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All right, so let us continue.



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So in the previous lecture, we have introduced the Hetson dipole and we have calculated the electrostatic



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potential fire and the vector potential, eh?



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So now we can go ahead and use these two equations that correspond to two equations and we can calculate



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our fields, E and B..



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So this is what we finally want.



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So we will start with this.



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B is equal to the rotation of E that is highlighted here because it is the more simple operation.



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So we just take our vector potential here with this -- time argument and we calculate the rotation.



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So basically we just have to calculate the rotation of p dot vector divided by R, but here it's again



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really important to remember that this time argument is also R dependent.



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So this derivative here also acts on the P.



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So what we get here is we have to use yet this identity here for the rotation, so if we calculate the



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rotation of a scalar function times a vector.



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So in our case, this would be one of our times the pitot vector.



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We get these two terms here and as I have written down here, so we get to scale our function and then



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the rotation of the vector minus the vector and then the the vector product with the gradient of the



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scale of function.



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So now we have to calculate the gradient of one of our and we have to calculate the rotation of pitot.



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So here this gives us these two terms, the gradient of one of our this we have already calculated a



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couple of times.



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This is minus our vector divided by our to the power of three.



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And then we get here for the rotation of P, we get first the outer derivative, which is the time derivative



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of P dot, which is P double dots.



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And then we also have to consider the inner derivative which is um yeah.



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Here this t does not give us anything because it's, it's not dependent on R but this term here again



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gives us something.



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So we have to calculate minus one of the C times the gradient of our gradient of are we already had.



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This is our vector divided by R so we get this expression here and once again consider at least the



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time argument with the retardation.



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So now we can simplify, we can.



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Holders are divided by r out of the brackets because we have it here and here.



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And so our final result for the magnetic field looks like this looks kind of simple.



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And we have here a second all the time derivative of P and the first all the time, the time derivative



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of P, which is our electric dipole moment.



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So while this calculation of B was kind of kind of simple, just like four or five steps, the calculation



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of E will be a bit more difficult because you can see we need here the time derivative of A and we need



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the gradient of.



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So first we have to calculate these two things.



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So the time derivative of A is actually really, really simple because the only time dependent thing



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in A is P dot.



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So a dot is just dependent on pedulla dot.



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So that's really all you have to know.



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It's just straightforward writing here.



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Another dot.



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OK, but the calculation of the gradient, a fight is more difficult.



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So here this is because we already had two terms and is fine and now we have to calculate the gradient



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of this whole expression here.



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So what we have to calculate is first we have to calculate the gradient of our dot dots divided by our



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square.



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And then we have to calculate the gradient of our dot p divided by R to the power of three.



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So you can see already we have to gradients of functions that have very similar shapes.



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So it turns out when we calculate this one here, it's done really simple to calculate only also this



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one because we have to apply exactly, exactly the same steps.



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We just have to be careful because here we have P dots.



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Here we have only P and here we have a path to and here we have a panel of three.



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But still now what we do is we use this identity for the gradient, so if we calculate the gradient



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of such a scale product of two vectors, we actually get four terms.



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And since we have here to two terms already, we get in total eight terms.



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So I have used now this law here to write down all of these eight terms.



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And here I have used that our AI is P dot or P P Dot and Pete and our B is our vector divided by our



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square or our vector divided by R to the power of three.



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All right.



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So now we have to go ahead and calculate all of these eight terms and we will.



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So I will I will start with the very first term here.



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And we calculate it in Cartesian coordinates.



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And you will see that this takes a lot of effort and is quite difficult.



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So for all of the other terms, we will then turn to a circle coordinates and make our lives very much



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easier.



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So this will then take not so much time.



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But I think for this term, it's it's just a good practice to just tried to calculate it once and Cartesian



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coordinates to really see what is happening here, because this is quite an unusual term that you don't



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see so often, because if you would not have these brackets here, you would have the gradient of a



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vector.



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And that's a bit a bit strange because the gradient is a vector and then you calculate a vector vector.



