Theta vector?"},{"Start":"03:58.205 ","End":"04:00.530","Text":"First of all, it\u0027s a d Theta,"},{"Start":"04:00.530 ","End":"04:03.935","Text":"and then it\u0027s going to have a vector direction."},{"Start":"04:03.935 ","End":"04:06.790","Text":"What is its direction?"},{"Start":"04:06.790 ","End":"04:11.630","Text":"We know that this Theta is in relation to the x-axis,"},{"Start":"04:11.630 ","End":"04:14.660","Text":"and that\u0027s generally how it will be defined."},{"Start":"04:14.660 ","End":"04:18.540","Text":"We know that Theta is opening up in this direction."},{"Start":"04:18.650 ","End":"04:21.920","Text":"Again, if we use the right-hand rule,"},{"Start":"04:21.920 ","End":"04:26.050","Text":"if we curl our fingers in the direction of our Theta,"},{"Start":"04:26.050 ","End":"04:33.470","Text":"so what we\u0027ll get is that our thumb is therefore pointing in the z direction."},{"Start":"04:33.470 ","End":"04:38.555","Text":"If we curl our fingers in the direction that Theta opens up in,"},{"Start":"04:38.555 ","End":"04:41.995","Text":"our thumb is pointing in the positive z direction,"},{"Start":"04:41.995 ","End":"04:45.855","Text":"we can write this like this,"},{"Start":"04:45.855 ","End":"04:50.575","Text":"so for our Theta we\u0027ll have our thumb pointing in this direction,"},{"Start":"04:50.575 ","End":"04:59.380","Text":"and all of our fingers will be curling in the direction of Theta,"},{"Start":"04:59.380 ","End":"05:02.280","Text":"in this anticlockwise direction."},{"Start":"05:02.840 ","End":"05:08.390","Text":"Before I carry on another way of explaining this is by,"},{"Start":"05:08.390 ","End":"05:10.970","Text":"if you remember in previous lessons and especially in"},{"Start":"05:10.970 ","End":"05:14.615","Text":"mechanics when we spoke about circular motion."},{"Start":"05:14.615 ","End":"05:20.575","Text":"We always defined a positive direction for Theta and it was usually in this direction,"},{"Start":"05:20.575 ","End":"05:25.140","Text":"that this is the positive Theta direction."},{"Start":"05:25.140 ","End":"05:29.515","Text":"This is the direction of rd Theta vector."},{"Start":"05:29.515 ","End":"05:34.490","Text":"We usually define it as being negative as the anticlockwise direction."},{"Start":"05:34.490 ","End":"05:36.440","Text":"On the other hand,"},{"Start":"05:36.440 ","End":"05:40.240","Text":"we see that this has a dipole moment p,"},{"Start":"05:40.240 ","End":"05:43.595","Text":"and it\u0027s trying to line up with this vector over here,"},{"Start":"05:43.595 ","End":"05:47.270","Text":"E. A dipole moment,"},{"Start":"05:47.270 ","End":"05:49.850","Text":"or a moment,"},{"Start":"05:49.850 ","End":"05:54.340","Text":"or torque is trying to move in this direction."},{"Start":"05:54.340 ","End":"05:57.375","Text":"This is our Tau vector direction."},{"Start":"05:57.375 ","End":"05:59.000","Text":"What we can see is that"},{"Start":"05:59.000 ","End":"06:04.260","Text":"our Tau vector direction is in the opposite direction to rd Theta."},{"Start":"06:05.360 ","End":"06:08.220","Text":"It\u0027s in the negative direction."},{"Start":"06:08.220 ","End":"06:12.580","Text":"That will show you that women to get 1 negative sign over here,"},{"Start":"06:12.580 ","End":"06:15.470","Text":"because we have something rotating in the positive direction,"},{"Start":"06:15.470 ","End":"06:22.759","Text":"but another thing trying to move in the negative direction."},{"Start":"06:22.759 ","End":"06:30.270","Text":"That\u0027s why we\u0027ll get a negative sign over here instead of writing this negative z."},{"Start":"06:30.310 ","End":"06:34.160","Text":"We have this negative sign over here that we can"},{"Start":"06:34.160 ","End":"06:38.170","Text":"just straight away right in without doing all of this."},{"Start":"06:38.170 ","End":"06:40.475","Text":"Of course over here,"},{"Start":"06:40.475 ","End":"06:43.290","Text":"we\u0027re meant to have a dot-product."},{"Start":"06:44.300 ","End":"06:48.590","Text":"This minus and this minus will cancel out."},{"Start":"06:48.590 ","End":"06:52.295","Text":"Then we have z hat, dot z hat."},{"Start":"06:52.295 ","End":"06:55.340","Text":"That\u0027s going to be equal to 1."},{"Start":"06:55.340 ","End":"07:04.565","Text":"As we know, what we\u0027re going to have is the integral on pE sin(Theta), d Theta."},{"Start":"07:04.565 ","End":"07:08.170","Text":"Then what we know is that pE are just constants,"},{"Start":"07:08.170 ","End":"07:12.420","Text":"their values, so we can write this as pE."},{"Start":"07:12.420 ","End":"07:17.310","Text":"Then the integral on sin(Theta), d Theta,"},{"Start":"07:17.310 ","End":"07:20.670","Text":"which is equal to pE,"},{"Start":"07:20.670 ","End":"07:24.215","Text":"the negative cos(Theta),"},{"Start":"07:24.215 ","End":"07:25.730","Text":"because the integral of sin(Theta),"},{"Start":"07:25.730 ","End":"07:28.580","Text":"d Theta is negative cos(Theta)."},{"Start":"07:28.580 ","End":"07:34.355","Text":"Because here\u0027s specifically, we did an integral without bounds,"},{"Start":"07:34.355 ","End":"07:43.040","Text":"so we\u0027ll add in this integrating constant k plus C. PE cos(Theta),"},{"Start":"07:43.040 ","End":"07:48.085","Text":"is simply equal to p vector dot E vector."},{"Start":"07:48.085 ","End":"07:50.685","Text":"We have still the minus,"},{"Start":"07:50.685 ","End":"07:53.730","Text":"so p cos(Theta) is simply equal to,"},{"Start":"07:53.730 ","End":"07:56.760","Text":"p vector dot E vector."},{"Start":"07:56.760 ","End":"07:59.315","Text":"That\u0027s the definition of this dot-product."},{"Start":"07:59.315 ","End":"08:03.380","Text":"Then we have plus C. In order to calibrate this,"},{"Start":"08:03.380 ","End":"08:04.940","Text":"we usually just sub in,"},{"Start":"08:04.940 ","End":"08:07.355","Text":"let\u0027s see over here is equal to 0,"},{"Start":"08:07.355 ","End":"08:11.090","Text":"because there\u0027s no reason for us to think that it\u0027s anything else at this stage,"},{"Start":"08:11.090 ","End":"08:14.460","Text":"so we calibrate it to equal 0."},{"Start":"08:14.540 ","End":"08:19.225","Text":"All this means is that when p dot E is equal to 0,"},{"Start":"08:19.225 ","End":"08:22.285","Text":"so the energy will be equal to 0."},{"Start":"08:22.285 ","End":"08:26.740","Text":"This is something that we can always do and dealing with potential energy,"},{"Start":"08:26.740 ","End":"08:29.770","Text":"just like when we are dealing with potential energy to"},{"Start":"08:29.770 ","End":"08:33.405","Text":"do with height and gravity, so MGH."},{"Start":"08:33.405 ","End":"08:37.060","Text":"We always define a certain height which is equal to 0."},{"Start":"08:37.060 ","End":"08:39.535","Text":"If we have a surface,"},{"Start":"08:39.535 ","End":"08:43.010","Text":"usually we define this area here,"},{"Start":"08:43.010 ","End":"08:46.110","Text":"over here as being of height zero."},{"Start":"08:46.110 ","End":"08:48.955","Text":"But sometimes we can also,"},{"Start":"08:48.955 ","End":"08:50.320","Text":"if we\u0027re looking at a cliff,"},{"Start":"08:50.320 ","End":"08:53.140","Text":"we can say that this height is still zero even though"},{"Start":"08:53.140 ","End":"08:56.125","Text":"we know that there\u0027s still this distance over here,"},{"Start":"08:56.125 ","End":"08:58.690","Text":"till the fall, and then we go into a minus."},{"Start":"08:58.690 ","End":"09:01.550","Text":"We can always do this calibration with potential energy,"},{"Start":"09:01.550 ","End":"09:05.825","Text":"where we just set an arbitrary value to be equal to 0."},{"Start":"09:05.825 ","End":"09:10.160","Text":"That\u0027s why we could define this constant of integration,"},{"Start":"09:10.160 ","End":"09:13.190","Text":"C is equal to 0. There we go."},{"Start":"09:13.190 ","End":"09:17.870","Text":"Now we got that our potential energy U is equal to negative p dot E,"},{"Start":"09:17.870 ","End":"09:19.830","Text":"which is what we were expecting to get."},{"Start":"09:19.830 ","End":"09:22.940","Text":"That is the end of this lesson."}],"ID":22332},{"Watched":false,"Name":"Force Acting on Dipole in an Electric Field","Duration":"14m 48s","ChapterTopicVideoID":21447,"CourseChapterTopicPlaylistID":99476,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.755","Text":"Hello. In this lesson,"},{"Start":"00:01.755 ","End":"00:06.000","Text":"we\u0027re going to be speaking about the force acting on a dipole in an electric field."},{"Start":"00:06.000 ","End":"00:09.240","Text":"Let\u0027s say that we have some axes,"},{"Start":"00:09.240 ","End":"00:16.155","Text":"and that somewhere over here we have our dipole moment"},{"Start":"00:16.155 ","End":"00:25.170","Text":"going like so in an arbitrary direction and in this direction we have our electric field."},{"Start":"00:25.580 ","End":"00:35.114","Text":"Let\u0027s consider that this dipole moment is just 2 point charges,"},{"Start":"00:35.114 ","End":"00:38.745","Text":"so 1 has a negative charge,"},{"Start":"00:38.745 ","End":"00:42.975","Text":"and the other charge is a positive charge,"},{"Start":"00:42.975 ","End":"00:47.830","Text":"and they\u0027re distance d away from 1 another."},{"Start":"00:48.500 ","End":"00:53.060","Text":"What I want to do is I want to find the total force acting on"},{"Start":"00:53.060 ","End":"00:57.290","Text":"this body or this system which has 2 charges,"},{"Start":"00:57.290 ","End":"01:00.420","Text":"a negative charge and a positive charge."},{"Start":"01:00.440 ","End":"01:04.354","Text":"I know that the sum of all the forces,"},{"Start":"01:04.354 ","End":"01:11.810","Text":"or the total forces acting on this system is going to be equal to the force acting on"},{"Start":"01:11.810 ","End":"01:20.330","Text":"the positive charge plus the force acting on the negative charge."},{"Start":"01:20.330 ","End":"01:25.190","Text":"Obviously we know that the force acting on the positive charge will"},{"Start":"01:25.190 ","End":"01:30.640","Text":"be in the opposite direction to the force acting on the negative charge."},{"Start":"01:31.640 ","End":"01:37.100","Text":"What we can say is that the force on"},{"Start":"01:37.100 ","End":"01:41.660","Text":"the positive charge is going to be pushing in this direction,"},{"Start":"01:41.660 ","End":"01:47.135","Text":"because our positive charge wants to move in this direction to line up with the"},{"Start":"01:47.135 ","End":"01:51.740","Text":"E. It\u0027s going to be in this direction and the force on"},{"Start":"01:51.740 ","End":"01:56.900","Text":"the negative charge is going to be in this direction,"},{"Start":"01:56.900 ","End":"02:04.800","Text":"because it\u0027s trying to move in this direction clockwise in order to have that"},{"Start":"02:04.800 ","End":"02:09.740","Text":"the negative charge is at the base of this line and"},{"Start":"02:09.740 ","End":"02:13.295","Text":"the positive charge is higher up in a straight line lined"},{"Start":"02:13.295 ","End":"02:17.580","Text":"up parallel to the E field lines."},{"Start":"02:17.580 ","End":"02:24.455","Text":"What we can see, both of these particles have an equal and opposite charge."},{"Start":"02:24.455 ","End":"02:30.605","Text":"The forces are going to be pulling with an equal force just in opposite directions."},{"Start":"02:30.605 ","End":"02:37.350","Text":"Therefore, we can say that this is approximately equal to 0."},{"Start":"02:38.480 ","End":"02:42.905","Text":"This is what happens when we have a uniform E field."},{"Start":"02:42.905 ","End":"02:44.510","Text":"We get that the total force on"},{"Start":"02:44.510 ","End":"02:50.060","Text":"this 2 body system is going to be approximately equal to 0."},{"Start":"02:50.060 ","End":"02:53.945","Text":"However, if we had a non-uniform electric field,"},{"Start":"02:53.945 ","End":"02:57.800","Text":"so the electric field acting over here would be"},{"Start":"02:57.800 ","End":"03:02.060","Text":"different to the electric field acting over here on the positive charge,"},{"Start":"03:02.060 ","End":"03:05.650","Text":"then we will have some force."},{"Start":"03:05.650 ","End":"03:08.285","Text":"This is for a uniform electric field."},{"Start":"03:08.285 ","End":"03:12.455","Text":"However, if we have a non-uniform electric field,"},{"Start":"03:12.455 ","End":"03:18.140","Text":"then our equation for the force is equal to in brackets p vector,"},{"Start":"03:18.140 ","End":"03:21.530","Text":"which is our dipole moment dot product with"},{"Start":"03:21.530 ","End":"03:28.320","Text":"the gradient and then multiplied by our E field."},{"Start":"03:28.820 ","End":"03:34.790","Text":"This gradient function, another way to call it is the Nabla."},{"Start":"03:34.790 ","End":"03:39.005","Text":"Let\u0027s open this up and this,"},{"Start":"03:39.005 ","End":"03:43.300","Text":"of course, I\u0027ve squared in red because this is an equation that you need to write down."},{"Start":"03:43.300 ","End":"03:45.240","Text":"I\u0027m just going to remind you,"},{"Start":"03:45.240 ","End":"03:48.020","Text":"Nabla function looks like this."},{"Start":"03:48.020 ","End":"03:51.530","Text":"The x component is d by dx,"},{"Start":"03:51.530 ","End":"03:54.395","Text":"the y component is d by dy,"},{"Start":"03:54.395 ","End":"03:57.950","Text":"and the z component is d by dz."},{"Start":"03:57.950 ","End":"04:01.100","Text":"We\u0027re taking the partial derivatives."},{"Start":"04:01.100 ","End":"04:07.040","Text":"Then if we have what we have inside the brackets, p dot Nabla,"},{"Start":"04:07.040 ","End":"04:15.360","Text":"so that is going to be equal to the x component of our p multiplied by d by dx,"},{"Start":"04:15.360 ","End":"04:21.840","Text":"so we\u0027ll have p_x d by dx plus,"},{"Start":"04:21.840 ","End":"04:23.120","Text":"so we\u0027re doing a dot product,"},{"Start":"04:23.120 ","End":"04:28.140","Text":"so we\u0027re getting a scalar function not a vector."},{"Start":"04:28.140 ","End":"04:36.720","Text":"P_x d by dx plus p_y multiplied by d by dy"},{"Start":"04:36.720 ","End":"04:40.260","Text":"plus p_z multiplied by"},{"Start":"04:40.260 ","End":"04:47.790","Text":"d by dz."},{"Start":"04:47.790 ","End":"04:51.785","Text":"All of this is going to be multiplied by our E field."},{"Start":"04:51.785 ","End":"04:53.780","Text":"We can see that our E field,"},{"Start":"04:53.780 ","End":"04:55.915","Text":"and we know it\u0027s a vector quantity,"},{"Start":"04:55.915 ","End":"05:01.770","Text":"so that means that we have the x component in the x direction plus"},{"Start":"05:01.770 ","End":"05:10.360","Text":"the y component in the y direction plus the z component in the z direction."},{"Start":"05:10.760 ","End":"05:13.800","Text":"Let\u0027s put this all together."},{"Start":"05:13.800 ","End":"05:17.525","Text":"What we\u0027re doing is we\u0027re multiplying what we have in the brackets,"},{"Start":"05:17.525 ","End":"05:21.800","Text":"which has p dot Nabla by our E vector."},{"Start":"05:21.800 ","End":"05:26.210","Text":"We\u0027re going to have all of this P_x d by dx, P_y d by dy,"},{"Start":"05:26.210 ","End":"05:32.085","Text":"P_z d by dz multiplied by our E_x in the x direction."},{"Start":"05:32.085 ","End":"05:33.960","Text":"Then all of this again,"},{"Start":"05:33.960 ","End":"05:39.690","Text":"this p dot Nabla multiplied by E_y in the y direction and then all of"},{"Start":"05:39.690 ","End":"05:43.615","Text":"this p dot Nabla multiplied by E_z"},{"Start":"05:43.615 ","End":"05:49.770","Text":"in the z direction or not multiplied but working on E_z."},{"Start":"05:50.510 ","End":"05:53.755","Text":"Let\u0027s write this out."},{"Start":"05:53.755 ","End":"05:56.470","Text":"I can write that the sum of all of my forces."},{"Start":"05:56.470 ","End":"06:00.815","Text":"I can write my x component of my force first."},{"Start":"06:00.815 ","End":"06:03.480","Text":"I\u0027m going to have this,"},{"Start":"06:03.480 ","End":"06:08.300","Text":"so I\u0027ll have P_x multiplied by dE_x,"},{"Start":"06:08.300 ","End":"06:17.900","Text":"so dE_x by dx plus P_y"},{"Start":"06:17.900 ","End":"06:26.480","Text":"multiplied by dE_x by dy"},{"Start":"06:26.480 ","End":"06:30.920","Text":"plus P_z multiplied"},{"Start":"06:30.920 ","End":"06:37.710","Text":"by dE_x by dz."},{"Start":"06:37.710 ","End":"06:39.770","Text":"All of this, I could either write this as"},{"Start":"06:39.770 ","End":"06:44.715","Text":"F in the x direction or I could write F vector x hat,"},{"Start":"06:44.715 ","End":"06:48.185","Text":"then I can write my F in the y direction,"},{"Start":"06:48.185 ","End":"06:50.285","Text":"which will just be this again."},{"Start":"06:50.285 ","End":"06:53.915","Text":"I have P_x d by E but this time,"},{"Start":"06:53.915 ","End":"07:03.410","Text":"y by dx plus P_y d by E_y by"},{"Start":"07:03.410 ","End":"07:09.800","Text":"dy plus P_z dE_y"},{"Start":"07:09.800 ","End":"07:15.230","Text":"by dz and of course,"},{"Start":"07:15.230 ","End":"07:17.495","Text":"my z component for the force,"},{"Start":"07:17.495 ","End":"07:24.935","Text":"which is P_x d by E_z dz plus"},{"Start":"07:24.935 ","End":"07:35.923","Text":"P_y d by E_z dy plus P_z dE_z by dz."