# Induced carge on parallel infinite plates

1. Oct 12, 2009

### Brian-san

1. The problem statement, all variables and given/known data
Two infinite grounded parallel conducting planes are separated by a distance d. A point charge q is placed between the planes. Determine the induced surface charge densities as well as the total charges on the two planes.

2. Relevant equations
Potential of a point charge:
$$\Phi=\frac{q}{r}$$

Electric Field/Potential relation:
$$\vec{E}=-\nabla\Phi$$

Electric Field discontinuity:
$$4\pi\sigma$$

3. The attempt at a solution
I chose the x-axis to be perpendicular to the plates, with the plates at ±d/2, and the point charge q at some location a, 0<a<d/2 (it doesn't matter where, but I wanted it on the positive side of the x-axis).

Using the method of images, since both plates are grounded (zero potential), we need a point charge -q at x=d-a to balance the right plate. But then to balance the left plate we need a charge q at x=-2d+a and a charge -q at x=-2d-a. This process needs to be repeated an infinite number of times for the image charges to recreate the boundary conditions. They form a pattern where charges of q are located at x=±2nd+a and charges of -q at x=±(2n+1)d-a, where n is a non-negative integer (0, 1, 2, ...).

So for a point (x,y z) between the two plates, I get the following expression for the potential:
$$\Phi(x,y,z)=q\sum_{n=-\infty}^{\infty}\left(\frac{1}{\sqrt{(x-2nd-a)^2+y^2+z^2}}-\frac{1}{\sqrt{(x-(2n+1)d+a)^2+y^2+z^2}}\right)$$

Since the discontinuity is related to the normal component of the electric field, which will be in the x direction, I'm only worried about the x component of the electric field. So using the gradient relation,
$$\vec{E_x}(x,y,z)=q\sum_{n=-\infty}^{\infty}\left(\frac{x-2nd-a}{((x-2nd-a)^2+y^2+z^2)^{\frac{3}{2}}}-\frac{x-(2n+1)d-a}{((x-(2n+1)d+a)^2+y^2+z^2)^{\frac{3}{2}}}\right)$$

To use the discontinuity fact, we need to know the value of the electric field as we come from the left and right of the plates. So for the plate at d/2, to the right E=0, and from the left, I chose point (d/2, 0, 0) for ease.
$$\vec{E_x}(x,y,z)=4q\sum_{n=-\infty}^{\infty}\left(\frac{1}{(d(1-4n)-2a)^2}-\frac{1}{(d(1+4n)-2a)^2}\right)$$

That expression has a lot of algebraic simplification done to it. On instinct, I believe the series will converge, but it is not of any form I am familiar with, perhaps more simplification is needed. I suppose a better question might be, should I approach the problem in this way with the method of images, or is there some other approach/trick that would work better?

2. Oct 14, 2009

### gabbagabbahey

Your series doesn't look quite right to me...

I this a typo, or did you really use this to calculate your series?

3. Oct 14, 2009

### Brian-san

Yeah, that was a type, I meant -q at x=-d-a, the -q charges are at odd multiples odd multiples of d.

I actually think the first series is wrong anyway, it should have four terms, since there are ± signs for the locations of the image charges, and I moved to cylindrical coordinates with x running along the axis of the cylinder
$$\Phi(x,r)=q\sum_{n=0}^{\infty}\left(\frac{1}{\sqrt{(x-2nd-a)^2+r^2}}+\frac{1}{\sqrt{(x+2nd-a)^2+r^2}}-\frac{1}{\sqrt{(x-(2n+1)d+a)^2+r^2}}-\frac{1}{\sqrt{(x+(2n+1)d+a)^2+r^2}}\right)-\frac{q}{\sqrt{(x-a)^2+r^2}}$$

The final term compensates for the fact that the potential from the actual point charge at x=a is counted twice in the sum when n=0.