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So that's really a bit difficult.



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So you need to be careful that you first calculate this DOT product here.



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So it means we get P extorts times, this partial derivative with respect to X, and then we get these



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two other terms where we have replaced the X with A, Y and Z.



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And now we have to sum here because the DOT product is of course a sum.



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And these three terms all have to act on this vector here, which is the vector X, Y, Z divided by



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our square.



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OK, so we must calculate now these three operations on all of these three terms.



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So this will give us nine independent calculations.



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But also you need to be careful because this hour vector is actually so our square is actually X squared



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plus Y squared plus C squared, because you need to be careful because this is also X, Y and Z dependent.



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So the derivatives will not be zero here.



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OK, so you get four.



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Let's maybe let's start with the first was the first operation where we have to calculate P, dot,



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x and then the X derivative of X divided by our square.



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So as I already mentioned, we have here in art that is also X dependent.



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So we have a product or quotient of two X dependant functions.



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So we need to once again apply the product rule.



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So the first term is I write down minus one of our square times to the X derivative of X, which is



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of course one.



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So this gives us one of our square and we have to multiply also by accident.



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That's this term here.



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And then we get the other term where we write X times the X dependence of one overall square.



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So now we have to calculate the exact derivative of one of our square and one of our square is one divided



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by X squared plus Y squared plus Z square.



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So we get then in the denominator, we have to increase the the power of the whole thing, which is



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which is now minus one.



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We have to decrease it by one.



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So we get minus two.



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So this gives us now an our vector here to the power of four because the vector is actually the square



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root.



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OK, and then we need to write this minus one also as a prefect here and we have to calculate the inner



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derivative, which is four X square, two X.



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This is where this one comes from here.



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So this may sound a bit confusing, but please go ahead, take a sheet of paper, write down right on



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this one here as X divided by X squared plus Y square, Plessey Square, and then do the chain rule



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yourself and you will see you will get this term.



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And don't forget the dot from here.



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Now for the other terms where this derivative X on this one and this one on this one, we do not have



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to apply, have to apply the product rule because X is of course not Y dependent.



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So we just get a single term, which is essentially the same one as this one.



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But here we get the Y because the inner derivative is here because of this y it's two Y, so we get



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to hear multiple terms that are very similar.



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Here we have a Z instead of the Y and the X, and then for the four, the second component, we get



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essentially the same thing for yapped for these three terms.



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But you have to, of course, consider the second term of the product rule by using disintegrate.



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This differentiation here.



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Yeah.



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So in total, we get four terms for all of these components and now we can simplify what is written



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down here.



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So these first things here, this is actually the p dot vector.



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So because we have pipsy, dot, dot, dot, dot, dot, dot vector divided by our square and then these



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things here, these are minus two in all cases, minus two times.



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And then we have here the OK, here we have.



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X times X, plus Y times P, Y dot, plus C times dot, and we have to send every single component and



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this X times dot plus Y Y plus this is actually just ah dot product P dot.



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And then we just have here is a pre vector x, y, z, x, y, z, x, y, z, x, y, z.



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And don't forget the denominator are two the power of four.



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So this is what we have and this is of course here the position vector.



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So we get as a final result, prad vector divided by our square minus two times our dot plot and our



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vector divided by our to the power of four.



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OK, wow, that was really difficult, but I think it's really purposeful practice if you try yourself



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and sort of recommended that you follow along these steps to see that this is really true.



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But of course we have seven other terms in front of us and we do not want to do such a difficult calculation



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seven more times.



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So for the only time in this course, I will turn to spherical coordinates and I will show you a nice



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trick.



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So if you really know about spherical coordinates, you can derive that enable operate and spherical



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coordinates.



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Looks like this one here.



146

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You in spherical coordinates.



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You do not have the X, Y and Z components, but you have the unit vectors along the radial direction



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along the as you move that direction and along the polar direction.



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And all of these three components will have of course, individual derivatives for this KNOBLER operator.