},{"Start":"07:35.923 ","End":"07:46.405","Text":"We can therefore write that the total force,"},{"Start":"07:46.405 ","End":"07:51.280","Text":"or the sum of the forces is going to be our F_x,"},{"Start":"07:51.280 ","End":"07:55.405","Text":"so all of this in the x direction plus our F_y,"},{"Start":"07:55.405 ","End":"08:06.140","Text":"so all of this in the y direction plus our F_z in the z direction."},{"Start":"08:06.990 ","End":"08:11.710","Text":"We can work out the force like so but"},{"Start":"08:11.710 ","End":"08:15.805","Text":"what we can see is that we get quite a lot of equations,"},{"Start":"08:15.805 ","End":"08:18.565","Text":"and it\u0027s quite complicated to solve this."},{"Start":"08:18.565 ","End":"08:21.885","Text":"What we can do is we can write,"},{"Start":"08:21.885 ","End":"08:25.320","Text":"and I\u0027m warning you this doesn\u0027t work in every case."},{"Start":"08:25.320 ","End":"08:26.925","Text":"Write when it does work."},{"Start":"08:26.925 ","End":"08:32.700","Text":"Then another equation to get the force in a non-uniform E-field is by"},{"Start":"08:32.700 ","End":"08:39.910","Text":"taking the negative gradient of our energy, U."},{"Start":"08:39.910 ","End":"08:45.640","Text":"The only time that you can use this equation for the force as equal"},{"Start":"08:45.640 ","End":"08:50.980","Text":"to negative the gradient of U of our energy,"},{"Start":"08:50.980 ","End":"08:53.110","Text":"so you can\u0027t use this all the time,"},{"Start":"08:53.110 ","End":"08:55.863","Text":"this equation is always correct,"},{"Start":"08:55.863 ","End":"08:58.330","Text":"and this 1 over here isn\u0027t always correct."},{"Start":"08:58.330 ","End":"09:04.060","Text":"You can only use this when the field is conservative."},{"Start":"09:04.060 ","End":"09:07.210","Text":"If the E-field is conservative, you can use this."},{"Start":"09:07.210 ","End":"09:16.165","Text":"That means that the E-field is formed from charges and another condition,"},{"Start":"09:16.165 ","End":"09:18.609","Text":"so the E-field has to be conservative,"},{"Start":"09:18.609 ","End":"09:20.065","Text":"but also p,"},{"Start":"09:20.065 ","End":"09:23.530","Text":"our dipole moment has to be constant."},{"Start":"09:23.530 ","End":"09:29.680","Text":"What is that mean? That means that it\u0027s independent of coordinates. What is that mean?"},{"Start":"09:29.680 ","End":"09:34.240","Text":"That means that if the dipole moment is located over here,"},{"Start":"09:34.240 ","End":"09:38.770","Text":"or if it\u0027s located over here,"},{"Start":"09:38.770 ","End":"09:43.405","Text":"we\u0027re going to have the same dipole moment."},{"Start":"09:43.405 ","End":"09:45.670","Text":"It\u0027s independent of coordinates,"},{"Start":"09:45.670 ","End":"09:50.425","Text":"it\u0027s independent of its position in space."},{"Start":"09:50.425 ","End":"09:55.525","Text":"If these 2 conditions take place,"},{"Start":"09:55.525 ","End":"09:57.498","Text":"then you can use this equation,"},{"Start":"09:57.498 ","End":"10:00.560","Text":"and then it makes the calculation a lot easier."},{"Start":"10:00.600 ","End":"10:03.025","Text":"Just a quick reminder,"},{"Start":"10:03.025 ","End":"10:05.365","Text":"so let\u0027s just write out this equation."},{"Start":"10:05.365 ","End":"10:13.120","Text":"We\u0027ll get that F is equal to negative gradient of U,"},{"Start":"10:13.120 ","End":"10:19.030","Text":"which is equal to negative Nabla and U if we remember from the previous lesson,"},{"Start":"10:19.030 ","End":"10:28.300","Text":"we saw that it\u0027s equal to negative p dot E. The negative and the negative go out."},{"Start":"10:28.300 ","End":"10:34.190","Text":"We\u0027ll get that F is equal to the gradient (p dot E)."},{"Start":"10:37.260 ","End":"10:39.805","Text":"If p is constant,"},{"Start":"10:39.805 ","End":"10:41.050","Text":"then we can use this equation."},{"Start":"10:41.050 ","End":"10:44.485","Text":"If it isn\u0027t, if it\u0027s dependent on its position,"},{"Start":"10:44.485 ","End":"10:47.688","Text":"then we also have to take the derivative of p,"},{"Start":"10:47.688 ","End":"10:52.940","Text":"and then it gets complicated and makes this equation incorrect."},{"Start":"10:53.910 ","End":"11:00.220","Text":"Over here we saw that F is equal to Nabla(p dot E)."},{"Start":"11:00.220 ","End":"11:02.635","Text":"Let\u0027s take a look over here,"},{"Start":"11:02.635 ","End":"11:05.620","Text":"and we can see that we have this identity,"},{"Start":"11:05.620 ","End":"11:10.075","Text":"Nabla A dot B is equal to all of this."},{"Start":"11:10.075 ","End":"11:19.135","Text":"What we\u0027re going to do, let\u0027s just write over here that F is equal to Nabla(p dot E)."},{"Start":"11:19.135 ","End":"11:22.180","Text":"What we\u0027re going to do is we\u0027re going to"},{"Start":"11:22.180 ","End":"11:26.065","Text":"write out this identity according to this equation."},{"Start":"11:26.065 ","End":"11:27.940","Text":"Let\u0027s scroll down."},{"Start":"11:27.940 ","End":"11:34.210","Text":"We can see that p is equal to A and E is equal to B in this equation."},{"Start":"11:34.210 ","End":"11:45.190","Text":"We\u0027ll have p cross multiplied by Nabla cross E"},{"Start":"11:45.190 ","End":"11:54.140","Text":"plus E cross multiplied by Nabla cross p"},{"Start":"11:55.470 ","End":"12:04.090","Text":"plus p dot Nabla E plus"},{"Start":"12:04.090 ","End":"12:14.170","Text":"E dot Nabla p. Let\u0027s just scroll back up."},{"Start":"12:14.170 ","End":"12:17.140","Text":"We\u0027ve already said that we can only use this equation."},{"Start":"12:17.140 ","End":"12:24.340","Text":"It\u0027s only correct if p is constant and if the E-field is conservative."},{"Start":"12:24.340 ","End":"12:29.665","Text":"Right now we\u0027re just going to show why that is true."},{"Start":"12:29.665 ","End":"12:33.475","Text":"Why we can only use this under these conditions."},{"Start":"12:33.475 ","End":"12:35.575","Text":"Let\u0027s go back here."},{"Start":"12:35.575 ","End":"12:38.440","Text":"If p is constant,"},{"Start":"12:38.440 ","End":"12:41.395","Text":"that means that if we take the derivative of it,"},{"Start":"12:41.395 ","End":"12:43.720","Text":"then it will be equal to 0."},{"Start":"12:43.720 ","End":"12:48.550","Text":"Here we\u0027re going to have E dot Nabla p. That"},{"Start":"12:48.550 ","End":"12:52.930","Text":"means that we\u0027re taking the derivative of p. Then if p is constant,"},{"Start":"12:52.930 ","End":"12:56.155","Text":"this will equal to 0 and also here,"},{"Start":"12:56.155 ","End":"12:58.540","Text":"when we do Nabla cross p,"},{"Start":"12:58.540 ","End":"13:04.920","Text":"that means that we\u0027re also taking the derivatives of p and if p is constant,"},{"Start":"13:04.920 ","End":"13:07.990","Text":"then the derivatives will equal to 0."},{"Start":"13:08.190 ","End":"13:12.085","Text":"Here we have our force."},{"Start":"13:12.085 ","End":"13:15.895","Text":"This is what we want to be getting to,"},{"Start":"13:15.895 ","End":"13:20.920","Text":"p dot Nabla multiplied by E,"},{"Start":"13:20.920 ","End":"13:25.130","Text":"this is the equation."},{"Start":"13:26.100 ","End":"13:34.180","Text":"What we want to do is we want to show that this over here is equal to 0,"},{"Start":"13:34.180 ","End":"13:36.985","Text":"that this element is equal to 0,"},{"Start":"13:36.985 ","End":"13:41.050","Text":"and that we\u0027re going to show by showing that the rotor of E,"},{"Start":"13:41.050 ","End":"13:45.010","Text":"so Nabla cross E is equal to 0."},{"Start":"13:45.010 ","End":"13:50.680","Text":"What do we know if we take the rotor of a vector quantity?"},{"Start":"13:50.680 ","End":"13:54.025","Text":"If the vector quantity is conservative,"},{"Start":"13:54.025 ","End":"13:57.145","Text":"then the rotor of it is equal to 0."},{"Start":"13:57.145 ","End":"14:00.925","Text":"Then the rotor of E will equal to 0."},{"Start":"14:00.925 ","End":"14:05.080","Text":"Therefore, this is also going to be equal to 0."},{"Start":"14:05.080 ","End":"14:15.925","Text":"Then we\u0027ll get that are only non-zero elements and this whole equation is p dot Nabla E,"},{"Start":"14:15.925 ","End":"14:20.170","Text":"which is exactly what we have over here for"},{"Start":"14:20.170 ","End":"14:26.260","Text":"the force in a non-uniform E-field on an electric dipole."},{"Start":"14:26.260 ","End":"14:31.750","Text":"What we can see is that we proved why this equation can only"},{"Start":"14:31.750 ","End":"14:38.720","Text":"work if E is conservative and if p is constant."},{"Start":"14:38.760 ","End":"14:43.810","Text":"Now we saw that this equation is correct as long as"},{"Start":"14:43.810 ","End":"14:49.280","Text":"these conditions are met and that is the end of this lesson."}],"ID":22333},{"Watched":false,"Name":"Exercise 4","Duration":"17m 40s","ChapterTopicVideoID":21297,"CourseChapterTopicPlaylistID":99476,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.235","Text":"Hello. In this lesson,"},{"Start":"00:02.235 ","End":"00:04.530","Text":"we\u0027re going to be answering the following question,"},{"Start":"00:04.530 ","End":"00:12.150","Text":"an electric dipole p is placed at a point r. A charge q is at the origin."},{"Start":"00:12.150 ","End":"00:13.860","Text":"We can assume that p,"},{"Start":"00:13.860 ","End":"00:15.765","Text":"r, and q are given to us."},{"Start":"00:15.765 ","End":"00:17.145","Text":"Question number 1,"},{"Start":"00:17.145 ","End":"00:18.585","Text":"which we\u0027ll start with is,"},{"Start":"00:18.585 ","End":"00:21.630","Text":"what is the torque acting on the dipole?"},{"Start":"00:21.630 ","End":"00:28.565","Text":"We\u0027ve already seen that the equation for torque is equal to p"},{"Start":"00:28.565 ","End":"00:38.450","Text":"cross E. Our electric field comes from the fact that we have a point charge over here."},{"Start":"00:38.450 ","End":"00:44.720","Text":"We can write that as p vector cross multiplied by the electric field from a point charge."},{"Start":"00:44.720 ","End":"00:48.440","Text":"We know that we can write the electric field from a point charge as"},{"Start":"00:48.440 ","End":"00:53.050","Text":"kq divided by r^2 in the r hat direction,"},{"Start":"00:53.050 ","End":"00:59.160","Text":"or kq r vector divided by r^3."},{"Start":"00:59.160 ","End":"01:05.279","Text":"Because remember r vector is equal to r hat multiplied"},{"Start":"01:05.279 ","End":"01:11.900","Text":"by r. Then we can put all of the constants to 1 side."},{"Start":"01:11.900 ","End":"01:15.125","Text":"We can write this as kq divided by"},{"Start":"01:15.125 ","End":"01:23.880","Text":"r cubed multiplied by p cross r vector."},{"Start":"01:24.170 ","End":"01:28.785","Text":"Because we are assuming that we know what P and r vector is,"},{"Start":"01:28.785 ","End":"01:30.350","Text":"that we are giving it in the question,"},{"Start":"01:30.350 ","End":"01:33.110","Text":"this is the onset to question number 1."},{"Start":"01:33.110 ","End":"01:36.095","Text":"Let\u0027s answer question number 2."},{"Start":"01:36.095 ","End":"01:40.049","Text":"What is the energy of the dipole?"},{"Start":"01:41.000 ","End":"01:51.480","Text":"The equation for energy U is equal to negative p dot E. We saw this a few lessons ago."},{"Start":"01:51.480 ","End":"01:54.720","Text":"That is equal to negative and then we"},{"Start":"01:54.720 ","End":"01:58.970","Text":"can write E again like this and take out all the constants,"},{"Start":"01:58.970 ","End":"02:04.035","Text":"so we have negative kq divided by r^3,"},{"Start":"02:04.035 ","End":"02:10.690","Text":"and then we have multiplied by p vector dot r vector."},{"Start":"02:11.000 ","End":"02:14.410","Text":"This is the answer to question 2."},{"Start":"02:14.410 ","End":"02:17.555","Text":"Let\u0027s take a look at question 3."},{"Start":"02:17.555 ","End":"02:23.975","Text":"We are being asked to show that the force acting on the dipole is equal to k"},{"Start":"02:23.975 ","End":"02:31.605","Text":"multiplied by p vector dot r^2 minus p vector dot r vector,"},{"Start":"02:31.605 ","End":"02:36.405","Text":"dot r vector divided by r^5."},{"Start":"02:36.405 ","End":"02:38.090","Text":"In the previous lesson,"},{"Start":"02:38.090 ","End":"02:42.950","Text":"we saw that we can use 2 equations to calculate the force,"},{"Start":"02:42.950 ","End":"02:48.110","Text":"and 1 of the equations that we saw was that F"},{"Start":"02:48.110 ","End":"02:53.420","Text":"is equal to the negative gradient of the energy U."},{"Start":"02:53.420 ","End":"03:00.315","Text":"We said that we can only use this equation if our dipole moment p is constant,"},{"Start":"03:00.315 ","End":"03:03.595","Text":"it\u0027s independent on its position."},{"Start":"03:03.595 ","End":"03:06.530","Text":"In the question, we are not being told that it\u0027s"},{"Start":"03:06.530 ","End":"03:09.230","Text":"dependent on anything and we are told that it\u0027s placed"},{"Start":"03:09.230 ","End":"03:14.550","Text":"at a random point r. We can see that p is independent of"},{"Start":"03:14.550 ","End":"03:21.620","Text":"coordinates and we can only use this equation if our E field is conservative,"},{"Start":"03:21.620 ","End":"03:27.790","Text":"and we know that the electric field of a point charge is a conservative field."},{"Start":"03:27.790 ","End":"03:33.180","Text":"Therefore, we can use this version of our equation."},{"Start":"03:33.520 ","End":"03:37.410","Text":"We can write this like so."},{"Start":"03:37.410 ","End":"03:41.685","Text":"We take negative Nabla of U,"},{"Start":"03:41.685 ","End":"03:46.470","Text":"where U is equal to negative kq divided by"},{"Start":"03:46.470 ","End":"03:52.290","Text":"r^3 P dot r. First of all,"},{"Start":"03:52.290 ","End":"03:58.750","Text":"these 2 minus signs can cancel out and we can take out these constants."},{"Start":"03:58.750 ","End":"04:07.360","Text":"What we are going to get is we are going to have kq multiplied by"},{"Start":"04:07.360 ","End":"04:17.080","Text":"Nabla(p dot r vector divided by r^3)."},{"Start":"04:18.800 ","End":"04:21.975","Text":"What is r vector?"},{"Start":"04:21.975 ","End":"04:24.735","Text":"Let\u0027s write a note over here."},{"Start":"04:24.735 ","End":"04:32.985","Text":"Our vector is equal to x in the x-direction plus y in the y-direction,"},{"Start":"04:32.985 ","End":"04:36.780","Text":"plus z in the z-direction."},{"Start":"04:36.780 ","End":"04:46.440","Text":"Therefore, we can rewrite this as kq Nabla, then in brackets."},{"Start":"04:46.440 ","End":"04:50.610","Text":"We have our x-component, P_x,"},{"Start":"04:50.610 ","End":"04:56.610","Text":"x plus P_y multiplied by y"},{"Start":"04:56.610 ","End":"05:04.750","Text":"plus P_z multiplied by z divided by r^3."},{"Start":"05:05.930 ","End":"05:08.160","Text":"The x, y,"},{"Start":"05:08.160 ","End":"05:12.730","Text":"and z hats cancel off because we have P_x"},{"Start":"05:12.730 ","End":"05:20.030","Text":"in the x-direction and then we have that dot-product with,"},{"Start":"05:20.030 ","End":"05:21.710","Text":"let\u0027s say over here,"},{"Start":"05:21.710 ","End":"05:24.330","Text":"x in the x-direction,"},{"Start":"05:24.330 ","End":"05:26.550","Text":"so these 2 will become 1."},{"Start":"05:26.550 ","End":"05:29.565","Text":"Therefore, it will have P_x X,"},{"Start":"05:29.565 ","End":"05:32.390","Text":"that\u0027s where we get that and then we can do it"},{"Start":"05:32.390 ","End":"05:35.585","Text":"for all the other elements in the equation,"},{"Start":"05:35.585 ","End":"05:38.486","Text":"and we are left with this."},{"Start":"05:38.486 ","End":"05:41.705","Text":"What I want to do is I want to write this,"},{"Start":"05:41.705 ","End":"05:45.185","Text":"what\u0027s in here in the brackets as 2 functions,"},{"Start":"05:45.185 ","End":"05:48.439","Text":"and then I will take the gradient."},{"Start":"05:48.439 ","End":"05:54.165","Text":"I\u0027m going to write this as kq Nabla,"},{"Start":"05:54.165 ","End":"06:02.445","Text":"and then my first function is going to be P_x X plus P_y Y"},{"Start":"06:02.445 ","End":"06:06.710","Text":"plus P_z Z and"},{"Start":"06:06.710 ","End":"06:12.660","Text":"this is going to be multiplied by the second function, which is r^negative 3."},{"Start":"06:16.060 ","End":"06:24.410","Text":"I\u0027m going to call this function over here, f(x,"},{"Start":"06:24.410 ","End":"06:26.320","Text":"y, z),"},{"Start":"06:26.320 ","End":"06:33.645","Text":"and this function over here will be g(x, y, z)."},{"Start":"06:33.645 ","End":"06:43.000","Text":"If we remember what our r size is, the magnitude of"},{"Start":"06:52.280 ","End":"06:56.400","Text":"r is equal to (x^2 plus y^2 plus z^2)^1/2 We can"},{"Start":"06:56.400 ","End":"06:59.640","Text":"see that r magnitude is the function of x,"},{"Start":"06:59.640 ","End":"07:03.195","Text":"y, z. Let\u0027s scroll down."},{"Start":"07:03.195 ","End":"07:11.530","Text":"What we are doing is we\u0027re taking the derivative of 2 scalar functions."