Then the electric field in the x direction is just
$$E_x(x,r)=q\sum_{n=0}^{\infty}\left(\frac{(x-2nd-a)}{((x-2nd-a)^2+r^2)^{\frac{3}{2}}}+\frac{(x+2nd-a)}{((x+2nd-a)^2+r^2)^{\frac{3}{2}}}-\frac{(x-(2n+1)d+a)}{((x-(2n+1)d+a)^2+r^2)^{\frac{3}{2}}}-\frac{(x+(2n+1)d+a)}{((x+(2n+1)d+a)^2+r^2)^{\frac{3}{2}}}\right)-\frac{q}{((x-a)^2+r^2)^{\frac{3}{2}}}$$

From the discontinuity of the electric field, I know the surface charge density is just
$$\sigma_R=\frac{-E_x(d/2,r)}{4\pi}, \sigma_L=\frac{E_x(-d/2,r)}{4\pi}$$

All terms in the charge densities are of the form
$$\frac{c}{(c^2+r^2)^{\frac{3}{2}}}$$

Where c is constant, so when we integrate over the surface, we find
$$2\pi\int_0^{\infty}\frac{crdr}{(c^2+r^2)^{\frac{3}{2}}}=\frac{-2\pi c}{\sqrt{c^2+r^2}}_0^{\infty}=2\pi$$

When I use this to integrate the terms in the series for the two surface charge densities, it turns into an infinite sum of zeroes (assuming I can integrate the series term by term), so the only term that contributes is the final one I used to correct for the double counting in the sum. Then end result is induced charges of
$$Q_{left}=-\frac{q}{2}, Q_{right}=\frac{q}{2}$$

4. Oct 14, 2009

### gabbagabbahey

I'm still not too sure abut your series....if the plates are at $x=\pm\frac{d}{2}$, shouldn't you have image charges of alternating sign at $x=d-a,\;\frac{3}{2}d-a,\;2d-a,\;\text{etc.}$ to balance the right-hand plate and image charges of alternating sign at $x=-d+a,\;-\frac{3}{2}d+a,\;-2d+a,\;\text{etc.}$ to balance the left-hand plate, and of course your original charge at $x=a$ leaving you with something like:

$$\Phi(x,y,z)=q\sum_{n=0}^{\infty}\left[\frac{(-1)^{n+1}}{\sqrt {(x+\frac{n+2}{2}d-a)^2+y^2+z^2}}+\frac{(-1)^{n+1}}{\sqrt{(x-\frac{n+2}{2}d+a)^2+y^2+z^2}}\right]+\frac{q}{\sqrt{(x-a)^2+y^2+z^2}}$$

?

5. Oct 14, 2009

### Brian-san

I'm pretty sure on it, I drew a picture of how I reasoned the locations. Blue dots are -q charges and red dots are +q charges, the plates are the thick black lines (sorry for the bad mspaint picture).

#### Attached Files:

• ###### plates.JPG
File size:
10.4 KB
Views:
198
6. Oct 14, 2009

### gabbagabbahey

Can you host it on imageshack.us instead? Attachments usually take a while to get approved.

7. Oct 16, 2009

### gabbagabbahey

Yes, I suppose you will actually have image charges of alternating sign at $x=d-a,\;2d+a,\;3d-a,\;\text{etc.}$ to balance the right-hand plate and image charges of alternating sign at $x=-d-a,\;-2d+a,\;-3d-a,\;\text{etc.}$ to balance the left-hand plate. Giving,

$$\Phi(x,y,z)=q\sum_{n=0}^{\infty}\left[\frac{(-1)^{n+1}}{\sqrt {(x-(n+1)d+(-1)^na)^2+y^2+z^2}}+\frac{(-1)^{n+1}}{\sqrt{(x+(n+1)d+(-1)^na)^2+y^2+z^2}}\right]+\frac{q}{\sqrt{(x-a)^2+y^2+z^2}}$$

Which is equivalent to your series in post #3.