00:01
Hello, everyone.
00:01
In this problem, we're asked to find a solution for the posa equation, which is written over here.
00:09
This is generally the form that you find a posa equation in for a electrostatic situation where you have some charge distribution, denoted by a row of r, depicted by this little blob off of charge.
00:25
And you're interested in the potential or the field that it produces, the electric field that it produces at a distance r minus r away from that source.
00:36
So here, i've just set up a coordinate system, general coordinate system, where i have some origin somewhere, and there is a vector r prime that is pointing to an element of the charge distribution.
00:49
And there's an other vector r, which is pointing to the position of interest where we want to find a field.
00:56
And so we know that, you know, by kulam's law, et cetera, etc., that the potential is going to go like 1 over r.
01:03
And so we actually know that the way that this distribution affects the electric field over here at this point, let's call it p, is through the distance or is related in some form to the distance of r minus r prime.
01:22
So it's the distance between the charge element and the position of interest.
01:28
Okay, so this is the scenario that we have, and so we want to find a solution 5r somehow.
01:34
So the way we do this is through greens function method, and so the greens function is the solution or the function on which if you apply the differential operator whose green function you're looking for in case, in this case, the lephasian, gives you a delta function.
01:54
So imagine that i have some generic operator instead of the laplasium.
01:59
If i apply the operator to its screens function, then i always get back to the alpha function.
02:07
Okay.
02:08
So the reason this works is because if i propose that the solution, phi of r, is equal to minus the integral of dr prime, grr prime, row of r prime over epsilon zero.
02:22
So this is just, i'm just imagining integrating this somehow through the greens function or like wading the right -hand side by the greens function and then integrating.
02:35
Then what i'm going to get back is that taking the laplacian of 5r with respect to the r coordinates is going to be equal to minus the loplasian of this whole integral over here.
02:47
But over here on the inside the integral, nothing depends on r except for g of r.
02:52
R and r prime.
02:54
Everything else depends on r prime.
02:56
And so i can pull a lapsian operator into the integral and just apply it to g of r and r prime.
03:02
But we know that that is a delta function.
03:04
And so then the delta function collapses the integral and we get back the solution, or we get back the differential equation, right? so we have that grete square of r5r is equal to minus a row of r.
03:18
Excuse me, so this should be a row of r.
03:20
So row of r over epsilon zero.
03:23
So we can use the greens function to solve this differential equation.
03:30
And although i'm not going to show it here, but you can find many explanations for this online, the greens function for the lassian in three -dimensional space is this function, which is minus 1 over 4 pi times 1 over r minus r prime magnitude.
03:46
So this, this notation, r minus r prime, is just the length.
03:53
Length of the vector.
03:54
So that's the length of this red vector that i've drawn here in the beginning.
03:59
Okay, it's the distance between the charge distribution element and the position of interest.
04:04
And so with that in mind, we now know or we find that the solution to the laplacian or the the poisson equation is going to be phi of r is equal to 1 over 4x0 0 times the integral of d3r prime times row of r prime over the magnitude of r minus.
04:23
Is r prime and vector.
04:25
So this is the solution in its integral form.
04:31
Okay, so this is the first part.
04:33
So we check that.
04:35
And so for the next part, we are asked to find, or we are asked to show that this magnitude of this vector, or the magnitude of this vector can be expanded in a series of r prime over r provided that r is much greater than r prime.
04:50
So the magnitude of the vector is the vector squared, square rooted.
04:56
So that means that i'm just taking the vector, the vector's dot product with itself, and then taking the square root of the result.
05:05
And that's going to give me the magnitude of the vector.
05:07
So that's exactly what i'm doing over here.
05:10
So taking this dot product over here, you find r.
05:14
.r prime, that r prime and minus 2 r.
05:17
Sorry, r.
05:19
.r prime.
05:20
And all of that under the square root or to the power of 1 now, r .r .r .r is r squared and r prime .r prime is r prime squared.
05:30
So then what i do is a factorized r squared because i'm proposing that r prime over r is going to be very small, as we will see later.
05:41
So in this situation, then it is a good idea to take out the big factor, so take out r in order to have a series in the small quantity, which is going to be a good idea.
05:54
R prime over r...