00:01
Hi there, so for this problem we are told that a uniform line charge lambda is placed on an infinite straight wire at a distance d above a grounded conducting plane, as is shown in this figure.
00:16
And for part a of this problem, we need to find the potential in the region above the plane.
00:27
So what we are going to do in here is to transform this into an image problem.
00:34
So that is lambda above and minus lambda below.
00:40
So that we can transform this into this following.
00:46
I'm just writing here.
00:48
This is the y direction.
00:51
This is the c direction.
00:53
So we're writing here positive lambda.
00:58
And a negative lambda.
01:01
And then we want the potential in a point in here that are coordinates y and c.
01:15
And this separation between this and this is the distance d that we are told, this distance right here, d.
01:23
So this is also d.
01:25
So as you can see, we'll have this distance right here.
01:29
In this other distance right here.
01:33
This distance, so we know that the potential can be written at that point is equal to 2 times lambda, because we have two values in here, minus lambda and lambda.
01:59
And this divided by 4 times pi times epsilon subsereal, the neparion logarithm of let's call this, x minus and this adds plus these two distances.
02:13
So it is the neparion logarithm of the distance x minus divided by the distance x positive.
02:22
So what we can do in here is to introduce this two into here by properties of the neparian logarithm.
02:31
So with in lambda divided by four times pi times epsilon sub zero times the neparian logarithm of x minus to the square.
02:40
Divided by x plus to the square.
02:43
And then we can write these distances x minus.
02:48
So we know that x minus, as you can see from the figure, this distance is equal to y square plus c squared plus t and that to the square.
03:04
Of course, we are treating in here this x square...