# Homework Help: Method of Image Charges

1. Oct 11, 2015

### BOAS

Hi,

I am learning about the method of image charges, but am struggling to sufficiently justify my answer. It appears to be correct, though it quite possibly isn't...

1. The problem statement, all variables and given/known data

Two semi-infinite grounded conducting planes (the shaded region in the figure, corresponding to $x ≤ 0$ or $y ≤ 0)$ meet at right angles. In the region between them (i.e. $x > 0$ and $y > 0$), there is a point charge, at a distance $d$ from each plane, as indicated in the figure. Set the z axis such that the charge q is at $z = 0$.

(a) To compute the potential between the planes you need three image charges. What is their magnitude and where should they be located?

(b) Compute the electric potential between the planes.

(c) Compute the electric charge density σ on the planes.

2. Relevant equations

3. The attempt at a solution

My answer to part (a) is essentially a guess, based on creating a symmetrical situation. I don't know how to justify it properly, and would love some help regarding this.

I have said that I need 2 charges of -q and one of charge +q.

-q (-d, d)
-q (d, -d)
+q (-d, -d)

Are the coordinates that I think they should be placed at.

I then look at the potential, and see that $V(x,y) = \frac{1}{4 \pi \epsilon_{0}} [\frac{q}{\sqrt{(x-d)^{2} + (y-d)^{2}}} - \frac{q}{\sqrt{(x+d)^{2} + (y-d)^{2}}} - \frac{q}{\sqrt{(x-d)^{2} + (y+d)^{2}}} + \frac{q}{\sqrt{(x+d)^{2} + (y+d)^{2}}}]$

which satisfies the conditions that at $V(0,y) = 0$ and $V(x,0) = 0$

Am I on the right track here?

Thanks!

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2. Oct 11, 2015

### BvU

Looks good to me !

3. Oct 11, 2015

### BOAS

That is good to hear! I do feel like I have cheated somewhat though. Is there a rigorous method to go about this, or is a guess generally acceptable provided it can be shown to satisfy the conditions?

For part (c) I need to find the charge distributions on each plane. I have done that in the same way I would for a single plane, and the expressions look like they make physical sense to me.

$\sigma_{x} = - \epsilon_{0} \frac{\partial V}{\partial y}|_{y = 0} = - \frac{q}{2 \pi} [\frac{d}{((x - d)^{2} + d^{2})^{3/2}} - \frac{d}{((x + d)^{2} + d^{2})^{3/2}} ]$

and $\sigma_{y} = - \epsilon_{0} \frac{\partial V}{\partial x}|_{x = 0} = - \frac{q}{2 \pi} [\frac{d}{((y - d)^{2} + d^{2})^{3/2}} - \frac{d}{((y + d)^{2} + d^{2})^{3/2}} ]$

I think this makes sense because as the charge distribution would not be constant along the planes

Last edited: Oct 11, 2015
4. Oct 11, 2015

### BvU

Doesn't look symmetric around y = x !?
try x = d and you'll see the error
 yes the $(x-d)^2 -d^2$ should have been $(x-d)^2+d^2$

Last edited: Oct 11, 2015
5. Oct 11, 2015

### BOAS

Sorry, what doesn't look symmetric around y = x?

Edit - Fixed a typo in my $\sigma_{x}$ expression