},{"Start":"07:11.530 ","End":"07:13.885","Text":"There\u0027s an identity for this."},{"Start":"07:13.885 ","End":"07:16.330","Text":"I\u0027m going to write this in blue."},{"Start":"07:16.330 ","End":"07:19.915","Text":"It\u0027s useful to remember or in green rather."},{"Start":"07:19.915 ","End":"07:25.290","Text":"If we have Nabla(f dot g),"},{"Start":"07:25.290 ","End":"07:29.140","Text":"where both f and g are scalar functions like we have over here."},{"Start":"07:29.140 ","End":"07:38.670","Text":"Then we can write this as g multiplied by the gradient of f plus f"},{"Start":"07:38.670 ","End":"07:49.285","Text":"multiplied by the gradient of g. Let\u0027s see what grad f is equal to."},{"Start":"07:49.285 ","End":"07:54.715","Text":"We\u0027re taking the x derivative of all of this."},{"Start":"07:54.715 ","End":"07:57.085","Text":"What we\u0027re going to be left with,"},{"Start":"07:57.085 ","End":"07:58.975","Text":"so here this is y and z."},{"Start":"07:58.975 ","End":"08:03.265","Text":"The derivative of P_y Y and P_z Z will equal 0."},{"Start":"08:03.265 ","End":"08:08.410","Text":"We will be left with in the x direction P_x."},{"Start":"08:08.410 ","End":"08:10.690","Text":"Then we take the right derivative."},{"Start":"08:10.690 ","End":"08:13.000","Text":"Only this element here has y."},{"Start":"08:13.000 ","End":"08:17.605","Text":"We have plus P_y in the y direction,"},{"Start":"08:17.605 ","End":"08:19.660","Text":"and only this element has z,"},{"Start":"08:19.660 ","End":"08:25.610","Text":"so plus P_z in the z direction."},{"Start":"08:25.710 ","End":"08:30.655","Text":"First of all, we can see that we get this vector quantity."},{"Start":"08:30.655 ","End":"08:33.760","Text":"We\u0027ve just taken the derivative according to x,"},{"Start":"08:33.760 ","End":"08:36.250","Text":"y, and z, and this is all we have left."},{"Start":"08:36.250 ","End":"08:42.655","Text":"Notice that this is exactly equal to our p vector, a dipole moment."},{"Start":"08:42.655 ","End":"08:46.150","Text":"We have the x component of the dipole moment in the x direction,"},{"Start":"08:46.150 ","End":"08:48.340","Text":"the y component in the y direction,"},{"Start":"08:48.340 ","End":"08:51.370","Text":"and the z component of the dipole moment in the z direction,"},{"Start":"08:51.370 ","End":"08:54.350","Text":"which is exactly a p vector."},{"Start":"08:54.750 ","End":"08:58.780","Text":"Let\u0027s take the gradient of g. The gradient"},{"Start":"08:58.780 ","End":"09:02.020","Text":"of g is going to be slightly less straightforward,"},{"Start":"09:02.020 ","End":"09:07.135","Text":"g is this r^negative 3."},{"Start":"09:07.135 ","End":"09:12.710","Text":"We are taking the gradient of r^negative 3."},{"Start":"09:13.650 ","End":"09:20.185","Text":"What we can see is right now before we move into these cartesian coordinates,"},{"Start":"09:20.185 ","End":"09:24.715","Text":"we have this r which has spherical coordinates."},{"Start":"09:24.715 ","End":"09:28.000","Text":"Maybe it\u0027s better if we take the gradient of"},{"Start":"09:28.000 ","End":"09:32.750","Text":"the spherical coordinates and then we can convert into cartesian."},{"Start":"09:32.850 ","End":"09:36.430","Text":"If we take the gradient of r when"},{"Start":"09:36.430 ","End":"09:39.775","Text":"spherical coordinates and we can see that we only have r,"},{"Start":"09:39.775 ","End":"09:41.680","Text":"we don\u0027t have Theta or Phi."},{"Start":"09:41.680 ","End":"09:48.020","Text":"The gradient of just r is simply going to be d"},{"Start":"09:48.020 ","End":"09:55.130","Text":"by dr of our function which is r to the negative 3 in the r direction."},{"Start":"09:55.860 ","End":"10:02.740","Text":"This is our gradient in spherical coordinates if we just have the r coordinate."},{"Start":"10:02.740 ","End":"10:05.785","Text":"If we take the derivative of this,"},{"Start":"10:05.785 ","End":"10:14.420","Text":"we\u0027re going to be left with negative 3r^negative 4 in the r direction."},{"Start":"10:15.930 ","End":"10:26.230","Text":"We can also write this as negative 3r hat divided by r^4."},{"Start":"10:26.230 ","End":"10:34.975","Text":"If we remember, our r vector is equal to r r hat."},{"Start":"10:34.975 ","End":"10:36.850","Text":"Here we have r hat,"},{"Start":"10:36.850 ","End":"10:39.970","Text":"but I don\u0027t know what this was equal to,"},{"Start":"10:39.970 ","End":"10:43.254","Text":"but I have my r vector given to me in the question,"},{"Start":"10:43.254 ","End":"10:45.820","Text":"we can scroll up."},{"Start":"10:45.820 ","End":"10:49.960","Text":"We\u0027re being told what r vector is equal to."},{"Start":"10:49.960 ","End":"10:58.070","Text":"Therefore, I can write this as negative 3r vector divided by r^5."},{"Start":"11:01.200 ","End":"11:05.860","Text":"What we want to do is we want to put everything together."},{"Start":"11:05.860 ","End":"11:10.030","Text":"I\u0027m going back to this equation over here,"},{"Start":"11:10.030 ","End":"11:13.105","Text":"so we\u0027re carrying on from here."},{"Start":"11:13.105 ","End":"11:18.280","Text":"What we have is kq multiplied by"},{"Start":"11:18.280 ","End":"11:23.845","Text":"gradient of f dot g. We\u0027re going to be using this identity."},{"Start":"11:23.845 ","End":"11:26.230","Text":"We have kq of g,"},{"Start":"11:26.230 ","End":"11:34.585","Text":"so g is r^negative 3 or 1 divided by r^3 multiplied by grad f. As we saw,"},{"Start":"11:34.585 ","End":"11:36.790","Text":"grad f is equal to all of this,"},{"Start":"11:36.790 ","End":"11:39.550","Text":"which is just equal to a p vector,"},{"Start":"11:39.550 ","End":"11:41.290","Text":"which is equal to a dipole moment."},{"Start":"11:41.290 ","End":"11:45.130","Text":"So that\u0027s that. plus we\u0027re adding on"},{"Start":"11:45.130 ","End":"11:50.620","Text":"f multiplied by grad g. F is this function over here,"},{"Start":"11:50.620 ","End":"12:01.120","Text":"so it\u0027s P_x X plus P_y Y plus P_z Z."},{"Start":"12:01.120 ","End":"12:04.135","Text":"All of this is multiplied by grad g,"},{"Start":"12:04.135 ","End":"12:05.680","Text":"which we saw over here,"},{"Start":"12:05.680 ","End":"12:10.210","Text":"is equal to negative 3r vector divided"},{"Start":"12:10.210 ","End":"12:15.985","Text":"by r^5."},{"Start":"12:15.985 ","End":"12:17.830","Text":"All of this, what we have over here,"},{"Start":"12:17.830 ","End":"12:21.580","Text":"P_x X, P_y Y and all of this,"},{"Start":"12:21.580 ","End":"12:26.515","Text":"so our f we know is simply equal to"},{"Start":"12:26.515 ","End":"12:32.050","Text":"p dot r. We saw that over here we had p dot r divided by r^3."},{"Start":"12:32.050 ","End":"12:37.420","Text":"What we can see over here on our numerator is that this is equal to p dot r,"},{"Start":"12:37.420 ","End":"12:41.800","Text":"and this is exactly f. Our f function,"},{"Start":"12:41.800 ","End":"12:43.255","Text":"we can write it over here,"},{"Start":"12:43.255 ","End":"12:49.120","Text":"is equal to p dot r, from over here."},{"Start":"12:49.120 ","End":"12:53.515","Text":"I can write all of this as equal to p"},{"Start":"12:53.515 ","End":"13:02.770","Text":"dot r. Let\u0027s try and get this under the same denominator."},{"Start":"13:03.740 ","End":"13:06.330","Text":"Let\u0027s carry it on over here."},{"Start":"13:06.330 ","End":"13:10.620","Text":"We\u0027re going to have kq multiplying everything out."},{"Start":"13:10.620 ","End":"13:13.025","Text":"Then we have p divided by r^3,"},{"Start":"13:13.025 ","End":"13:15.300","Text":"but we want to divide it by r^5."},{"Start":"13:15.300 ","End":"13:25.240","Text":"We can write p multiplied by r^2 divided by r^5 plus,"},{"Start":"13:25.240 ","End":"13:34.960","Text":"and then we have here negative 3 multiplied by p dot r,"},{"Start":"13:34.960 ","End":"13:45.050","Text":"all of this multiplied by r vector divided by r^5."},{"Start":"13:47.790 ","End":"13:58.449","Text":"We can just join this up together and what we get is kq multiplied by p vector multiplied"},{"Start":"13:58.449 ","End":"14:06.850","Text":"by r^2 minus 3p vector dot r vector"},{"Start":"14:06.850 ","End":"14:14.180","Text":"multiplied by r vector and all of this divided by r^5."},{"Start":"14:15.060 ","End":"14:17.170","Text":"This is what we get,"},{"Start":"14:17.170 ","End":"14:20.425","Text":"which is equal to the force."},{"Start":"14:20.425 ","End":"14:22.990","Text":"If we scroll back up,"},{"Start":"14:22.990 ","End":"14:24.760","Text":"remember what we have written here."},{"Start":"14:24.760 ","End":"14:27.040","Text":"If we scroll back up,"},{"Start":"14:27.040 ","End":"14:32.810","Text":"we see that it\u0027s the exact same equation that we got over here."},{"Start":"14:33.090 ","End":"14:39.620","Text":"Just notice this equation was missing a 3 over here."},{"Start":"14:39.990 ","End":"14:42.925","Text":"We saw this way, it\u0027s a bit long,"},{"Start":"14:42.925 ","End":"14:46.720","Text":"but let\u0027s look at another way that we could solve it."},{"Start":"14:46.720 ","End":"14:52.570","Text":"Here we have our diagram and we can see that the force being applied to"},{"Start":"14:52.570 ","End":"14:55.540","Text":"the dipole is going to be equal and opposite to"},{"Start":"14:55.540 ","End":"14:59.650","Text":"the force that the dipole is applying to the charge."},{"Start":"14:59.650 ","End":"15:08.964","Text":"This was our first way of showing the force."},{"Start":"15:08.964 ","End":"15:14.245","Text":"Now let\u0027s look at the second way."},{"Start":"15:14.245 ","End":"15:23.035","Text":"We know that the force of the charge on the dipole,"},{"Start":"15:23.035 ","End":"15:24.580","Text":"what we just said,"},{"Start":"15:24.580 ","End":"15:26.980","Text":"is going to be equal and opposite,"},{"Start":"15:26.980 ","End":"15:35.540","Text":"here we have a minus to the force of the dipole on the charge."},{"Start":"15:37.080 ","End":"15:45.545","Text":"The force of the dipole on the charge is just like regular Coulomb\u0027s law."},{"Start":"15:45.545 ","End":"15:48.185","Text":"That\u0027s going to be equal to negative."},{"Start":"15:48.185 ","End":"15:57.780","Text":"Then we have q multiplied by the electric field caused by the dipole."},{"Start":"15:59.410 ","End":"16:10.355","Text":"In that case we have here negative q multiplied by the electric field of the dipole."},{"Start":"16:10.355 ","End":"16:20.350","Text":"If you remember that was k multiplied by 3p dot r in"},{"Start":"16:20.350 ","End":"16:25.465","Text":"the r vector direction minus p dot r"},{"Start":"16:25.465 ","End":"16:32.525","Text":"squared and all of this is divided by r^5."},{"Start":"16:32.525 ","End":"16:38.395","Text":"We can see that we have the exact same answer that we got over here."},{"Start":"16:38.395 ","End":"16:43.330","Text":"We just have to play around a little bit with the brackets, but it\u0027s the same thing."},{"Start":"16:43.330 ","End":"16:46.870","Text":"The only difference is that we have this minus over here."},{"Start":"16:46.870 ","End":"16:53.440","Text":"Notice here we\u0027re working out the electric field of the dipole on the charge."},{"Start":"16:53.440 ","End":"16:59.995","Text":"That means the electric field and the dipole on the charge is going in this direction,"},{"Start":"16:59.995 ","End":"17:07.515","Text":"which is in the negative direction to the electric field from the charge to the dipole."},{"Start":"17:07.515 ","End":"17:11.710","Text":"We can see that our radius is in the opposite direction."},{"Start":"17:11.710 ","End":"17:15.830","Text":"If we substitute n over here minus r will"},{"Start":"17:15.830 ","End":"17:20.480","Text":"get the exact same equation to what we had over here."},{"Start":"17:20.480 ","End":"17:23.300","Text":"This minus will cancel out with"},{"Start":"17:23.300 ","End":"17:27.570","Text":"minus r because our radius is pointing in the opposite direction."},{"Start":"17:27.570 ","End":"17:32.240","Text":"This is the 2nd method for solving this question and"},{"Start":"17:32.240 ","End":"17:38.345","Text":"showing that this is the force of a dipole interacting with a point charge."},{"Start":"17:38.345 ","End":"17:41.280","Text":"That\u0027s the end of this lesson."}],"ID":21377},{"Watched":false,"Name":"Force Equation Proof","Duration":"12m 56s","ChapterTopicVideoID":21448,"CourseChapterTopicPlaylistID":99476,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.310","Text":"Hello. In this lesson we\u0027re going to be looking at the proof for the force equation,"},{"Start":"00:05.310 ","End":"00:07.695","Text":"so I\u0027m reminding you it\u0027s the force equation"},{"Start":"00:07.695 ","End":"00:12.615","Text":"relating the force on a dipole when it\u0027s placed inside an E field."},{"Start":"00:12.615 ","End":"00:16.650","Text":"If you remember the equation goes like this,"},{"Start":"00:16.650 ","End":"00:25.660","Text":"p dot Nabla multiplied by the E field."},{"Start":"00:26.270 ","End":"00:34.046","Text":"Let\u0027s say that we have some axes going like so."},{"Start":"00:34.046 ","End":"00:37.640","Text":"We have a dipole placed somewhere in here,"},{"Start":"00:37.640 ","End":"00:39.260","Text":"so I\u0027m not going to draw the dipole moment."},{"Start":"00:39.260 ","End":"00:44.240","Text":"I\u0027m just going to draw the 2 charges that make up the dipole,"},{"Start":"00:44.240 ","End":"00:48.140","Text":"so like this and we have this vector"},{"Start":"00:48.140 ","End":"00:53.225","Text":"d. Remember it always goes from the negative to the positive."},{"Start":"00:53.225 ","End":"00:57.560","Text":"We know that the dipole moment p is given"},{"Start":"00:57.560 ","End":"01:01.850","Text":"by q which is the absolute value of 1 of the charges."},{"Start":"01:01.850 ","End":"01:04.595","Text":"Q multiplied by d,"},{"Start":"01:04.595 ","End":"01:09.210","Text":"the vector distance between the 2 charges."},{"Start":"01:09.740 ","End":"01:15.880","Text":"We can say that in this direction over here"},{"Start":"01:15.880 ","End":"01:21.350","Text":"we have an E field as a function of this vector r,"},{"Start":"01:21.350 ","End":"01:26.765","Text":"and over here on the positive charge we have a different E field which"},{"Start":"01:26.765 ","End":"01:33.580","Text":"is E as a function of our r vector plus d,"},{"Start":"01:33.580 ","End":"01:38.575","Text":"this distance between the 2 charges."},{"Start":"01:38.575 ","End":"01:42.440","Text":"Of course our r vector is going from"},{"Start":"01:42.440 ","End":"01:50.023","Text":"the origin up until the negative charge,"},{"Start":"01:50.023 ","End":"01:57.780","Text":"and then of course the r plus d vector is the vector that points to the positive charge."},{"Start":"01:58.300 ","End":"02:03.030","Text":"Let\u0027s write out the force equation."},{"Start":"02:03.320 ","End":"02:07.164","Text":"We have the sum of all of the forces or in other words"},{"Start":"02:07.164 ","End":"02:13.645","Text":"the total force is going to be equal to the force on the positive charge,"},{"Start":"02:13.645 ","End":"02:22.170","Text":"so F plus, plus the force on the negative charge, so f minus."},{"Start":"02:22.640 ","End":"02:25.880","Text":"As we know the force,"},{"Start":"02:25.880 ","End":"02:33.410","Text":"we learned that the force is equal to q multiplied by the electric field."},{"Start":"02:33.410 ","End":"02:38.780","Text":"The force on the positive charge is its charge q which"},{"Start":"02:38.780 ","End":"02:43.597","Text":"is a positive charge q multiplied by the E field acting on it"},{"Start":"02:43.597 ","End":"02:46.970","Text":"which as we know we\u0027re given that it is equal"},{"Start":"02:46.970 ","End":"02:54.965","Text":"to the E field at r plus D at this position over here."},{"Start":"02:54.965 ","End":"02:58.190","Text":"Then let\u0027s add in the force for the negative charge,"},{"Start":"02:58.190 ","End":"03:00.915","Text":"so that\u0027s going to be the charge."},{"Start":"03:00.915 ","End":"03:03.770","Text":"Here we have a negative charge of minus q,"},{"Start":"03:03.770 ","End":"03:11.750","Text":"so we have minus q multiplied by the E field on the negative charge which is E at"},{"Start":"03:11.750 ","End":"03:21.110","Text":"point r. We can just rewrite this as q,"},{"Start":"03:21.110 ","End":"03:31.365","Text":"and then in brackets we have E at point r plus d minus E"},{"Start":"03:31.365 ","End":"03:39.630","Text":"at point r. We"},{"Start":"03:39.630 ","End":"03:48.834","Text":"know that our p vector is equal to q multiplied by d vector."},{"Start":"03:48.834 ","End":"03:59.090","Text":"Our r vector is equal to x in the x-direction plus y in the y-direction,"},{"Start":"03:59.090 ","End":"04:04.355","Text":"plus z in the z-direction and our d vector,"},{"Start":"04:04.355 ","End":"04:08.780","Text":"so vector between the 2 charges is equal to d_x."},{"Start":"04:08.780 ","End":"04:13.890","Text":"Some value d_x in the x-direction,"},{"Start":"04:13.890 ","End":"04:16.815","Text":"so that\u0027s the x-component and the y-component is d_y"},{"Start":"04:16.815 ","End":"04:24.970","Text":"In the y-direction and some d_z in the z-direction."