# Homework Help: Potential Energy of the charge

1. May 20, 2010

### Redsummers

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

The potential energy between charges q and q' at distance r is given by:

$$E_p=\frac{1}{4\pi\epsilon_0}\frac{qq'}{r}$$

A regular plane arrangement of alternate positive and negative charges of the same magnitude is obtained by placing the charges at the centre of squares of side a.

Calculate the potential energy of charge A in the electric field created by the others charges for the finite grid (see the attached image).

3. The attempt at a solution

So, we know that the distance between two charges is going to be 2a, as described in the problem. So then, if we work out this equation:

$$U_E=\frac{1}{4\pi\epsilon_0}(\frac{q_1q_2}{r_{1,2}}+\frac{q_1q_3}{r_{1,3}}+...)$$

according to our problem, we get:

$$U_E=\frac{q^2}{4\pi\epsilon_0}(\frac{1}{r_{1,2}}-\frac{1}{r_{1,3}}\pm...)$$

and so on, alternating the signs.

And at this point, I don't know what should I do, it looks rather like a geometrical problem involving a finite series (i.e. converge). It looks like most of the distances are going to cancel, but it's not trivially seen how they cancel, because it's not that we're going to sum the 48 different distances anyway, neither that it looks to give 0 as a result. I guess it should have something to do with Pythagoras' theorem, but I don't see how to apply it here. Any help would be appreciated.

Note: For some reason the equations are shown up as if I used the slash /, but in fact I used \, I don't know why the LaTeX is not working... Hope you can see the equations correctly.

#### Attached Files:

• ###### Field.png
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2. May 20, 2010

### Redbelly98

Staff Emeritus
I don't see any cancellations, as I don't see any opposite-sign charges that have the same distance to A. Looks like you need to slog through this and do the summation.

One thing I do notice is that many charges are the same distance from A. So you don't need to calculate all 48 separate, distinct terms.

The equations look fine.

3. May 21, 2010

### Redsummers

Yeah, thanks! I have actually figured it out already, and as you said nothing cancels so far.
If you actually put the finite field in i,j coordinates system, where for instance, we would have a negative charge at (0,1) and a positive one at (0,2), you came up with a sum like this:

$$E_p= \frac{q^2}{4a\pi\epsilon_0}\sum_{i,j=-n}^{n} \frac{(-1)^{i+j+1}}{\sqrt{i^2+j^2}}$$

Where n will depend on how big is our field. Oh! and we should exclude i,j = 0 from the sum, i.e. not counting the point (0,0), since we have to calculate the potential energy at that point.

Last edited: May 21, 2010