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However, now it's really important to realize that all of our terms here depend only on R and we have



151

00:12:47,300 --> 00:12:49,280

no theta and no fi dependent's.



152

00:12:49,820 --> 00:12:57,030

So this means when we let this novela operate to act on on our terms here, these will always give zero.



153

00:12:57,650 --> 00:13:04,040

So this means we can just consider our Nabala operator to be e r times this partial derivative or respect



154

00:13:04,190 --> 00:13:10,910

to this radius r and you can write of course E are also as the correct divided by R.



155

00:13:12,800 --> 00:13:16,970

All right, so now we can use this and go ahead and calculate the other terms.



156

00:13:17,600 --> 00:13:20,350

For example, we can calculate now this one here.



157

00:13:20,780 --> 00:13:21,760

Yeah, this one here.



158

00:13:22,460 --> 00:13:28,420

So we write down instead of Arnab operator are divided by our times, this derivative here.



159

00:13:29,150 --> 00:13:37,040

So we have now our times, our vector divided by R-squared, divided by R and R, the times our vector



160

00:13:37,280 --> 00:13:38,810

is just R squared.



161

00:13:39,620 --> 00:13:44,150

So we get one of our in total then this are derivative of p dot.



162

00:13:45,370 --> 00:13:53,280

And this is, of course, yeah, because this is now not not the gradient here, but it's really just



163

00:13:53,290 --> 00:13:59,290

a partial derivative with respect to our we get the outer derivative, which is our double dots, and



164

00:13:59,290 --> 00:14:07,060

the inner derivative is just minus one over three times partial derivative of R with respect to our



165

00:14:07,180 --> 00:14:09,370

which is one to our total result.



166

00:14:09,370 --> 00:14:10,260

Is that one here.



167

00:14:12,010 --> 00:14:18,080

Now for the next term, we need to apply a bit, a bit more thought.



168

00:14:18,400 --> 00:14:28,510

So here we have here we have are divided by our square then the vector product of the operator and vector



169

00:14:28,510 --> 00:14:29,680

product of P Dot.



170

00:14:30,520 --> 00:14:38,110

And so once again we just take this one here instead of an apple operator and put it in so we can pull



171

00:14:38,110 --> 00:14:39,820

out the one of ours.



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00:14:39,820 --> 00:14:48,700

We get here one of our three, then we get our vector product with our vector product with this our



173

00:14:48,700 --> 00:14:50,930

derivative of P Dot and this.



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00:14:50,950 --> 00:14:57,610

In fact, we have already calculated here in this step there, we have found out that and this derivative



175

00:14:57,610 --> 00:15:03,280

of pi dot dot is just minus one of the C times P double dot.



176

00:15:04,180 --> 00:15:05,490

So this is what we get.



177

00:15:05,500 --> 00:15:06,220

And here.



178

00:15:07,870 --> 00:15:17,380

Here, you know, we can just use another law for this double vector product where we can write it down



179

00:15:17,520 --> 00:15:20,260

in terms of B, A, C minus CAIB.



180

00:15:20,300 --> 00:15:21,790

This is one of the famous rules.



181

00:15:22,300 --> 00:15:30,940

And since we have here actually A equal to B, we can also write it down as a times dot product of A



182

00:15:30,940 --> 00:15:37,270

C minus C times dot product of A, which is just the absolute value of a square.



183

00:15:38,800 --> 00:15:46,870

And since here are a is are we get for this one the absolute value of our square.



184

00:15:47,320 --> 00:15:52,360

So this is why in this term here we have all squared divided by R to the power of three, which is just



185

00:15:52,360 --> 00:15:53,200

one of our.



186

00:15:54,720 --> 00:16:01,860

OK, so this is what we get for this term and know the last term, that is the most easy one we use



187

00:16:01,860 --> 00:16:05,120

again, this one here in terms of our KNOBLER operator.



188

00:16:05,430 --> 00:16:13,950

And now you can see we have here the vector product of our and some things and are and some vectors.