},{"Start":"04:25.460 ","End":"04:31.760","Text":"In that case, we can substitute in all of this into this equation."},{"Start":"04:31.760 ","End":"04:38.090","Text":"Let\u0027s start writing it a bit more to this direction."},{"Start":"04:38.090 ","End":"04:42.710","Text":"We have q and then we have multiplied by"},{"Start":"04:42.710 ","End":"04:49.335","Text":"the E field at point r plus d. Point r is this plus d,"},{"Start":"04:49.335 ","End":"04:57.140","Text":"so we can write this as x plus d_x is the x-component,"},{"Start":"04:57.140 ","End":"05:00.830","Text":"y plus d_y is the y-component,"},{"Start":"05:00.830 ","End":"05:05.885","Text":"and z plus d_z is the z-component."},{"Start":"05:05.885 ","End":"05:12.185","Text":"That\u0027s r plus d minus E at point r,"},{"Start":"05:12.185 ","End":"05:17.220","Text":"so that is just x, y, z."},{"Start":"05:19.490 ","End":"05:27.970","Text":"Let\u0027s assume that I want the sum of all of the forces in the x-direction."},{"Start":"05:27.970 ","End":"05:31.600","Text":"That\u0027s going to be equal to q multiplied by"},{"Start":"05:31.600 ","End":"05:36.795","Text":"the x-component of my E field which is at this point."},{"Start":"05:36.795 ","End":"05:42.715","Text":"The x-component of the E field at this point will just be E_x(x plus d_x,"},{"Start":"05:42.715 ","End":"05:45.010","Text":"y plus d_y,"},{"Start":"05:45.010 ","End":"05:52.470","Text":"z plus d_z) where of course this x,"},{"Start":"05:52.470 ","End":"05:56.700","Text":"y, and this z are of course subscripts."},{"Start":"05:56.700 ","End":"06:00.120","Text":"We\u0027re not doing derivatives over here,"},{"Start":"06:00.120 ","End":"06:03.705","Text":"they are subscripts for this d vector."},{"Start":"06:03.705 ","End":"06:06.450","Text":"D_x, d_y, d_z that\u0027s what we have here."},{"Start":"06:06.450 ","End":"06:13.190","Text":"We have this minus the x-component of this E field at this point x,"},{"Start":"06:13.190 ","End":"06:20.650","Text":"y z, so minus E_x at point x, y, z."},{"Start":"06:22.040 ","End":"06:31.585","Text":"If we assume that our d vector is very small,"},{"Start":"06:31.585 ","End":"06:34.490","Text":"so then we can do the following."},{"Start":"06:34.490 ","End":"06:38.300","Text":"I\u0027m not going to prove how we move from this step to"},{"Start":"06:38.300 ","End":"06:42.380","Text":"this step because it\u0027s a bit complicated,"},{"Start":"06:42.380 ","End":"06:44.585","Text":"but let\u0027s just take it as it is."},{"Start":"06:44.585 ","End":"06:52.895","Text":"We can write that q is equal to the partial derivative of the x-component of the E field."},{"Start":"06:52.895 ","End":"06:59.945","Text":"This is the sum of the forces in the x-direction by d_x,"},{"Start":"06:59.945 ","End":"07:07.830","Text":"d_x plus the partial derivative of E_x according to y d_y"},{"Start":"07:07.830 ","End":"07:17.580","Text":"plus the partial derivative of the x according to z d_z."},{"Start":"07:17.580 ","End":"07:23.039","Text":"How we got from all of this"},{"Start":"07:23.039 ","End":"07:30.600","Text":"to this requires a proof that we\u0027re not going to look at now."},{"Start":"07:31.970 ","End":"07:34.655","Text":"Before we move on from here,"},{"Start":"07:34.655 ","End":"07:37.505","Text":"let\u0027s just take a look at our p vector,"},{"Start":"07:37.505 ","End":"07:42.545","Text":"so it\u0027s q multiplied by our d vector. What does it mean?"},{"Start":"07:42.545 ","End":"07:44.960","Text":"It\u0027s q and our d vector,"},{"Start":"07:44.960 ","End":"07:50.345","Text":"so it\u0027s q multiplied by d_x in the x-direction"},{"Start":"07:50.345 ","End":"07:56.030","Text":"plus q multiplied by d_y in the y-direction,"},{"Start":"07:56.030 ","End":"08:02.940","Text":"plus q multiplied by d_z in the z-direction."},{"Start":"08:02.940 ","End":"08:06.615","Text":"This is what our p vector is,"},{"Start":"08:06.615 ","End":"08:09.825","Text":"so now let\u0027s carry on over here."},{"Start":"08:09.825 ","End":"08:17.645","Text":"What we get is that we have q and then we can take out this constant over here."},{"Start":"08:17.645 ","End":"08:27.305","Text":"We have q d_x multiplied by the partial derivative of E_x according to x plus q,"},{"Start":"08:27.305 ","End":"08:28.626","Text":"and this is a constant."},{"Start":"08:28.626 ","End":"08:33.585","Text":"So d_y multiplied by the partial derivative of E_x"},{"Start":"08:33.585 ","End":"08:39.945","Text":"according to y plus q. D_z is also a constant."},{"Start":"08:39.945 ","End":"08:42.860","Text":"D_z, d_y, and, d_x,"},{"Start":"08:42.860 ","End":"08:47.180","Text":"I\u0027m reminding are you just directions that we move in"},{"Start":"08:47.180 ","End":"08:51.080","Text":"the different directions to show this vector"},{"Start":"08:51.080 ","End":"08:55.400","Text":"the distance between these 2 charged particles."},{"Start":"08:55.400 ","End":"08:59.840","Text":"This distance is a constant distance multiplied"},{"Start":"08:59.840 ","End":"09:05.470","Text":"by the partial derivative of E_x with respect to z."},{"Start":"09:05.810 ","End":"09:10.200","Text":"What we have over here is we have qd_x,"},{"Start":"09:10.440 ","End":"09:15.750","Text":"so here we have qd_x in the x-direction."},{"Start":"09:15.750 ","End":"09:23.393","Text":"Qd_x in the x-direction is simply our x-component of our p,"},{"Start":"09:23.393 ","End":"09:26.325","Text":"so let\u0027s write this out."},{"Start":"09:26.325 ","End":"09:30.870","Text":"This over here is our p_x."},{"Start":"09:30.870 ","End":"09:38.795","Text":"Qd_y is simply our y-component of our p. Qd_y is P_y,"},{"Start":"09:38.795 ","End":"09:44.660","Text":"and qd_z is simply our z-component of our"},{"Start":"09:44.660 ","End":"09:51.360","Text":"P. Let\u0027s take a look at what we had,"},{"Start":"09:51.360 ","End":"09:54.100","Text":"so let\u0027s scroll down even more."},{"Start":"09:54.100 ","End":"09:57.395","Text":"What we see is that we have P_x, P_y,"},{"Start":"09:57.395 ","End":"10:01.880","Text":"P_z which is simply our P vector;"},{"Start":"10:01.880 ","End":"10:07.540","Text":"dot product with our Nabla function,"},{"Start":"10:07.540 ","End":"10:11.535","Text":"and that\u0027s what gives us this d by d_x, d by d_y,"},{"Start":"10:11.535 ","End":"10:16.665","Text":"d by d_z, and then all of this with"},{"Start":"10:16.665 ","End":"10:23.155","Text":"our E_x because we\u0027re taking the partial derivative of our E_x each time."},{"Start":"10:23.155 ","End":"10:28.025","Text":"Let\u0027s just show what this is."},{"Start":"10:28.025 ","End":"10:33.620","Text":"P dot Nabla, so p is P_x in the x-direction,"},{"Start":"10:33.620 ","End":"10:35.705","Text":"P_y in the y,"},{"Start":"10:35.705 ","End":"10:38.435","Text":"and P_z in the z-direction."},{"Start":"10:38.435 ","End":"10:45.058","Text":"Dot-product with the Nabla function."},{"Start":"10:45.058 ","End":"10:48.500","Text":"The x-component of Nabla is d by d_x,"},{"Start":"10:48.500 ","End":"10:51.080","Text":"the y-component is d by d_y,"},{"Start":"10:51.080 ","End":"10:55.510","Text":"and the z-component is d by dz."},{"Start":"10:55.510 ","End":"11:05.530","Text":"If we do the dot-product we\u0027re going to have P_x with d by d_x and then E_x over here,"},{"Start":"11:05.530 ","End":"11:07.935","Text":"that\u0027s what we have."},{"Start":"11:07.935 ","End":"11:12.900","Text":"Then we\u0027ll have P_y d by d_y,"},{"Start":"11:12.900 ","End":"11:16.275","Text":"so P_y d by d_y (E_x),"},{"Start":"11:16.275 ","End":"11:18.930","Text":"and then P_z d by d_z,"},{"Start":"11:18.930 ","End":"11:22.890","Text":"so P_z, d by d_z E_x."},{"Start":"11:22.890 ","End":"11:28.540","Text":"This is what we have over here."},{"Start":"11:29.780 ","End":"11:36.535","Text":"We see that this is the x-component of the force acting,"},{"Start":"11:36.535 ","End":"11:40.790","Text":"so therefore if we want the force acting in the x,"},{"Start":"11:40.790 ","End":"11:42.589","Text":"y, and z direction;,"},{"Start":"11:42.589 ","End":"11:46.590","Text":"so that is the sum of all of the forces,"},{"Start":"11:46.590 ","End":"11:56.610","Text":"that\u0027s just going to be equal to P dot Nabla and this will be multiplied by"},{"Start":"11:56.610 ","End":"12:00.455","Text":"E_x in the x-direction plus"},{"Start":"12:00.455 ","End":"12:09.065","Text":"p dot Nabla multiplied by E_y in the y-direction,"},{"Start":"12:09.065 ","End":"12:19.720","Text":"plus P dot Nabla multiplied by E_z in the z-direction."},{"Start":"12:20.210 ","End":"12:27.330","Text":"We get these next 2 terms just how we did with our x term."},{"Start":"12:27.330 ","End":"12:32.075","Text":"Then in other words we can write all over this simply as"},{"Start":"12:32.075 ","End":"12:38.708","Text":"P dot Nabla multiplied by our entire E vector,"},{"Start":"12:38.708 ","End":"12:41.110","Text":"and this E vector we have the x,"},{"Start":"12:41.110 ","End":"12:44.225","Text":"y, and z-components."},{"Start":"12:44.225 ","End":"12:49.160","Text":"This is the exact equation that we wanted"},{"Start":"12:49.160 ","End":"12:53.420","Text":"to prove for the force on a dipole in an E field,"},{"Start":"12:53.420 ","End":"12:56.340","Text":"and that is the end of this lesson."}],"ID":22334},{"Watched":false,"Name":"Exercise 5","Duration":"1h 1m 33s","ChapterTopicVideoID":21449,"CourseChapterTopicPlaylistID":99476,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.920","Text":"Hello. In this lesson,"},{"Start":"00:01.920 ","End":"00:04.545","Text":"we\u0027re going to be answering the following question."},{"Start":"00:04.545 ","End":"00:07.830","Text":"We\u0027re being told that we have dipole p_1,"},{"Start":"00:07.830 ","End":"00:12.180","Text":"which is located at this position r_1 over here."},{"Start":"00:12.180 ","End":"00:18.505","Text":"We have dipole p_2 located at this position r_2 over here."},{"Start":"00:18.505 ","End":"00:21.584","Text":"Question number 1, which is what we\u0027ll start with,"},{"Start":"00:21.584 ","End":"00:25.095","Text":"is to show that the energy of p_2,"},{"Start":"00:25.095 ","End":"00:31.485","Text":"when it\u0027s placed in the field of p_1 is equal to this expression over here."},{"Start":"00:31.485 ","End":"00:34.261","Text":"We can see it\u0027s a little bit of a complicated expression."},{"Start":"00:34.261 ","End":"00:36.915","Text":"We\u0027ll go over it right now."},{"Start":"00:36.915 ","End":"00:48.180","Text":"We can see that we have this r hat vector that has a wave or a Tilda above it."},{"Start":"00:48.520 ","End":"00:52.549","Text":"What we can see is when it\u0027s in vector form,"},{"Start":"00:52.549 ","End":"00:59.610","Text":"so this r vector Tilda is equal to r_2 minus r_1."},{"Start":"00:59.610 ","End":"01:06.515","Text":"This vector goes from here to here in this direction,"},{"Start":"01:06.515 ","End":"01:11.520","Text":"and this is our r Tilda vector."},{"Start":"01:11.590 ","End":"01:18.045","Text":"The size or the magnitude of our r Tilda vector is this,"},{"Start":"01:18.045 ","End":"01:26.180","Text":"and the direction, the unit vector for it is our r Tilda vector divided by the magnitude."},{"Start":"01:26.180 ","End":"01:30.570","Text":"This is r hat Tilda."},{"Start":"01:31.250 ","End":"01:37.220","Text":"What we\u0027re trying to prove is that the energy of p_2,"},{"Start":"01:37.220 ","End":"01:42.770","Text":"when it\u0027s in the field of p_1 is equal to k divided by the magnitude of"},{"Start":"01:42.770 ","End":"01:50.495","Text":"our r Tilda cubed multiplied by p_1.p_2 minus 3 times p_1."},{"Start":"01:50.495 ","End":"01:54.675","Text":"r hat Tilda multiplied by p_2."},{"Start":"01:54.675 ","End":"01:56.505","Text":"r hat Tilda."},{"Start":"01:56.505 ","End":"01:59.479","Text":"Here what we have are square brackets."},{"Start":"01:59.479 ","End":"02:02.090","Text":"For some reason we can\u0027t see the top."},{"Start":"02:02.090 ","End":"02:05.840","Text":"Let\u0027s begin answering question number 1."},{"Start":"02:05.840 ","End":"02:09.215","Text":"First of all, we know that the energy is going to"},{"Start":"02:09.215 ","End":"02:13.015","Text":"be equal to what we\u0027ve seen before, negative P.E."},{"Start":"02:13.015 ","End":"02:16.490","Text":"This is the general equation for the energy."},{"Start":"02:16.490 ","End":"02:18.110","Text":"What we want this to be,"},{"Start":"02:18.110 ","End":"02:19.580","Text":"is equal to negative p,"},{"Start":"02:19.580 ","End":"02:21.215","Text":"so which p are we looking at?"},{"Start":"02:21.215 ","End":"02:23.630","Text":"We\u0027re looking for the energy of p_2."},{"Start":"02:23.630 ","End":"02:29.375","Text":"Negative p_2.E."},{"Start":"02:29.375 ","End":"02:31.940","Text":"Which E are we looking at?"},{"Start":"02:31.940 ","End":"02:34.895","Text":"When p_2 is in the field of p_1,"},{"Start":"02:34.895 ","End":"02:42.195","Text":"so we\u0027re looking at the electric field that p_1 exerts on p_2."},{"Start":"02:42.195 ","End":"02:47.185","Text":"Now, let\u0027s write out the electric field equation."},{"Start":"02:47.185 ","End":"02:49.690","Text":"Of course this is the electric field of a dipole,"},{"Start":"02:49.690 ","End":"02:51.684","Text":"so we\u0027ve already seen this equation."},{"Start":"02:51.684 ","End":"02:57.500","Text":"We have negative p_2 multiplied by the E field."},{"Start":"02:57.500 ","End":"03:02.740","Text":"For our dipole, and because we\u0027re looking at the field of p_1,"},{"Start":"03:02.740 ","End":"03:05.050","Text":"so that\u0027s what we\u0027re going to do."},{"Start":"03:05.050 ","End":"03:10.825","Text":"The electric field of dipole p_1=k multiplied by"},{"Start":"03:10.825 ","End":"03:17.100","Text":"3 multiplied by p_1 vector.r_i vector."},{"Start":"03:17.100 ","End":"03:22.350","Text":"Usually we\u0027re going from the origin if"},{"Start":"03:22.350 ","End":"03:24.620","Text":"our dipole is located at the origin to"},{"Start":"03:24.620 ","End":"03:28.159","Text":"some r in space where we\u0027re measuring the electric field."},{"Start":"03:28.159 ","End":"03:33.990","Text":"However, here our p_1 is located over here,"},{"Start":"03:34.040 ","End":"03:38.950","Text":"and we\u0027re measuring the electric field over here at p_2."},{"Start":"03:38.950 ","End":"03:46.085","Text":"We know that the vector going between these 2 points is r i Tilda vector."},{"Start":"03:46.085 ","End":"03:50.959","Text":"Here specifically, we\u0027re using the r hat version,"},{"Start":"03:50.959 ","End":"03:58.545","Text":"so it\u0027s going to be r Tilda hat or r hat Tilda,"},{"Start":"03:58.545 ","End":"04:00.545","Text":"that\u0027s how we wrote it over here."},{"Start":"04:00.545 ","End":"04:07.895","Text":"Then this is going to be in the r hat Tilda direction,"},{"Start":"04:07.895 ","End":"04:11.645","Text":"and then minus our p_1 vector."},{"Start":"04:11.645 ","End":"04:17.070","Text":"All of these is divided by r Tilda cubed."},{"Start":"04:17.070 ","End":"04:21.760","Text":"The magnitude of our r Tilda vector cubed."},{"Start":"04:23.000 ","End":"04:29.025","Text":"Now we just have to do some rearranging and we\u0027ve proven this."},{"Start":"04:29.025 ","End":"04:35.355","Text":"We can take our k outside and we can put our p_2 in here."},{"Start":"04:35.355 ","End":"04:37.860","Text":"Here we have a minus and here we have a minus,"},{"Start":"04:37.860 ","End":"04:40.730","Text":"so what we can do is have this as a plus."},{"Start":"04:40.730 ","End":"04:44.180","Text":"Then here we\u0027ll have minus and here we\u0027ll have plus."},{"Start":"04:44.180 ","End":"04:49.880","Text":"Let\u0027s do p_2 dot product with this p_1 over here."},{"Start":"04:49.880 ","End":"04:53.780","Text":"Or we can just write p_1.p_2,"},{"Start":"04:53.780 ","End":"04:56.035","Text":"it means the same thing."},{"Start":"04:56.035 ","End":"05:04.970","Text":"Then we have negative 3 and then we have p_1. r hat."},{"Start":"05:04.970 ","End":"05:12.410","Text":"P_1.r hat Tilda, dot,"},{"Start":"05:12.410 ","End":"05:22.090","Text":"then here we have.p_2 dot this r hat Tilda."},{"Start":"05:22.090 ","End":"05:27.280","Text":"Then all of this is divided by r^3."},{"Start":"05:27.350 ","End":"05:34.109","Text":"Then we got k divided by this r Tilda cubed and then we have p_1.p_2"},{"Start":"05:34.109 ","End":"05:41.355","Text":"minus 3 multiplied by p_1.r hat multiplied by p_2.r hat."},{"Start":"05:41.355 ","End":"05:45.450","Text":"Now we have the answer to Question 1."},{"Start":"05:46.300 ","End":"05:53.260","Text":"Question number 2 is saying that this energy that we got,"},{"Start":"05:53.260 ","End":"05:55.969","Text":"or that we proved, is the energy of a dipole,"},{"Start":"05:55.969 ","End":"05:58.865","Text":"dipole system, so a system with 2 dipoles."},{"Start":"05:58.865 ","End":"06:05.540","Text":"Show that if we were to calculate the energy this time of p_1 in the field of p_2,"},{"Start":"06:05.540 ","End":"06:08.345","Text":"so the opposite of what we did in the previous question,"},{"Start":"06:08.345 ","End":"06:12.180","Text":"we would get the exact same answer."