00:01
So in this question we have a square loop of side length a, and we have a point p at a distance z above the centre of the loop.
00:14
And we have a uniform line charge lambda.
00:19
So we have lambda is the line charge.
00:28
Okay, so we want to work out the electric field.
00:32
So what we can do is we can notice that each of the...
00:36
Lines so each line will give an electric field with a z component and another component in the xy plane but the xy components will cancel out so we only need to keep track of the z component and since each of them is symmetrical under a rotation of 90 degrees they will each give the same z component, said component same for each.
01:34
So we only need to calculate one, multiply it by four, and we're done.
01:38
So let's consider a line and a point.
01:44
The line is of length a, it's at a position x equals minus a over two.
01:51
Our point p is at a position x equals y equals 0, z equals z, and we have a line charge lambda.
02:01
This goes from y equals minus a over 2 to y equals a over 2.
02:09
So we can parameterize this in terms of y.
02:14
And what's the distance from here to here? what's r of y? well r of y squared is equal to y squared plus x squared plus z squared.
02:34
X is equal to minus a over two.
02:36
So this is a squared over four plus z squared plus y squared.
02:47
So now what's the what's the component of the electric field going to be? well, the component of the electric field at a position y is going to be.
03:03
So we're going to have divided by 4 pi epsilon 0, r of y squared, so a squared over 4 plus z squared plus y squared, and then it's in the r direction.
03:25
So what is that going to be? well, it's going to have, so what is this vector r of y? so r of y is p minus the point here.
03:44
So it's going to be 0 ,0 ,0, z minus a over 2, y, 0.
03:54
So that's a over 2 minus y, z.
04:03
So what we can do is we can just look at the z component of this, and this is going to give us z over root a squared over 4 plus z squared plus y squared in the z direction.
04:20
And then there's going to be other bits in the x and y direction, but we don't care about those, because only the z direction is going to stay once we've integrated everything out.
04:33
So then the ez is going to be, now remember we had to multiply by four, because we've got four different lines.
04:45
So four sides of the loop, and then the integral from minus a over 2 to a over 2 of d, e, z by d, y, which is lambda z over 4 pi epsilon nought, a squared over 4 plus z squared plus y squared to the three halves, z hat.
05:16
So let's take the constants out.
05:18
We get lambda over pi epsilon naught, and we can take the z out as well, and an integral from minus a over 2 to a over 2 of 1 over a squared over 4 plus z squared plus y squared to the three halves, d, y.
05:42
So now let's take out a factor of a squared over 4 plus z squared to the three halves.
05:48
Lambda z over pi, epsilon 0, a squared over 4 plus z squared to the three halves.
05:55
Then we integrate from minus a over two to a over two, d .y, one plus y over y squared over four plus z squared to the three halves.
06:11
Now let's write that, because we're looking at one plus something, let's write that y is equal to root a squared over 4 plus z squared times tan theta so that d y is equal to root a squared over 4 plus z squared sec squared theta and then 1 plus y squared over a squared over 4 plus z squared is 1 plus tan squared theta, which is sec squared theta.
07:05
So now we can write that the ez is equal to, we've got this factor of root a squared over four plus z squared, cancel is one of these, lambda z over pi, epsilon n, a squared over four plus z squared, uh, just the power of one...