189

00:16:14,520 --> 00:16:21,420

And so we have a vector product of R and R, and since these two are of course parallel.



190

00:16:21,660 --> 00:16:26,670

So they because they are the same today, of course parallel, then the vector product will always be



191

00:16:26,670 --> 00:16:27,100

zero.



192

00:16:27,340 --> 00:16:29,190

And so this whole thing will be zero.



193

00:16:30,650 --> 00:16:37,670

OK, so this is what we get, we have our four terms now and now we can calculate the other four terms



194

00:16:37,670 --> 00:16:43,700

in the exact same way and really not much changes here.



195

00:16:44,030 --> 00:16:50,330

Of course, for all the peace, you will have to remove one of the dots.



196

00:16:50,330 --> 00:16:56,760

So you will get here p p dot, dot, dot, where you have here, p p double, double, double that.



197

00:16:57,470 --> 00:17:04,610

And you also have to take care of this different exponent here in this R, so here you have three three,



198

00:17:04,610 --> 00:17:07,230

three three and here you had two everywhere.



199

00:17:08,180 --> 00:17:14,460

So this means you will typically have to increase here the power of all of these one of ours.



200

00:17:15,200 --> 00:17:18,580

So yeah, it's basically everywhere you can you can see.



201

00:17:19,130 --> 00:17:22,070

And then here for this term, you need to be really careful.



202

00:17:22,370 --> 00:17:26,090

Maybe you could you you write it down yourself and take your time.



203

00:17:26,090 --> 00:17:31,340

But of course you will get to hear different prefecture because here you have now two, three instead



204

00:17:31,340 --> 00:17:33,410

of the two instead of this two.



205

00:17:33,410 --> 00:17:33,830

I'm sorry.



206

00:17:35,660 --> 00:17:36,080

All right.



207

00:17:36,090 --> 00:17:37,160

That was quite an effort.



208

00:17:37,160 --> 00:17:38,450

But we have finally done it.



209

00:17:38,780 --> 00:17:46,010

We have written down we have calculated all of the eight terms so we can now write down what our gradient



210

00:17:46,010 --> 00:17:48,550

of the electrostatic potential looks like.



211

00:17:49,160 --> 00:17:52,400

So we have these six terms because these two give us zero.



212

00:17:52,730 --> 00:17:56,840

And you can also see that some of the terms look exactly the same.



213

00:17:56,990 --> 00:17:58,220

So they will.



214

00:17:58,250 --> 00:17:59,570

So you can simplify them.



215

00:18:00,320 --> 00:18:04,250

For example, you have this term year and this term here.



216

00:18:05,750 --> 00:18:12,680

Yeah, because here you see you also have a one of a C, so this term and this one can be transformed



217

00:18:12,680 --> 00:18:13,400

to.



218

00:18:13,760 --> 00:18:13,880

Yeah.



219

00:18:14,000 --> 00:18:15,590

Basically this term with a three.



220

00:18:16,740 --> 00:18:21,260

And when you do this really carefully, you can write down all of these terms and you will see that



221

00:18:21,260 --> 00:18:25,790

you have five terms that have different functional dependencies.



222

00:18:27,720 --> 00:18:34,050

OK, so here we are, we have the electrostatic potential, we have developed a potential, we have



223

00:18:34,770 --> 00:18:43,190

the gradient of fire, we have a dot, and now we have already cut what we had already calculated be



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00:18:43,200 --> 00:18:44,070

the magnetic field.



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And now we can finally calculate the electric field.



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And this is basically just this expression here with a minus sign.



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And then we have to take into account also this single term because we have also minus a dot.



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So this means now, instead of five terms for gradient five, we get six terms.



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And here I have written all of them down and I have ordered them with respect to p p p dot, dot, dot,



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dot, dot, dot.



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And now in the next session, we will discuss the shapes and the dependencies of these two fields,



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B and E, and we'll look at the fields and I will show you what they look like.