},{"Start":"06:12.920 ","End":"06:19.294","Text":"What this question is trying to show is that the energy that we got in question number"},{"Start":"06:19.294 ","End":"06:25.810","Text":"1 is not the energy just of p_2 in the field of p_1,"},{"Start":"06:25.810 ","End":"06:32.120","Text":"but rather it\u0027s the energy of the system of 2 dipoles."},{"Start":"06:32.120 ","End":"06:39.839","Text":"What this means is that if over here in the first question our p_1 was fixed and was"},{"Start":"06:39.839 ","End":"06:48.710","Text":"causing some electric field that p_2 felt and p_2 is free to move in any direction,"},{"Start":"06:48.710 ","End":"06:52.340","Text":"or rather to rotate in any direction, then as we know,"},{"Start":"06:52.340 ","End":"06:57.679","Text":"this dipole is going to want to lineup to be parallel to the electric field,"},{"Start":"06:57.679 ","End":"07:02.380","Text":"which has a fixed electric field because we\u0027re holding p_1."},{"Start":"07:02.380 ","End":"07:09.515","Text":"If we would hold the dipole p_2 such that it is exerting a fixed electric field,"},{"Start":"07:09.515 ","End":"07:11.480","Text":"then our p_1 dipole,"},{"Start":"07:11.480 ","End":"07:15.335","Text":"which is free this time to rotate any which direction at once,"},{"Start":"07:15.335 ","End":"07:18.380","Text":"will rotate itself in order to line up"},{"Start":"07:18.380 ","End":"07:22.230","Text":"with the electric fields that is being exerted by p_2."},{"Start":"07:22.430 ","End":"07:25.880","Text":"What we can see is that it\u0027s the energy of the system."},{"Start":"07:25.880 ","End":"07:31.580","Text":"If I would have both p_1 and p_2 freely being able to rotate,"},{"Start":"07:31.580 ","End":"07:34.850","Text":"so what we will see is that both will still"},{"Start":"07:34.850 ","End":"07:38.915","Text":"rotate to line up with the electric field of one or the other,"},{"Start":"07:38.915 ","End":"07:41.955","Text":"but each one will rotate a bit less."},{"Start":"07:41.955 ","End":"07:45.270","Text":"Because p_1 will align slightly with p_2,"},{"Start":"07:45.270 ","End":"07:49.055","Text":"whilst p_2 is aligning simultaneously with p_1."},{"Start":"07:49.055 ","End":"07:52.340","Text":"Eventually, they will line up with one another and"},{"Start":"07:52.340 ","End":"07:57.020","Text":"each one will have to move or rotate slightly less because"},{"Start":"07:57.020 ","End":"08:02.330","Text":"the energy of the system will be split between the 2 dipoles rather"},{"Start":"08:02.330 ","End":"08:08.510","Text":"than in the first case where it\u0027s split between just 1 dipole p_2 being able to move."},{"Start":"08:08.510 ","End":"08:11.735","Text":"That\u0027s why it\u0027s the energy of the system."},{"Start":"08:11.735 ","End":"08:13.205","Text":"In Question 2,"},{"Start":"08:13.205 ","End":"08:17.824","Text":"we\u0027re showing that the energy of the system is still the same energy"},{"Start":"08:17.824 ","End":"08:23.690","Text":"if this time p_2 is held fixed and p_1 can rotate to line up."},{"Start":"08:23.690 ","End":"08:30.970","Text":"If both of them were allowed to rotate due to each of the other\u0027s electric field,"},{"Start":"08:30.970 ","End":"08:33.125","Text":"we\u0027ll still have the same energy,"},{"Start":"08:33.125 ","End":"08:35.360","Text":"but it will be split between 2 dipoles."},{"Start":"08:35.360 ","End":"08:38.280","Text":"Each dipole will rotate less."},{"Start":"08:38.360 ","End":"08:42.950","Text":"This question, we would do it exactly like in the previous question."},{"Start":"08:42.950 ","End":"08:45.575","Text":"We have that u = negative PE."},{"Start":"08:45.575 ","End":"08:48.170","Text":"This time our p is p_1,"},{"Start":"08:48.170 ","End":"08:51.320","Text":"so negative p_1 multiplied by E,"},{"Start":"08:51.320 ","End":"08:58.050","Text":"but this time it\u0027s E of our dipole p_2 on our dipole p_1."},{"Start":"08:58.050 ","End":"09:00.135","Text":"Then we\u0027ll have negative."},{"Start":"09:00.135 ","End":"09:06.390","Text":"Then wherever we see p_2 we put p_1 and wherever we see p_1 we put p_2."},{"Start":"09:06.390 ","End":"09:11.105","Text":"I\u0027ll have negative p_1 multiplied by,"},{"Start":"09:11.105 ","End":"09:16.195","Text":"so then we\u0027ll have k multiplied by 3,"},{"Start":"09:16.195 ","End":"09:20.325","Text":"and then we have p_2 dot."},{"Start":"09:20.325 ","End":"09:22.650","Text":"Now, here we have to look."},{"Start":"09:22.650 ","End":"09:26.205","Text":"Here we were going from 1-2,"},{"Start":"09:26.205 ","End":"09:27.970","Text":"the electric field from 1-2,"},{"Start":"09:27.970 ","End":"09:30.455","Text":"which is in this I vector direction."},{"Start":"09:30.455 ","End":"09:34.835","Text":"However, now we\u0027re looking at the electric field from 2-1."},{"Start":"09:34.835 ","End":"09:37.114","Text":"We\u0027re going in the negative r direction."},{"Start":"09:37.114 ","End":"09:39.915","Text":"We\u0027re going in this direction."},{"Start":"09:39.915 ","End":"09:46.083","Text":"This will be dot-product with negative r hat Tilda."},{"Start":"09:46.083 ","End":"09:48.670","Text":"Then dot product again,"},{"Start":"09:48.670 ","End":"09:52.880","Text":"so we have the negative r hat tilde over here,"},{"Start":"09:52.880 ","End":"10:01.135","Text":"because again we\u0027re going in the opposite direction plus p_2 vector."},{"Start":"10:01.135 ","End":"10:09.849","Text":"All of this is divided by r tilde cubed."},{"Start":"10:09.849 ","End":"10:11.230","Text":"Here there\u0027s no minus,"},{"Start":"10:11.230 ","End":"10:17.185","Text":"because r tilde is simply the magnitude of r tilde vector."},{"Start":"10:17.185 ","End":"10:21.805","Text":"If it\u0027s magnituded, obviously doesn\u0027t have direction."},{"Start":"10:21.805 ","End":"10:25.195","Text":"This is our equation."},{"Start":"10:25.195 ","End":"10:31.030","Text":"Again, we\u0027ll see that this minus and this minus cancel out."},{"Start":"10:31.030 ","End":"10:38.530","Text":"Then again, we just multiply what\u0027s in here by negative 1 so then we\u0027ll get here plus,"},{"Start":"10:38.530 ","End":"10:39.819","Text":"here negative 3,"},{"Start":"10:39.819 ","End":"10:42.459","Text":"and here this plus over here."},{"Start":"10:42.459 ","End":"10:44.482","Text":"Before it was meant to be a minus,"},{"Start":"10:44.482 ","End":"10:46.829","Text":"but now it\u0027s a plus because I multiplied over here,"},{"Start":"10:46.829 ","End":"10:52.459","Text":"and then we just play around and we\u0027ll get this exact same answer."},{"Start":"10:53.310 ","End":"10:58.105","Text":"Now, let\u0027s move on to question number 3."},{"Start":"10:58.105 ","End":"10:59.500","Text":"Question number 3 is,"},{"Start":"10:59.500 ","End":"11:05.080","Text":"what is the force acting on p_2 and p_1?"},{"Start":"11:05.080 ","End":"11:10.255","Text":"First, let\u0027s work out the force on p_2, so F_2."},{"Start":"11:10.255 ","End":"11:14.814","Text":"We saw that 1 version for this equation is taking"},{"Start":"11:14.814 ","End":"11:19.975","Text":"the negative gradient of the energy, u."},{"Start":"11:19.975 ","End":"11:24.264","Text":"We saw that we can only use this version of the equation,"},{"Start":"11:24.264 ","End":"11:27.640","Text":"if we have 2 conditions that are met."},{"Start":"11:27.640 ","End":"11:31.824","Text":"1, our dipole moment is independent"},{"Start":"11:31.824 ","End":"11:36.564","Text":"of coordinate system or independent of its position in space."},{"Start":"11:36.564 ","End":"11:43.825","Text":"We have no reason for us to believe that it is dependent on its position in space."},{"Start":"11:43.825 ","End":"11:48.610","Text":"The second thing is that the E field is a conservative field."},{"Start":"11:48.610 ","End":"11:53.312","Text":"We can see that the E field is being caused by either 1 of the dipole,"},{"Start":"11:53.312 ","End":"11:59.485","Text":"so here specifically on F_2 what\u0027s been caused by p_1, dipole number 1."},{"Start":"11:59.485 ","End":"12:02.680","Text":"A dipole is simply 2 charges."},{"Start":"12:02.680 ","End":"12:05.635","Text":"It\u0027s an E field being formed by 2 charges,"},{"Start":"12:05.635 ","End":"12:09.500","Text":"which means that it is a conservative field."},{"Start":"12:10.440 ","End":"12:13.750","Text":"We can use this equation, so let\u0027s do this."},{"Start":"12:13.750 ","End":"12:15.090","Text":"Negative nabla, so nabla is d by."},{"Start":"12:15.090 ","End":"12:21.504","Text":"The first component is the x component, so it\u0027s d by dx."},{"Start":"12:21.504 ","End":"12:26.140","Text":"But notice we\u0027re looking at the force on dipole number 2."},{"Start":"12:26.140 ","End":"12:29.860","Text":"It\u0027s d by dx_2,"},{"Start":"12:29.860 ","End":"12:32.110","Text":"the x coordinate of this,"},{"Start":"12:32.110 ","End":"12:35.935","Text":"then we have d by dy_2,"},{"Start":"12:35.935 ","End":"12:39.745","Text":"and d by dz_2."},{"Start":"12:39.745 ","End":"12:44.919","Text":"All of this is multiplied by u, our energy."},{"Start":"12:44.919 ","End":"12:47.919","Text":"We can use either equation,"},{"Start":"12:47.919 ","End":"12:49.960","Text":"so let\u0027s just use this."},{"Start":"12:49.960 ","End":"12:52.670","Text":"This is going to be multiplied by k,"},{"Start":"12:52.670 ","End":"13:00.145","Text":"and then we have p_2.p_1 minus"},{"Start":"13:00.145 ","End":"13:08.110","Text":"3p_1.r hat tilde multiplied"},{"Start":"13:08.110 ","End":"13:13.640","Text":"by p_2.r hat tilde."},{"Start":"13:14.460 ","End":"13:26.450","Text":"Then all of this is divided by r tilde cubed or r tilde^negative 3."},{"Start":"13:28.470 ","End":"13:32.680","Text":"Now, what we\u0027re going to do is we\u0027re going to start writing out"},{"Start":"13:32.680 ","End":"13:38.410","Text":"an extremely long equation. Get ready."},{"Start":"13:38.410 ","End":"13:40.509","Text":"We have negative,"},{"Start":"13:40.509 ","End":"13:47.746","Text":"and then this k is a constant so we can take it out also over here. You know what?"},{"Start":"13:47.746 ","End":"13:53.470","Text":"What I\u0027m going to write is I\u0027m going to write the x component of this force,"},{"Start":"13:53.470 ","End":"13:56.455","Text":"so let\u0027s write F_2x."},{"Start":"13:56.455 ","End":"13:58.210","Text":"Just the x component,"},{"Start":"13:58.210 ","End":"14:01.520","Text":"and then we\u0027ll have the same for the y and the z component."},{"Start":"14:01.590 ","End":"14:06.490","Text":"As I said, we have the minus from here and the k from here,"},{"Start":"14:06.490 ","End":"14:08.442","Text":"and then the x component,"},{"Start":"14:08.442 ","End":"14:17.620","Text":"so we\u0027re taking d by dx_2 of."},{"Start":"14:17.620 ","End":"14:19.689","Text":"First of all, p_2.p_1,"},{"Start":"14:19.689 ","End":"14:23.865","Text":"so these are constants."},{"Start":"14:23.865 ","End":"14:27.240","Text":"Remember we said our dipole is a constant,"},{"Start":"14:27.240 ","End":"14:34.750","Text":"so there\u0027s no variables over here that could be dependent on x or x_2 specifically."},{"Start":"14:34.750 ","End":"14:39.369","Text":"Therefore when I take the derivative this p_2.p_1 is going to cancel out,"},{"Start":"14:39.369 ","End":"14:42.729","Text":"so I\u0027m just going to keep that written like this."},{"Start":"14:42.729 ","End":"14:47.260","Text":"Let\u0027s just write p_1.p_2."},{"Start":"14:47.260 ","End":"14:49.934","Text":"Like so, this will cancel out very soon when we take the derivative,"},{"Start":"14:49.934 ","End":"14:55.315","Text":"and then I have negative 3 multiplied by."},{"Start":"14:55.315 ","End":"15:00.679","Text":"Here I have p_1.r tilde hat."},{"Start":"15:00.950 ","End":"15:03.750","Text":"Let\u0027s just write in blue."},{"Start":"15:03.750 ","End":"15:09.340","Text":"First of all, our r tilde vector is equal to."},{"Start":"15:09.340 ","End":"15:13.330","Text":"We saw above, but we can also see from here,"},{"Start":"15:13.330 ","End":"15:19.240","Text":"r_2 vector minus our r_1 vector. What is that?"},{"Start":"15:19.240 ","End":"15:24.009","Text":"The x component will be x_2 minus x_1,"},{"Start":"15:24.009 ","End":"15:29.049","Text":"the y component will be y_2 minus y_1,"},{"Start":"15:29.049 ","End":"15:34.165","Text":"and the z component will be z_2 minus z_1."},{"Start":"15:34.165 ","End":"15:38.785","Text":"Then the r hat tilde,"},{"Start":"15:38.785 ","End":"15:47.245","Text":"so this is the unit vector of this is simply going to be our r vector."},{"Start":"15:47.245 ","End":"15:51.920","Text":"That is x_2 minus x_1,"},{"Start":"15:52.200 ","End":"15:54.475","Text":"y_2 minus y_1,"},{"Start":"15:54.475 ","End":"16:00.894","Text":"and z_2 minus z_1 divided by the magnitude of this vector,"},{"Start":"16:00.894 ","End":"16:09.324","Text":"which is simply the square root of x_2 minus x_1 squared plus"},{"Start":"16:09.324 ","End":"16:18.969","Text":"y_2 minus y_1 squared plus z_2 minus z_1 squared."},{"Start":"16:18.969 ","End":"16:20.434","Text":"The square root of all of this,"},{"Start":"16:20.434 ","End":"16:23.658","Text":"so we can say that this is to the power of 1/2."},{"Start":"16:23.658 ","End":"16:33.279","Text":"Therefore we can say that r tilde magnitude to the power of negative 3,"},{"Start":"16:33.279 ","End":"16:38.545","Text":"so that\u0027s what we have over here in the denominator is"},{"Start":"16:38.545 ","End":"16:44.770","Text":"equal to just what we have over here in our r hat tilde as denominator."},{"Start":"16:44.770 ","End":"16:47.725","Text":"That was, as we said,"},{"Start":"16:47.725 ","End":"16:54.939","Text":"x_2 minus x_1 squared plus y_2,"},{"Start":"16:54.939 ","End":"16:56.905","Text":"sorry,"},{"Start":"16:56.905 ","End":"17:07.029","Text":"plus y_2 minus y_1 squared plus z_2 minus z_1 squared."},{"Start":"17:07.029 ","End":"17:11.740","Text":"All this is to the power of 1/2 for r magnitude,"},{"Start":"17:11.740 ","End":"17:14.140","Text":"but if it\u0027s to the power of negative 3,"},{"Start":"17:14.140 ","End":"17:23.449","Text":"so this is going to be to the power of negative 3/2."},{"Start":"17:23.449 ","End":"17:27.370","Text":"Now we\u0027ve seen what these vectors are."},{"Start":"17:27.370 ","End":"17:28.750","Text":"So we can carry on."},{"Start":"17:28.750 ","End":"17:33.984","Text":"We have negative 3 and then we have our p1 vector,"},{"Start":"17:33.984 ","End":"17:37.465","Text":"but of course we\u0027re only taking right now the x component."},{"Start":"17:37.465 ","End":"17:43.165","Text":"We\u0027ll have p1x dot product with this r hat Tilde."},{"Start":"17:43.165 ","End":"17:44.860","Text":"Let\u0027s write this out."},{"Start":"17:44.860 ","End":"17:53.170","Text":"We have p1 with the x component multiplied by r hat vector."},{"Start":"17:53.170 ","End":"17:56.755","Text":"So our r hat vector is simply this."},{"Start":"17:56.755 ","End":"18:03.860","Text":"We have x2 minus x1"},{"Start":"18:04.290 ","End":"18:10.844","Text":"plus our p1 vector multiplied by the y component."},{"Start":"18:10.844 ","End":"18:20.350","Text":"So plus p1x multiplied by y2 minus y1 plus p1x multiplied by"},{"Start":"18:20.350 ","End":"18:25.959","Text":"z2 minus z1 because this is the dot product and all of"},{"Start":"18:25.959 ","End":"18:32.889","Text":"this is divided by the rest of our r hat vector,"},{"Start":"18:32.889 ","End":"18:36.149","Text":"which is simply this."},{"Start":"18:36.149 ","End":"18:42.965","Text":"So X2 minus x1 squared plus y2 minus y1"},{"Start":"18:42.965 ","End":"18:50.889","Text":"squared plus z2 minus z1 squared and all of this, the square roots."},{"Start":"18:50.889 ","End":"18:54.619","Text":"So to the power of 1/2."},{"Start":"18:54.930 ","End":"18:59.035","Text":"This is our p1.y hat vector."},{"Start":"18:59.035 ","End":"19:06.805","Text":"Now we have to multiply this by p2 dot r hat vector."},{"Start":"19:06.805 ","End":"19:12.940","Text":"P2x multiplied by, so the x component."},{"Start":"19:12.940 ","End":"19:20.109","Text":"So x2 minus x1 plus p2x multiplied by"},{"Start":"19:20.109 ","End":"19:30.019","Text":"y2 minus y1 plus p2x multiplied by z2 minus z1."},{"Start":"19:30.870 ","End":"19:38.635","Text":"Also all of this is divided by this over here."},{"Start":"19:38.635 ","End":"19:45.099","Text":"We have x2 minus x1 squared plus y2"},{"Start":"19:45.099 ","End":"19:56.185","Text":"minus y1 squared plus z2 minus z1 squared and all of this is to the power of 1/2."},{"Start":"19:56.185 ","End":"20:00.639","Text":"Then I told you this was going to be very, very long,"},{"Start":"20:00.639 ","End":"20:08.410","Text":"multiplied by this r Tilde magnitude to the power of negative 3."},{"Start":"20:08.410 ","End":"20:10.255","Text":"That\u0027s what we had over here."},{"Start":"20:10.255 ","End":"20:18.786","Text":"We\u0027re going to be multiplying all of this by this over here."},{"Start":"20:18.786 ","End":"20:22.945","Text":"Here we\u0027re meant to have another bracket."},{"Start":"20:22.945 ","End":"20:33.129","Text":"All of this is multiplied by x2 minus x1 squared plus y2"},{"Start":"20:33.129 ","End":"20:39.880","Text":"minus y1 squared plus z2"},{"Start":"20:39.880 ","End":"20:48.290","Text":"minus z1 squared and all of this to the power of negative 3 over 2."},{"Start":"20:48.840 ","End":"20:54.414","Text":"Of course we have the square bracket because we\u0027re taking the derivative of all of this,"},{"Start":"20:54.414 ","End":"20:58.915","Text":"including to this r magnitude here."},{"Start":"20:58.915 ","End":"21:03.655","Text":"We have to close off the square bracket over here."},{"Start":"21:03.655 ","End":"21:11.634","Text":"Now, inside the square brackets we have p1 dot p2 minus 3 multiplied by all of this."},{"Start":"21:11.634 ","End":"21:21.239","Text":"Everything that is in over here is multiplied by this over here,"},{"Start":"21:21.239 ","End":"21:25.040","Text":"which is r to the negative 3."},{"Start":"21:25.560 ","End":"21:29.560","Text":"What we\u0027re going to do is we\u0027re just going to multiply"},{"Start":"21:29.560 ","End":"21:35.930","Text":"everything within these brackets by this r to the negative 3."},{"Start":"21:36.660 ","End":"21:44.095","Text":"What we\u0027ll get is that f of 2x is equal to"},{"Start":"21:44.095 ","End":"21:53.293","Text":"negative k multiplied by d by dx2 and then we have our p1 dot p2."},{"Start":"21:53.293 ","End":"22:02.695","Text":"So p1 dot p2 and all of this is divided by our r Tilde cubed."},{"Start":"22:02.695 ","End":"22:05.882","Text":"Sorry. It\u0027s not even r Tilde cubed."},{"Start":"22:05.882 ","End":"22:10.795","Text":"We can see it\u0027s to the power of 3 over 2,"},{"Start":"22:10.795 ","End":"22:13.480","Text":"or negative 3 over 2."},{"Start":"22:13.480 ","End":"22:15.970","Text":"We can write it like so."},{"Start":"22:15.970 ","End":"22:25.450","Text":"We have x2 minus x1 squared plus y2 minus y1 squared"},{"Start":"22:25.450 ","End":"22:31.065","Text":"plus z2 minus z1"},{"Start":"22:31.065 ","End":"22:36.905","Text":"squared and then all of this is to the power of 3 over 2."},{"Start":"22:36.905 ","End":"22:45.880","Text":"Then we have negative 3 times and then everything that we have over here."},{"Start":"22:46.200 ","End":"22:49.520","Text":"Here we have all of this."},{"Start":"22:51.270 ","End":"22:55.720","Text":"What we had is this."},{"Start":"22:55.720 ","End":"22:58.405","Text":"We have the magnitude of r Tilde."},{"Start":"22:58.405 ","End":"23:01.675","Text":"We have all of this to the power of 1/2,"},{"Start":"23:01.675 ","End":"23:05.590","Text":"and then we have multiplied again by the magnitude of r Tilde."},{"Start":"23:05.590 ","End":"23:07.464","Text":"Again, we have the same thing,"},{"Start":"23:07.464 ","End":"23:09.430","Text":"again, to the power of a 1/2."},{"Start":"23:09.430 ","End":"23:13.480","Text":"Together, we\u0027re squaring them."},{"Start":"23:13.480 ","End":"23:15.670","Text":"We\u0027re multiplying this by itself."},{"Start":"23:15.670 ","End":"23:18.920","Text":"We\u0027ll get that this is to the power of 1."},{"Start":"23:19.710 ","End":"23:22.029","Text":"All of these denominators,"},{"Start":"23:22.029 ","End":"23:24.955","Text":"everything over here, x2 minus x1 squared,"},{"Start":"23:24.955 ","End":"23:26.725","Text":"y2 minus y1 squared,"},{"Start":"23:26.725 ","End":"23:31.255","Text":"z2 minus z1 squared to the power of 1."},{"Start":"23:31.255 ","End":"23:40.450","Text":"But then we\u0027re multiplying it by over here this power of negative 3 over 2."},{"Start":"23:40.450 ","End":"23:43.330","Text":"If we call everything that\u0027s in here,"},{"Start":"23:43.330 ","End":"23:45.280","Text":"let\u0027s say y,"},{"Start":"23:45.280 ","End":"23:50.425","Text":"so here we can say that either we have y to the power of negative 3 over 2,"},{"Start":"23:50.425 ","End":"23:55.345","Text":"which is equal to 1 divided by y to the power of 3 over 2."},{"Start":"23:55.345 ","End":"23:57.115","Text":"This we already know."},{"Start":"23:57.115 ","End":"24:05.746","Text":"We also know that if we have y to the power of n multiplied by y to the power of m,"},{"Start":"24:05.746 ","End":"24:12.850","Text":"then we can write this as y to the power of n plus m. In that case,"},{"Start":"24:12.850 ","End":"24:17.170","Text":"what we can do is we can start afresh."},{"Start":"24:17.170 ","End":"24:20.579","Text":"Let\u0027s say all of this inside the brackets is y."},{"Start":"24:20.579 ","End":"24:22.975","Text":"So we have y,"},{"Start":"24:22.975 ","End":"24:29.649","Text":"and here we have y to the power of 1/2 multiplied by all of these brackets again,"},{"Start":"24:29.649 ","End":"24:35.080","Text":"which is again y to the power of 1/2 multiplied by this,"},{"Start":"24:35.080 ","End":"24:38.170","Text":"which is the same y inside the brackets."},{"Start":"24:38.170 ","End":"24:43.150","Text":"This time, if we put it in this form as a denominator,"},{"Start":"24:43.150 ","End":"24:46.239","Text":"because over here our y is in the denominator."},{"Start":"24:46.239 ","End":"24:48.835","Text":"So this is to the power of 3 over 2."},{"Start":"24:48.835 ","End":"24:55.810","Text":"We have 1/2 plus 1/2 is 1 plus 3 over 2, which is 1.5."},{"Start":"24:55.810 ","End":"24:59.485","Text":"We have 1 plus 1.5 is 2.5."},{"Start":"24:59.485 ","End":"25:03.820","Text":"We can write that as y to the power of 5 over 2."},{"Start":"25:03.820 ","End":"25:09.670","Text":"Then we go back over here and we can write this as 5 over 2."},{"Start":"25:09.670 ","End":"25:12.999","Text":"This is to the power 5 over 2."},{"Start":"25:12.999 ","End":"25:16.921","Text":"Now we can close these brackets,"},{"Start":"25:16.921 ","End":"25:18.490","Text":"so this is just a note."},{"Start":"25:18.490 ","End":"25:23.720","Text":"Now let\u0027s go to solve this."},{"Start":"25:23.720 ","End":"25:28.185","Text":"Right now what we\u0027re doing is we\u0027re taking the derivative."},{"Start":"25:28.185 ","End":"25:31.305","Text":"Let\u0027s begin. This will be equal to,"},{"Start":"25:31.305 ","End":"25:36.775","Text":"so we have negative k and then we have 2 functions over here."},{"Start":"25:36.775 ","End":"25:40.975","Text":"We have the function in the numerator and the function in the denominator."},{"Start":"25:40.975 ","End":"25:45.115","Text":"When we\u0027re taking the derivative with respect to x_2,"},{"Start":"25:45.115 ","End":"25:49.060","Text":"we\u0027re going to use what we saw earlier in the lesson where we"},{"Start":"25:49.060 ","End":"25:52.885","Text":"take the derivative of 1 function and multiply it"},{"Start":"25:52.885 ","End":"25:56.784","Text":"by the second function and then we add onto it"},{"Start":"25:56.784 ","End":"26:01.629","Text":"the derivative of the second function multiplied by the first function."},{"Start":"26:01.629 ","End":"26:06.940","Text":"First let\u0027s take the derivative of a p_1, p_2."},{"Start":"26:06.940 ","End":"26:09.340","Text":"We already said that this is constant."},{"Start":"26:09.340 ","End":"26:10.960","Text":"If we take the derivative of this,"},{"Start":"26:10.960 ","End":"26:16.090","Text":"it\u0027s going to equal to 0 and then multiplied by the second function,"},{"Start":"26:16.090 ","End":"26:19.345","Text":"it\u0027s going to be equal to 0."},{"Start":"26:19.345 ","End":"26:26.154","Text":"First we have 0 and then what we\u0027re going to do is we\u0027re going to take"},{"Start":"26:26.154 ","End":"26:34.615","Text":"the derivative of the second function and then multiply it by p_1.p_2."},{"Start":"26:34.615 ","End":"26:36.679","Text":"The first function."},{"Start":"26:37.080 ","End":"26:42.370","Text":"Let\u0027s again consider everything that\u0027s in the brackets as"},{"Start":"26:42.370 ","End":"26:47.709","Text":"some function or some variable y."},{"Start":"26:47.709 ","End":"26:54.475","Text":"What we can see is that we have 1 over y^3/2,"},{"Start":"26:54.475 ","End":"26:59.800","Text":"which is equal to as we know, y^-3/2."},{"Start":"26:59.800 ","End":"27:02.826","Text":"Then if we take d by dy."},{"Start":"27:02.826 ","End":"27:10.210","Text":"We\u0027re taking the derivative with respect to y over y^-3/2."},{"Start":"27:10.210 ","End":"27:19.280","Text":"This will be equal to negative 3/2 y^-5/2."},{"Start":"27:20.100 ","End":"27:26.215","Text":"What we\u0027re going to have is this over here."},{"Start":"27:26.215 ","End":"27:32.169","Text":"We\u0027re going to have plus our first function."},{"Start":"27:32.169 ","End":"27:41.166","Text":"P_1.p_2 and then this is going to be multiplied by negative 3/2y."},{"Start":"27:41.166 ","End":"27:51.469","Text":"Our y over here is x_2 minus x_1^2 plus y_2 minus y_1^2"},{"Start":"27:51.469 ","End":"27:57.239","Text":"plus z_2 minus z_1^2"},{"Start":"27:57.239 ","End":"28:04.955","Text":"and all of this to the power of negative 5/2."},{"Start":"28:04.955 ","End":"28:08.169","Text":"That\u0027s the derivative of all of this,"},{"Start":"28:08.169 ","End":"28:11.665","Text":"but we\u0027re deriving according to x_2,"},{"Start":"28:11.665 ","End":"28:15.355","Text":"then we have to take the antiderivative."},{"Start":"28:15.355 ","End":"28:20.829","Text":"That means that we\u0027re going to be multiplying all of this by the anti-derivative."},{"Start":"28:20.829 ","End":"28:26.425","Text":"Here this time let\u0027s call it t. Let\u0027s say that"},{"Start":"28:26.425 ","End":"28:32.950","Text":"x_2 minus x_1= t. Therefore,"},{"Start":"28:32.950 ","End":"28:36.500","Text":"what we have over here is t^2."},{"Start":"28:36.630 ","End":"28:39.190","Text":"If we want to take the derivative,"},{"Start":"28:39.190 ","End":"28:44.260","Text":"so d by dt(t^2) is going to be equal to 2t."},{"Start":"28:44.260 ","End":"28:49.930","Text":"This is going to be multiplied by 2t,"},{"Start":"28:49.930 ","End":"28:53.994","Text":"which is x_2 minus x_1,"},{"Start":"28:53.994 ","End":"29:03.115","Text":"that\u0027s t. Then it\u0027s going to be multiplied by the extra anti-derivative of x_2,"},{"Start":"29:03.115 ","End":"29:06.160","Text":"because x_2 is our variables,"},{"Start":"29:06.160 ","End":"29:10.029","Text":"so we\u0027re going to take d by dx_2,"},{"Start":"29:10.029 ","End":"29:15.680","Text":"so that\u0027s just going to be equal to 1 and then we multiply this by 1."},{"Start":"29:16.650 ","End":"29:19.225","Text":"What we\u0027ve used here,"},{"Start":"29:19.225 ","End":"29:22.240","Text":"just to take the derivative of this is using"},{"Start":"29:22.240 ","End":"29:26.215","Text":"a combination of the product rule and of the chain rule."},{"Start":"29:26.215 ","End":"29:30.355","Text":"I\u0027m going to rub out this to make this a bit neater."},{"Start":"29:30.355 ","End":"29:35.680","Text":"Now before we move on to taking the derivative of this,"},{"Start":"29:35.680 ","End":"29:37.809","Text":"let\u0027s just sort this out,"},{"Start":"29:37.809 ","End":"29:40.690","Text":"so all of this,"},{"Start":"29:40.690 ","End":"29:43.985","Text":"let\u0027s show it in blue."},{"Start":"29:43.985 ","End":"29:49.690","Text":"All of this will simplify to this."},{"Start":"29:49.690 ","End":"29:51.310","Text":"We have 0 plus,"},{"Start":"29:51.310 ","End":"29:52.704","Text":"so we can take that out."},{"Start":"29:52.704 ","End":"29:57.730","Text":"What we have in the numerator is we have negative 3/2,"},{"Start":"29:57.730 ","End":"30:02.770","Text":"but this is multiplied by this 2 over here so we can cross out this 2 and this 2."},{"Start":"30:02.770 ","End":"30:10.224","Text":"We have negative 3 multiplied by p_1.p_2."},{"Start":"30:10.224 ","End":"30:17.755","Text":"Then we have all of what is in this bracket to the negative 5/2,"},{"Start":"30:17.755 ","End":"30:20.485","Text":"so we can put it in the denominator."},{"Start":"30:20.485 ","End":"30:30.020","Text":"We have x_2 minus x_1^2 plus y_2 minus y_1^2 plus"},{"Start":"30:30.020 ","End":"30:39.775","Text":"z_2 minus z_1^2 and all of this is to the power of 5/2."},{"Start":"30:39.775 ","End":"30:48.205","Text":"Then in the numerator we also have this x_2 minus x_1."},{"Start":"30:48.205 ","End":"30:51.175","Text":"Of course, if we multiply by 1,"},{"Start":"30:51.175 ","End":"30:52.810","Text":"we\u0027re not changing anything."},{"Start":"30:52.810 ","End":"30:57.619","Text":"Now we\u0027ve simplified all of this to this fraction over here."},{"Start":"30:58.020 ","End":"31:06.025","Text":"Up until now we\u0027ve taken the derivative with respect to x_2 of just this over here."},{"Start":"31:06.025 ","End":"31:08.170","Text":"We\u0027ve just done this step,"},{"Start":"31:08.170 ","End":"31:09.939","Text":"which we got over here,"},{"Start":"31:09.939 ","End":"31:13.479","Text":"and then we just simplified it to this over here."},{"Start":"31:13.479 ","End":"31:22.359","Text":"Now we\u0027re going to take the derivative of the second function that we have over here."},{"Start":"31:22.359 ","End":"31:27.369","Text":"Here we can consider that we have 3 functions."},{"Start":"31:27.369 ","End":"31:30.235","Text":"Let\u0027s show them."},{"Start":"31:30.235 ","End":"31:35.634","Text":"Our first function is this over here."},{"Start":"31:35.634 ","End":"31:41.049","Text":"Our second function, let\u0027s draw it in pink is this over here,"},{"Start":"31:41.049 ","End":"31:48.475","Text":"and our third function is gray over here in the denominator."},{"Start":"31:48.475 ","End":"31:53.830","Text":"We\u0027re going to again use the product rule and of course, the chain rule."},{"Start":"31:53.830 ","End":"31:57.040","Text":"What we\u0027re going to do is we\u0027re going to take the derivative of"},{"Start":"31:57.040 ","End":"32:01.779","Text":"the first function and then multiply it by the other 2 functions."},{"Start":"32:01.779 ","End":"32:03.550","Text":"Then we\u0027ll take the derivative of"},{"Start":"32:03.550 ","End":"32:07.105","Text":"the second function and multiply it by functions 1 and 3,"},{"Start":"32:07.105 ","End":"32:09.220","Text":"and then finally, we will take the derivative of"},{"Start":"32:09.220 ","End":"32:13.585","Text":"the third function and multiply it by functions 1 and 2."},{"Start":"32:13.585 ","End":"32:17.395","Text":"Let\u0027s do this 1 step at a time."},{"Start":"32:17.395 ","End":"32:20.289","Text":"Let\u0027s first find the derivative of the green function."},{"Start":"32:20.289 ","End":"32:22.405","Text":"We\u0027re going to write it on top."},{"Start":"32:22.405 ","End":"32:28.795","Text":"We\u0027re finding, I\u0027m reminding you d by dx_2 of this first function in green."},{"Start":"32:28.795 ","End":"32:32.079","Text":"Let\u0027s include this 3 in over here."},{"Start":"32:32.079 ","End":"32:36.190","Text":"It\u0027s of all of this,"},{"Start":"32:36.190 ","End":"32:38.125","Text":"I\u0027m not going to write it out again."},{"Start":"32:38.125 ","End":"32:41.635","Text":"Of all of this bracket,"},{"Start":"32:41.635 ","End":"32:44.965","Text":"from here until here."},{"Start":"32:44.965 ","End":"32:47.919","Text":"Because we\u0027re doing it according to x_2,"},{"Start":"32:47.919 ","End":"32:51.399","Text":"so we can see that we only have an x_2 over here."},{"Start":"32:51.399 ","End":"32:54.489","Text":"If we take the derivative of this,"},{"Start":"32:54.489 ","End":"32:58.105","Text":"we can see it\u0027s multiplied by p_1x."},{"Start":"32:58.105 ","End":"33:05.019","Text":"We\u0027re essentially taking the derivative of 3p_1 x, x_2."},{"Start":"33:05.019 ","End":"33:11.575","Text":"All we\u0027re going to be left with is with 3p_1x."},{"Start":"33:11.575 ","End":"33:16.180","Text":"Now let\u0027s take a look at the pink."},{"Start":"33:16.180 ","End":"33:19.370","Text":"I\u0027m going to rub out all of this that I have on top."},{"Start":"33:19.410 ","End":"33:25.075","Text":"The pink function, or the second function is from here until here."},{"Start":"33:25.075 ","End":"33:31.660","Text":"Again, we\u0027re taking the derivative with respect to x_2 of whatever is in these brackets,"},{"Start":"33:31.660 ","End":"33:33.220","Text":"I won\u0027t write it again."},{"Start":"33:33.220 ","End":"33:43.375","Text":"Again, we only have x_2 over here and we can see we\u0027ve already used this 3 over here."},{"Start":"33:43.375 ","End":"33:49.075","Text":"We\u0027re just taking the derivative essentially of p_2x x_2."},{"Start":"33:49.075 ","End":"33:51.790","Text":"Because all of this will become 0,"},{"Start":"33:51.790 ","End":"33:53.410","Text":"because there\u0027s no x_2 here."},{"Start":"33:53.410 ","End":"33:57.294","Text":"We\u0027ll get that this is just equal to p_2x."},{"Start":"33:57.294 ","End":"34:00.355","Text":"Then let\u0027s look in the gray."},{"Start":"34:00.355 ","End":"34:07.975","Text":"Essentially what we have in the gray is everything inside these brackets."},{"Start":"34:07.975 ","End":"34:12.310","Text":"But we can also write it to the power of negative 5/2."},{"Start":"34:12.310 ","End":"34:16.450","Text":"Again, it\u0027s according to dx_2."},{"Start":"34:16.450 ","End":"34:20.755","Text":"We can see that we only have this x_2 over here."},{"Start":"34:20.755 ","End":"34:23.934","Text":"We\u0027re going to have to use the chain rule,"},{"Start":"34:23.934 ","End":"34:28.300","Text":"because we can see that first we have x_2 minus x_1^2."},{"Start":"34:28.300 ","End":"34:30.249","Text":"By taking the derivative of that,"},{"Start":"34:30.249 ","End":"34:35.124","Text":"we\u0027re going to have 2x_2 minus x_1 and then"},{"Start":"34:35.124 ","End":"34:40.974","Text":"multiplied by the anti-derivative so the inner derivative of x_2 is simply going to be 1."},{"Start":"34:40.974 ","End":"34:47.409","Text":"We\u0027ll be left with just 2 multiplied by x_2 minus x_1."},{"Start":"34:47.409 ","End":"34:51.400","Text":"Now we have all of the derivatives."},{"Start":"34:51.400 ","End":"34:55.775","Text":"We\u0027re going to add on according to the product rule, what do we have."},{"Start":"34:55.775 ","End":"34:58.492","Text":"We\u0027re carrying on along this line."},{"Start":"34:58.492 ","End":"35:01.179","Text":"We have negative."},{"Start":"35:01.179 ","End":"35:05.065","Text":"Let\u0027s just break this up."},{"Start":"35:05.065 ","End":"35:08.500","Text":"We have negative from this negative over here."},{"Start":"35:08.500 ","End":"35:12.160","Text":"The derivative of our first function is"},{"Start":"35:12.160 ","End":"35:20.060","Text":"3p_1x multiplied by our pink function and our gray function."},{"Start":"35:20.790 ","End":"35:26.800","Text":"Multiplied by, so we have p_2x,"},{"Start":"35:26.800 ","End":"35:32.200","Text":"x2 minus x1 plus p_2x,"},{"Start":"35:32.200 ","End":"35:41.395","Text":"y_2 minus y_1 plus p_2x, z_2 minus z_1."},{"Start":"35:41.395 ","End":"35:45.410","Text":"Then multiplied by our gray function."},{"Start":"35:50.760 ","End":"35:55.795","Text":"Then we\u0027re going to add onto it."},{"Start":"35:55.795 ","End":"35:59.635","Text":"This is the derivative of this multiplied by this and this."},{"Start":"35:59.635 ","End":"36:02.740","Text":"Now we\u0027re going to take this 1,"},{"Start":"36:02.740 ","End":"36:07.285","Text":"so we\u0027re adding on the derivative of this multiplied by this and this."},{"Start":"36:07.285 ","End":"36:13.579","Text":"Again, so we have plus and then we have negative."},{"Start":"36:14.280 ","End":"36:16.599","Text":"We have negative."},{"Start":"36:16.599 ","End":"36:20.400","Text":"The derivative of the pink function is p_2x."},{"Start":"36:20.400 ","End":"36:25.620","Text":"But then it\u0027s also multiplied by the green function, without the derivative."},{"Start":"36:25.620 ","End":"36:27.090","Text":"We have the 3 again,"},{"Start":"36:27.090 ","End":"36:31.050","Text":"so we have negative 3 and then we can write over here"},{"Start":"36:31.050 ","End":"36:36.400","Text":"p_2x and then multiplied by the rest of the green function."},{"Start":"36:36.400 ","End":"36:39.085","Text":"We have p_1x,"},{"Start":"36:39.085 ","End":"36:41.830","Text":"x_2 minus x_1,"},{"Start":"36:41.830 ","End":"36:44.390","Text":"and then it\u0027ll carry this on."},{"Start":"36:44.700 ","End":"36:52.120","Text":"Here I have the rest of my green function and multiply it by my gray function."},{"Start":"36:52.120 ","End":"36:57.310","Text":"Now I\u0027m going to add on the derivative of my gray function,"},{"Start":"36:57.310 ","End":"37:00.520","Text":"which we have this derivative over here."},{"Start":"37:00.520 ","End":"37:02.469","Text":"I made a mistake earlier on."},{"Start":"37:02.469 ","End":"37:06.460","Text":"But the derivative is negative 5 divided by 2 multiplied by"},{"Start":"37:06.460 ","End":"37:10.990","Text":"everything inside the gray brackets."},{"Start":"37:10.990 ","End":"37:14.019","Text":"Everything inside here, and then"},{"Start":"37:14.019 ","End":"37:16.975","Text":"instead of writing it in the denominator to the power 5 over 2."},{"Start":"37:16.975 ","End":"37:20.890","Text":"We wrote it in the numerator to the power of negative 5 over 2."},{"Start":"37:20.890 ","End":"37:24.100","Text":"We have negative 5 divided by 2 multiplied by"},{"Start":"37:24.100 ","End":"37:26.469","Text":"everything in the brackets to the power of negative"},{"Start":"37:26.469 ","End":"37:32.350","Text":"7 over 2 multiplied by 2x_2 minus x_1 multiplied by 1."},{"Start":"37:32.350 ","End":"37:34.975","Text":"This is the derivative of that."},{"Start":"37:34.975 ","End":"37:39.590","Text":"We\u0027re going to add this in."},{"Start":"37:39.690 ","End":"37:44.935","Text":"We have over here negative 3."},{"Start":"37:44.935 ","End":"37:51.520","Text":"However, it\u0027s multiplied by this negative over here, of the 5/2."},{"Start":"37:51.520 ","End":"37:55.090","Text":"We can just write plus 3."},{"Start":"37:55.090 ","End":"37:58.855","Text":"Then we have our p_1x,"},{"Start":"37:58.855 ","End":"38:03.820","Text":"x_2 minus x_1 plus p_1x,"},{"Start":"38:03.820 ","End":"38:13.580","Text":"y_2 minus y_1 plus p_1x z_2 minus z_1."},{"Start":"38:13.650 ","End":"38:17.425","Text":"Then multiplied by our pink function."},{"Start":"38:17.425 ","End":"38:19.640","Text":"I\u0027ll write it in already."},{"Start":"38:19.830 ","End":"38:22.945","Text":"We have 3p_1x,"},{"Start":"38:22.945 ","End":"38:28.359","Text":"all of this is the green function up until here,"},{"Start":"38:28.359 ","End":"38:32.890","Text":"and then we have p_2x up until here."},{"Start":"38:32.890 ","End":"38:34.704","Text":"This is our pink function,"},{"Start":"38:34.704 ","End":"38:38.920","Text":"and all of this is multiplied by the derivative of our gray function."},{"Start":"38:38.920 ","End":"38:41.545","Text":"We already took into account the minus."},{"Start":"38:41.545 ","End":"38:47.140","Text":"Then we\u0027re just going to multiply it by 5 divided"},{"Start":"38:47.140 ","End":"38:52.869","Text":"by 2 multiplied by everything that\u0027s in the brackets."},{"Start":"38:52.869 ","End":"38:54.890","Text":"I\u0027ll write it out."},{"Start":"38:55.080 ","End":"39:01.750","Text":"We have 5/2 multiplied by everything inside the bracket to the power of negative 7/2"},{"Start":"39:01.750 ","End":"39:08.530","Text":"multiplied by 2_x2 minus x_1 and then multiplied by 1."},{"Start":"39:08.530 ","End":"39:11.575","Text":"This is a very long-expression."},{"Start":"39:11.575 ","End":"39:18.205","Text":"Now we\u0027re going to try and simplify this significantly."},{"Start":"39:18.205 ","End":"39:21.309","Text":"What do we have over here is, first of all,"},{"Start":"39:21.309 ","End":"39:25.420","Text":"we can see that we have negative k. Then here we have negative 3,"},{"Start":"39:25.420 ","End":"39:27.310","Text":"negative 3, negative 3,"},{"Start":"39:27.310 ","End":"39:29.140","Text":"and here we have plus 3."},{"Start":"39:29.140 ","End":"39:32.350","Text":"First of all, we can see that everything has a"},{"Start":"39:32.350 ","End":"39:37.015","Text":"negative 3 aside from this and k multiplying everything."},{"Start":"39:37.015 ","End":"39:39.009","Text":"We should actually scroll all the way"},{"Start":"39:39.009 ","End":"39:45.534","Text":"here and close all of this to signify that k multiplies that as well."},{"Start":"39:45.534 ","End":"39:53.550","Text":"What we can do is we can take out 3k,"},{"Start":"39:53.550 ","End":"39:56.955","Text":"so then these negatives will become positives."},{"Start":"39:56.955 ","End":"40:01.110","Text":"Then here we\u0027ll just remember to substitute in a negative to cancel that out."},{"Start":"40:01.110 ","End":"40:05.140","Text":"We have 3k multiplied by."},{"Start":"40:05.160 ","End":"40:08.890","Text":"We got rid of these 3, so k we put it over here."},{"Start":"40:08.890 ","End":"40:17.979","Text":"We have p_1.p_2 multiplied"},{"Start":"40:17.979 ","End":"40:20.920","Text":"by x_2 minus x_1."},{"Start":"40:20.920 ","End":"40:23.395","Text":"Then what do we have over here?"},{"Start":"40:23.395 ","End":"40:25.614","Text":"We have what\u0027s in brackets,"},{"Start":"40:25.614 ","End":"40:27.849","Text":"and then we take the square root of it,"},{"Start":"40:27.849 ","End":"40:32.875","Text":"and then we put it to the power of 5."},{"Start":"40:32.875 ","End":"40:34.449","Text":"What do we have here?"},{"Start":"40:34.449 ","End":"40:36.639","Text":"We have x_2 minus x_1 squared,"},{"Start":"40:36.639 ","End":"40:38.050","Text":"y_2 minus y_1 squared,"},{"Start":"40:38.050 ","End":"40:40.269","Text":"z_2 minus z_1 squared,"},{"Start":"40:40.269 ","End":"40:42.430","Text":"and then the square root of all of that."},{"Start":"40:42.430 ","End":"40:46.660","Text":"If we remember that was equal to,"},{"Start":"40:46.660 ","End":"40:49.569","Text":"scroll up over here."},{"Start":"40:49.569 ","End":"40:54.535","Text":"If you remember, that was equal to r magnitude."},{"Start":"40:54.535 ","End":"40:58.795","Text":"Imagine if it\u0027s not to the power of negative 3,"},{"Start":"40:58.795 ","End":"41:00.415","Text":"then we get rid of this."},{"Start":"41:00.415 ","End":"41:03.970","Text":"It\u0027s to the power of 1.5."},{"Start":"41:03.970 ","End":"41:06.805","Text":"That\u0027s r Tilde magnitude."},{"Start":"41:06.805 ","End":"41:14.750","Text":"We can say that all of this is equal to r Tilde to the power of 5."},{"Start":"41:15.390 ","End":"41:18.505","Text":"That\u0027s all we have over here."},{"Start":"41:18.505 ","End":"41:21.475","Text":"Then we\u0027re adding on."},{"Start":"41:21.475 ","End":"41:23.514","Text":"This 3 is over here,"},{"Start":"41:23.514 ","End":"41:28.150","Text":"and then we have p_1x multiplied by,"},{"Start":"41:28.150 ","End":"41:31.359","Text":"so we saw that I made a mistake earlier."},{"Start":"41:31.359 ","End":"41:32.785","Text":"We have p_2x,"},{"Start":"41:32.785 ","End":"41:37.190","Text":"p_2y and p_2z multiplying everything over here."},{"Start":"41:37.260 ","End":"41:42.880","Text":"This is simply going to be p_1x multiplied by,"},{"Start":"41:42.880 ","End":"41:50.410","Text":"so we have just a p_2 vector dot multiplied with x_2 minus x_1,"},{"Start":"41:50.410 ","End":"41:52.615","Text":"y_2 minus y_1, z_2 minus z_1,"},{"Start":"41:52.615 ","End":"41:57.775","Text":"which is just our r Tilde vector."},{"Start":"41:57.775 ","End":"42:02.030","Text":"That\u0027s exactly what that is."},{"Start":"42:02.610 ","End":"42:06.610","Text":"Then all of that is divided."},{"Start":"42:06.610 ","End":"42:08.919","Text":"We have all of this in the numerator,"},{"Start":"42:08.919 ","End":"42:11.065","Text":"and then this is in the denominator."},{"Start":"42:11.065 ","End":"42:15.070","Text":"This is again, all what\u0027s in the brackets with"},{"Start":"42:15.070 ","End":"42:19.660","Text":"this square root is just our r tilde, a magnitude."},{"Start":"42:19.660 ","End":"42:22.975","Text":"Then to the power of 5."},{"Start":"42:22.975 ","End":"42:28.435","Text":"All of this is divided by r Tilde ^5."},{"Start":"42:28.435 ","End":"42:36.880","Text":"Then we have plus again because this minus was with this minus over here and the 3."},{"Start":"42:36.880 ","End":"42:42.700","Text":"Now it\u0027s the same thing just we\u0027re switching our p_2x with our p_1."},{"Start":"42:42.700 ","End":"42:46.705","Text":"Now we just have p_2x."},{"Start":"42:46.705 ","End":"42:52.030","Text":"Then what we have is p_1x, p_1y, p_1z."},{"Start":"42:52.030 ","End":"43:00.130","Text":"That\u0027s just our p_1 vector dot product with x_2 minus x_1 and y_2 minus y_1,"},{"Start":"43:00.130 ","End":"43:01.540","Text":"and z_2 minus z_1."},{"Start":"43:01.540 ","End":"43:06.500","Text":"Dot-product again with our r vector Tilde."},{"Start":"43:06.870 ","End":"43:11.019","Text":"Then we have in the denominator this over here,"},{"Start":"43:11.019 ","End":"43:16.255","Text":"which is again, it\u0027s just our r magnitude."},{"Start":"43:16.255 ","End":"43:20.950","Text":"I told the magnitude to the power of 5."},{"Start":"43:20.950 ","End":"43:23.379","Text":"Then over here finally,"},{"Start":"43:23.379 ","End":"43:25.675","Text":"we have this positive."},{"Start":"43:25.675 ","End":"43:31.315","Text":"Because we took out this negative over here,"},{"Start":"43:31.315 ","End":"43:37.135","Text":"so we have to write this as negative and the 3 as a common factor outside the brackets."},{"Start":"43:37.135 ","End":"43:40.599","Text":"What we have is,"},{"Start":"43:40.599 ","End":"43:43.645","Text":"so here we have p_1x, p_1y, p_1z."},{"Start":"43:43.645 ","End":"43:49.690","Text":"Have our p_1 vector multiplied by this x,"},{"Start":"43:49.690 ","End":"43:51.730","Text":"this y, and these z components?"},{"Start":"43:51.730 ","End":"43:57.175","Text":"Dot-product with our r Tilde vector."},{"Start":"43:57.175 ","End":"44:03.720","Text":"Then all of this is multiplied by our p_2vector,"},{"Start":"44:03.720 ","End":"44:06.130","Text":"p_2y, p_2y, p_2z."},{"Start":"44:06.130 ","End":"44:08.830","Text":"We have our p_2 vector,"},{"Start":"44:08.830 ","End":"44:11.350","Text":"which is itself multiplied by the r vector,"},{"Start":"44:11.350 ","End":"44:14.140","Text":"the x, y, and z components."},{"Start":"44:14.140 ","End":"44:16.435","Text":"We have a r Tilde vector,"},{"Start":"44:16.435 ","End":"44:26.080","Text":"and then all of this is multiplied by 5 divided by 2."},{"Start":"44:26.080 ","End":"44:28.885","Text":"But we have a 2 over here."},{"Start":"44:28.885 ","End":"44:32.560","Text":"Here, we\u0027re multiplying by 2 and here we\u0027re dividing by 2."},{"Start":"44:32.560 ","End":"44:38.320","Text":"We have all of this multiplied by 5, x2 minus x1."},{"Start":"44:38.320 ","End":"44:40.480","Text":"Let\u0027s put the 5 over here."},{"Start":"44:40.480 ","End":"44:44.509","Text":"Then here we can have our x_2 minus x_1."},{"Start":"44:44.509 ","End":"44:47.159","Text":"Then we\u0027re just left with this,"},{"Start":"44:47.159 ","End":"44:49.604","Text":"which has just with the square root,"},{"Start":"44:49.604 ","End":"44:52.125","Text":"it\u0027s a r-tilde magnitude,"},{"Start":"44:52.125 ","End":"44:56.205","Text":"but to the power of 7 this time."},{"Start":"44:56.205 ","End":"45:02.110","Text":"All of this is divided by r-tilde^7."},{"Start":"45:03.170 ","End":"45:07.050","Text":"Somehow we managed to sort out"},{"Start":"45:07.050 ","End":"45:11.039","Text":"this very long equation to just this which"},{"Start":"45:11.039 ","End":"45:15.975","Text":"is more manageable and of course we have to close the brackets over here."},{"Start":"45:15.975 ","End":"45:20.093","Text":"Then of course we can write this also, k here,"},{"Start":"45:20.093 ","End":"45:24.030","Text":"remember way back when we started with the x component."},{"Start":"45:24.030 ","End":"45:28.680","Text":"Now we have to write the y-component for dipole number 2,"},{"Start":"45:28.680 ","End":"45:34.185","Text":"so it\u0027s F_2y and this will be equal to 3k."},{"Start":"45:34.185 ","End":"45:42.150","Text":"Then we\u0027re just going to have p_1.p_2 and then instead of x_2 minus x_1,"},{"Start":"45:42.150 ","End":"45:48.990","Text":"we have y_2 minus y_1, divided by r-tilda^5."},{"Start":"45:48.990 ","End":"45:53.280","Text":"Then we have plus now instead of p_1x,"},{"Start":"45:53.280 ","End":"45:57.450","Text":"so we have p_1y and then the same thing,"},{"Start":"45:57.450 ","End":"46:06.150","Text":"so p_2.r vector and then divided by r^5 plus."},{"Start":"46:06.150 ","End":"46:08.190","Text":"Now instead of p_2x,"},{"Start":"46:08.190 ","End":"46:15.854","Text":"we have p_2y multiplied by p_1 vector.r-tilde vector."},{"Start":"46:15.854 ","End":"46:18.720","Text":"Then we have negative"},{"Start":"46:18.720 ","End":"46:28.830","Text":"5p_1.r-tilde vector multiplied by p_2.r-tilde vector."},{"Start":"46:28.830 ","End":"46:31.230","Text":"Then instead of x_2 minus x_1,"},{"Start":"46:31.230 ","End":"46:39.960","Text":"we have y_2 minus y_1 and of course this is divided by r^5 over here."},{"Start":"46:39.960 ","End":"46:50.040","Text":"This is divided by r-tilde^7."},{"Start":"46:50.040 ","End":"46:56.430","Text":"Of course, we wouldn\u0027t be anywhere without the z component, so F_2z."},{"Start":"46:56.430 ","End":"46:59.580","Text":"Again, it\u0027s the exact same thing, 3k,"},{"Start":"46:59.580 ","End":"47:04.390","Text":"and then we have p_1.p_2."},{"Start":"47:05.210 ","End":"47:10.049","Text":"It\u0027s exactly the same as what we\u0027ve seen for the x and the y components."},{"Start":"47:10.049 ","End":"47:11.550","Text":"Just instead of x and y,"},{"Start":"47:11.550 ","End":"47:14.910","Text":"we have z_2 minus z_1 over here,"},{"Start":"47:14.910 ","End":"47:18.760","Text":"and z_2 minus z_1 over here."},{"Start":"47:20.360 ","End":"47:23.909","Text":"I know this is very difficult and long."},{"Start":"47:23.909 ","End":"47:24.990","Text":"Just bear with me,"},{"Start":"47:24.990 ","End":"47:27.000","Text":"we\u0027re not done yet."},{"Start":"47:27.000 ","End":"47:30.629","Text":"We\u0027re powering through if you need to take a break,"},{"Start":"47:30.629 ","End":"47:33.734","Text":"please, if you haven\u0027t already pause the video."},{"Start":"47:33.734 ","End":"47:38.279","Text":"Our total F vector is going to be"},{"Start":"47:38.279 ","End":"47:44.985","Text":"simply our F_2x in the x-direction,"},{"Start":"47:44.985 ","End":"47:50.249","Text":"plus our F_2y in the y-direction,"},{"Start":"47:50.249 ","End":"47:55.544","Text":"plus our F_2z in the z-direction."},{"Start":"47:55.544 ","End":"48:01.830","Text":"Let\u0027s write that out because we can still simplify even more."},{"Start":"48:01.830 ","End":"48:06.870","Text":"First of all, we can see that 3k is a common multiple over here,"},{"Start":"48:06.870 ","End":"48:08.520","Text":"so we put that out here."},{"Start":"48:08.520 ","End":"48:15.105","Text":"Then we can see that everywhere we have p_1.p_2 divided by r^5."},{"Start":"48:15.105 ","End":"48:16.320","Text":"Let\u0027s write that down,"},{"Start":"48:16.320 ","End":"48:23.610","Text":"p_1.p_2 divided by our r-tilde^5."},{"Start":"48:23.610 ","End":"48:31.270","Text":"Then here we have x_2 minus x_1 in the x-direction,"},{"Start":"48:38.240 ","End":"48:45.959","Text":"plus here we have y_2 minus y_1 in the y-direction plus"},{"Start":"48:45.959 ","End":"48:54.609","Text":"here we have z_2 minus z_1 in the z-direction."},{"Start":"48:55.070 ","End":"48:58.365","Text":"What exactly is this?"},{"Start":"48:58.365 ","End":"49:01.485","Text":"All of this is simply equal to"},{"Start":"49:01.485 ","End":"49:09.859","Text":"our r-tilde vector which is x_2 minus x_1 and the x this y in the y-direction,"},{"Start":"49:09.859 ","End":"49:13.169","Text":"in this set component in the z-direction."},{"Start":"49:13.450 ","End":"49:16.759","Text":"All of this as just our r-tilde vector."},{"Start":"49:16.759 ","End":"49:20.005","Text":"I\u0027m going to rub all of this out and substitute that in."},{"Start":"49:20.005 ","End":"49:23.159","Text":"Then the next column that we\u0027re dealing with."},{"Start":"49:23.159 ","End":"49:29.354","Text":"We have plus, so what do we have everywhere as p_2.r vector,"},{"Start":"49:29.354 ","End":"49:36.600","Text":"p_2.r vector, p_2.r vector and all of them are also divided by r-tilde^5."},{"Start":"49:36.600 ","End":"49:43.940","Text":"We have p_2.r vector divided by r-tilde^5."},{"Start":"49:43.940 ","End":"49:48.720","Text":"Then we have p_1x in the x-direction,"},{"Start":"49:48.720 ","End":"49:51.734","Text":"so p_1x in the x-direction,"},{"Start":"49:51.734 ","End":"49:55.379","Text":"p_1y in the y-direction."},{"Start":"49:55.379 ","End":"49:58.064","Text":"Plus p_1y in the y-direction."},{"Start":"49:58.064 ","End":"50:02.730","Text":"This is z and this is of course z,"},{"Start":"50:02.730 ","End":"50:06.869","Text":"just substitute in the zs over here."},{"Start":"50:06.869 ","End":"50:14.909","Text":"Plus p_1z in the z-direction."},{"Start":"50:15.160 ","End":"50:17.600","Text":"What exactly is this?"},{"Start":"50:17.600 ","End":"50:19.460","Text":"This is the x, y,"},{"Start":"50:19.460 ","End":"50:27.644","Text":"and z components of our p_1 dipole in the correct directions that it\u0027s meant to be."},{"Start":"50:27.644 ","End":"50:34.994","Text":"All of this is simply equal to my p_1 vector or my p_1 dipole."},{"Start":"50:34.994 ","End":"50:39.730","Text":"Now I\u0027m going to rub all of this out and substitute in p_1 vector."},{"Start":"50:39.800 ","End":"50:47.280","Text":"Now the next column has the exact same thing except instead of having p_2.r vector,"},{"Start":"50:47.280 ","End":"50:49.754","Text":"we have p_1.r vector."},{"Start":"50:49.754 ","End":"50:57.315","Text":"Let\u0027s add plus p_1.r vector divided by r^5."},{"Start":"50:57.315 ","End":"51:00.659","Text":"Then we have p_2x in the x-direction plus"},{"Start":"51:00.659 ","End":"51:04.530","Text":"p_2y in the y-direction plus p_2z in the z-direction."},{"Start":"51:04.530 ","End":"51:06.855","Text":"That\u0027s just our p_2 vector,"},{"Start":"51:06.855 ","End":"51:11.039","Text":"we get it the same as how we did it in the last step and we\u0027re almost done."},{"Start":"51:11.039 ","End":"51:13.380","Text":"This is so exciting kind of,"},{"Start":"51:13.380 ","End":"51:15.405","Text":"we have another question to answer."},{"Start":"51:15.405 ","End":"51:18.735","Text":"Here again, we have negative 5 everywhere."},{"Start":"51:18.735 ","End":"51:28.905","Text":"We have p_1.r vector and p_2.r vector and everything is also divided by r^5."},{"Start":"51:28.905 ","End":"51:31.997","Text":"We can just write this like so."},{"Start":"51:31.997 ","End":"51:34.415","Text":"We have p_1."},{"Start":"51:34.415 ","End":"51:38.615","Text":"r vector and then p_2.r"},{"Start":"51:38.615 ","End":"51:45.570","Text":"vector-tilde of course and then divide it by r^7."},{"Start":"51:45.570 ","End":"51:49.544","Text":"Then we have x_2 minus x_1 in the x-direction,"},{"Start":"51:49.544 ","End":"51:51.719","Text":"plus y_2 minus y_1 in the y-direction,"},{"Start":"51:51.719 ","End":"51:53.699","Text":"plus z_2 minus z_1 in"},{"Start":"51:53.699 ","End":"51:59.744","Text":"the z-direction which is exactly what we saw when we did our first column."},{"Start":"51:59.744 ","End":"52:04.659","Text":"That is just our r-tilde vector."},{"Start":"52:04.700 ","End":"52:10.275","Text":"You\u0027ll be surprised to know that we can further simplify this."},{"Start":"52:10.275 ","End":"52:15.390","Text":"What we\u0027re going to see is that we can get the same denominator and everything."},{"Start":"52:15.390 ","End":"52:22.424","Text":"What we\u0027re going to do is we\u0027re going to change r vector into r-hat."},{"Start":"52:22.424 ","End":"52:28.889","Text":"We\u0027ll again have p_1.p_2 divided by,"},{"Start":"52:28.889 ","End":"52:33.164","Text":"so if we have over here r-hat-tilde,"},{"Start":"52:33.164 ","End":"52:38.760","Text":"then here we have r-tilde^4 plus the same thing over here,"},{"Start":"52:38.760 ","End":"52:43.029","Text":"so p_2 vector dot."},{"Start":"52:43.160 ","End":"52:46.604","Text":"Then here instead of r vector,"},{"Start":"52:46.604 ","End":"52:51.240","Text":"we\u0027re going to have r-hat-tilda divided by"},{"Start":"52:51.240 ","End":"52:58.799","Text":"r^4 multiplied by p_1 vector plus p_1 vector dot."},{"Start":"52:58.799 ","End":"53:08.580","Text":"Again r-hat, so r-hat-tilde divided by r^4p_2 vector minus 5."},{"Start":"53:08.580 ","End":"53:10.470","Text":"Here we have an r vector,"},{"Start":"53:10.470 ","End":"53:11.745","Text":"here we have an r vector,"},{"Start":"53:11.745 ","End":"53:13.289","Text":"and here we have an r vector."},{"Start":"53:13.289 ","End":"53:16.260","Text":"In the numerator we have 3r vectors,"},{"Start":"53:16.260 ","End":"53:18.624","Text":"so let\u0027s change it to r-hat."},{"Start":"53:18.624 ","End":"53:28.650","Text":"We have this multiplied by p_2 vector.r-hat-tilde,"},{"Start":"53:28.650 ","End":"53:32.880","Text":"all of this in the direction of r-hat-tilde."},{"Start":"53:32.880 ","End":"53:36.390","Text":"Then because we got rid of 3r vectors,"},{"Start":"53:36.390 ","End":"53:38.145","Text":"we get rid of 3 powers,"},{"Start":"53:38.145 ","End":"53:40.064","Text":"so we have r^4."},{"Start":"53:40.064 ","End":"53:43.400","Text":"Of cause 7 minus 3 is 4."},{"Start":"53:43.400 ","End":"53:46.705","Text":"If this wasn\u0027t enough,"},{"Start":"53:46.705 ","End":"53:48.654","Text":"now that we have this common denominator,"},{"Start":"53:48.654 ","End":"53:52.490","Text":"we can take it outside of the brackets."},{"Start":"53:54.510 ","End":"53:57.400","Text":"This is what it looks like,"},{"Start":"53:57.400 ","End":"54:00.755","Text":"and this is an answer."},{"Start":"54:00.755 ","End":"54:05.160","Text":"Let\u0027s just go back 1 second to the question."},{"Start":"54:05.160 ","End":"54:09.424","Text":"Our original question was,"},{"Start":"54:09.424 ","End":"54:13.510","Text":"what is the force acting on P_2 and P_1?"},{"Start":"54:13.510 ","End":"54:16.060","Text":"Let\u0027s scroll back down."},{"Start":"54:16.060 ","End":"54:22.240","Text":"This is the force of course, acting on P_2."},{"Start":"54:22.240 ","End":"54:26.545","Text":"Therefore, the force acting on P_1,"},{"Start":"54:26.545 ","End":"54:32.425","Text":"so it\u0027s just going to be equal and opposite from Newton\u0027s third law."},{"Start":"54:32.425 ","End":"54:39.100","Text":"We\u0027re going to have that F_2 is equal to negative F_1,"},{"Start":"54:39.100 ","End":"54:43.070","Text":"so negative the force acting on the first dipole."},{"Start":"54:43.380 ","End":"54:48.654","Text":"Ladies and gentlemen, this was the answer to the penultimate question."},{"Start":"54:48.654 ","End":"54:55.400","Text":"We have 1 more question to solve and it\u0027s going to be shorter than this one, I promise."},{"Start":"54:56.010 ","End":"55:00.565","Text":"Question 4 goes like so,"},{"Start":"55:00.565 ","End":"55:03.340","Text":"what is the force on P_2,"},{"Start":"55:03.340 ","End":"55:05.065","Text":"the dipole P_2,"},{"Start":"55:05.065 ","End":"55:09.355","Text":"if P_2 and P_1 are parallel,"},{"Start":"55:09.355 ","End":"55:15.835","Text":"and it\u0027s also parallel to the direction of our r Tilde vector."},{"Start":"55:15.835 ","End":"55:17.680","Text":"The vector is P_2,"},{"Start":"55:17.680 ","End":"55:20.215","Text":"P_1 and r Tilde vector, all parallel."},{"Start":"55:20.215 ","End":"55:23.090","Text":"What\u0027s the force on P_2?"},{"Start":"55:23.580 ","End":"55:25.749","Text":"This is our first case,"},{"Start":"55:25.749 ","End":"55:27.400","Text":"and then we have another case."},{"Start":"55:27.400 ","End":"55:31.610","Text":"We\u0027re dealing with 4, a."},{"Start":"55:32.670 ","End":"55:35.680","Text":"Let\u0027s draw this out."},{"Start":"55:35.680 ","End":"55:41.530","Text":"Here we\u0027re being told that P_1 is"},{"Start":"55:41.530 ","End":"55:50.080","Text":"parallel to P_2 and it\u0027s also parallel to our r Tilde vector."},{"Start":"55:50.080 ","End":"55:54.085","Text":"If we draw our axes like so,"},{"Start":"55:54.085 ","End":"55:57.684","Text":"so what we\u0027re going to have is,"},{"Start":"55:57.684 ","End":"56:03.395","Text":"let\u0027s say we have P_1 in this direction,"},{"Start":"56:03.395 ","End":"56:08.910","Text":"P_2 in the same parallel direction,"},{"Start":"56:08.910 ","End":"56:17.400","Text":"and our r_1 also all in the same direction like so."},{"Start":"56:17.400 ","End":"56:19.035","Text":"Sorry, not r_1,"},{"Start":"56:19.035 ","End":"56:21.184","Text":"our r Tilde vector."},{"Start":"56:21.184 ","End":"56:28.839","Text":"Then we can just simply write this out as P_1 vector being equal to the magnitude"},{"Start":"56:28.839 ","End":"56:36.145","Text":"of P_1 vector and in the r Tilde vector direction."},{"Start":"56:36.145 ","End":"56:40.075","Text":"Or we can just write it with a hat,"},{"Start":"56:40.075 ","End":"56:45.039","Text":"and that our P_2 vector is equal to the magnitude of our P_2 vector,"},{"Start":"56:45.039 ","End":"56:47.860","Text":"also in this r Tilde direction."},{"Start":"56:47.860 ","End":"56:53.934","Text":"That r Tilde a vector is equal to the magnitude of"},{"Start":"56:53.934 ","End":"57:00.530","Text":"r Tilde vector in the r Tilde direction."},{"Start":"57:01.410 ","End":"57:06.625","Text":"Now we just have to plug all of this in to our equation over here."},{"Start":"57:06.625 ","End":"57:09.804","Text":"What we\u0027ll get is that let\u0027s write it over here,"},{"Start":"57:09.804 ","End":"57:12.835","Text":"that F_2 is equal to,"},{"Start":"57:12.835 ","End":"57:20.995","Text":"so we have 3k divided by r Tilde to the power 4 multiplied by P_1 vector dot P_2 vector."},{"Start":"57:20.995 ","End":"57:24.475","Text":"We\u0027ll have P_1, P_2 magnitude."},{"Start":"57:24.475 ","End":"57:28.285","Text":"Then the r hats will drop-off and then multiply by r hat."},{"Start":"57:28.285 ","End":"57:29.620","Text":"We have P_1,"},{"Start":"57:29.620 ","End":"57:33.680","Text":"P_2, r hat Tilde."},{"Start":"57:35.010 ","End":"57:42.115","Text":"Then we\u0027ll have plus P_2 dot P_1."},{"Start":"57:42.115 ","End":"57:45.820","Text":"Again, we\u0027re just going to be left with P_1,"},{"Start":"57:45.820 ","End":"57:52.220","Text":"P_2, and then we get the r hat Tilde."},{"Start":"57:52.650 ","End":"57:57.070","Text":"Sorry, from this. plus again P_1,"},{"Start":"57:57.070 ","End":"57:59.470","Text":"P_2 dot r hat,"},{"Start":"57:59.470 ","End":"58:01.015","Text":"so it\u0027s the same as what we had here."},{"Start":"58:01.015 ","End":"58:02.860","Text":"We can multiply this by 2,"},{"Start":"58:02.860 ","End":"58:07.030","Text":"and then what we\u0027ll have is negative 5."},{"Start":"58:07.030 ","End":"58:10.619","Text":"Then we have p1 dot i-hat."},{"Start":"58:10.619 ","End":"58:12.390","Text":"So we\u0027re left with P_1,"},{"Start":"58:12.390 ","End":"58:13.920","Text":"P_2 dot r hat,"},{"Start":"58:13.920 ","End":"58:15.300","Text":"we\u0027re left with P_2,"},{"Start":"58:15.300 ","End":"58:18.045","Text":"and then here we have the r hat."},{"Start":"58:18.045 ","End":"58:23.000","Text":"Again, we\u0027re going to have P_1,"},{"Start":"58:23.000 ","End":"58:26.505","Text":"P_2 r hat Tilde."},{"Start":"58:26.505 ","End":"58:28.720","Text":"Then we can just simplify this,"},{"Start":"58:28.720 ","End":"58:34.555","Text":"so we have here 1 plus 2 minus 5."},{"Start":"58:34.555 ","End":"58:39.010","Text":"What we have is negative 3 multiplied by this 3,"},{"Start":"58:39.010 ","End":"58:44.875","Text":"so negative 9k divided by r Tilde to the power of 4,"},{"Start":"58:44.875 ","End":"58:47.650","Text":"multiplied by P_1,"},{"Start":"58:47.650 ","End":"58:52.180","Text":"P_2 r hat Tilde."},{"Start":"58:52.190 ","End":"58:55.470","Text":"This is our answer to the first case."},{"Start":"58:55.470 ","End":"58:58.605","Text":"The second case is where P,"},{"Start":"58:58.605 ","End":"59:01.335","Text":"we\u0027re finding again the force on P_2,"},{"Start":"59:01.335 ","End":"59:05.350","Text":"but this time P_2 and P_1 are parallel,"},{"Start":"59:05.350 ","End":"59:10.180","Text":"but they\u0027re perpendicular to the r Tilde vector."},{"Start":"59:10.180 ","End":"59:13.030","Text":"Of course here we had 3 minus 5,"},{"Start":"59:13.030 ","End":"59:15.430","Text":"okay, so this is meant to be, sorry,"},{"Start":"59:15.430 ","End":"59:16.885","Text":"this is minus 5,"},{"Start":"59:16.885 ","End":"59:20.350","Text":"but here it\u0027s meant to be in total inside these brackets,"},{"Start":"59:20.350 ","End":"59:23.110","Text":"negative 2 multiplied by 3."},{"Start":"59:23.110 ","End":"59:26.155","Text":"Here we\u0027re meant to have negative 6k."},{"Start":"59:26.155 ","End":"59:31.690","Text":"Now let\u0027s go into our second case. Here we have b."},{"Start":"59:31.690 ","End":"59:37.600","Text":"Here we have that P_1 vector is parallel to P_2 vector."},{"Start":"59:37.600 ","End":"59:44.480","Text":"However, they\u0027re both perpendicular to r Tilde vector."},{"Start":"59:45.390 ","End":"59:51.039","Text":"If we draw out our axes like so,"},{"Start":"59:51.039 ","End":"59:58.570","Text":"so we can have again that our P_1 vector is like so,"},{"Start":"59:58.570 ","End":"00:02.740","Text":"that our P_2 vector is like so."},{"Start":"00:02.740 ","End":"00:08.920","Text":"But our r Tilde vector cuts at exactly 90 degrees."},{"Start":"00:08.920 ","End":"00:16.705","Text":"What we\u0027ll see is that we\u0027ll get that the force on 2 is equal to,"},{"Start":"00:16.705 ","End":"00:23.020","Text":"again 3k divided by r Tilde to the power of 4."},{"Start":"00:23.020 ","End":"00:30.220","Text":"Then here we have P_1 dot product P_2 where they\u0027re parallel, so that\u0027s fine."},{"Start":"00:30.220 ","End":"00:33.580","Text":"We\u0027ll be left with P_1, P_2."},{"Start":"00:33.580 ","End":"00:38.930","Text":"Then we have in the perpendicular direction."},{"Start":"00:39.060 ","End":"00:41.559","Text":"We don\u0027t know what direction that is,"},{"Start":"00:41.559 ","End":"00:45.594","Text":"but it\u0027s going to be in the r hat Tilde direction."},{"Start":"00:45.594 ","End":"00:49.975","Text":"Then here we have P_2 dot r hat."},{"Start":"00:49.975 ","End":"00:53.950","Text":"Of course we know that that\u0027s going to be equal to 0 because the dot-product"},{"Start":"00:53.950 ","End":"00:58.389","Text":"between 2 vectors which are perpendicular to one another is equal to 0."},{"Start":"00:58.389 ","End":"01:01.630","Text":"Then here again we have P_1 dot r hat,"},{"Start":"01:01.630 ","End":"01:05.200","Text":"so they are also perpendicular."},{"Start":"01:05.200 ","End":"01:09.369","Text":"The dot-product between 2 perpendicular vectors is equal to 0,"},{"Start":"01:09.369 ","End":"01:11.005","Text":"so that also becomes 0."},{"Start":"01:11.005 ","End":"01:18.190","Text":"Then here again we have P_1 and r are perpendicular and P_2 and r are perpendicular,"},{"Start":"01:18.190 ","End":"01:21.320","Text":"so this is also going to be equal to 0."},{"Start":"01:22.470 ","End":"01:27.264","Text":"That is the answer to our second case and I cannot believe it,"},{"Start":"01:27.264 ","End":"01:29.260","Text":"we finished this lesson."},{"Start":"01:29.260 ","End":"01:33.500","Text":"Well done. It was truly a difficult lesson."}],"ID":22335}],"Thumbnail":null,"ID":99476}]

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