# Electric field from infinite charges

1. Sep 8, 2006

### DieCommie

Here is my problem: A charge of -q is located at x=a. Another charge, +q, is at x=2a. This sequence of alternating charges continues indefinitely in the +x direction. What is the electric field at the origin?

So I figure I need an infinite alternating sum. The equation to be used is $$\frac{q}{4\pi\epsilon_0r^2}$$ So the sum would be $$\sum \frac{(-1)^nq}{4\pi\epsilon_0(na)^2}$$. Which can be factored to $$\frac{q}{4\pi\epsilon_0a^2} \sum \frac{(-1)^n}{n^2}$$. (sum from n=one to n=infinity)

Is that correct so far? Im not sure how to do the sum..... Any help please, Thx!

Last edited: Sep 8, 2006
2. Sep 9, 2006

### DieCommie

With some help from the math forum, I believe I have the answer..

$$-\frac{q\pi}{48\epsilon_0a^2}$$

Last edited: Sep 9, 2006
3. Sep 9, 2006

### Galileo

Be careful about the signs here. A negative charge at x=a will create a field that points in the positive x-direction at the origin. Otherwise what you got is correct.

The sum is very famous. I think they'd expect you to know/memorize it and leave the derivation for a math class:
$$\sum_{n=1}^{\infty}\frac{\pi^2}{6}$$

EDIT: You just beat me to it. But it should be 4*6=24 in the denominator.

Last edited: Sep 9, 2006
4. Sep 9, 2006

### DieCommie

So, the field goes to a negative charge and because the first charge is strongest, the field will terminate there. Meaning at the origion it points to x=a.

But I did get a negative answer in the end, which makes me think that the field is coming to the origin.

I can see that you are right the field points in the +x, but mathmatically shouldnt i get a positive value for the field then?

EDIT- also, I got the sum as $$-\frac{\pi^2}{12}$$ (with help from the math forum). Your value i do memorize, but what i needed was slightly different... right? I needed an alternating sum, which is why I have a 48 in the denominator.

Last edited: Sep 9, 2006
5. Sep 9, 2006

### Galileo

Remember that the electric field is a vector. For a point charge:
$$\frac{q}{4\pi\epsilon_0r^2}\hat r$$ where $\hat r$ points radially away from the charge. That would be in the negative x-direction if the charge is at x=a. So E points in the positive x direction if q is negative. You got it just the other way around.

Erm, right.

6. Sep 11, 2006

### DieCommie

Well I got an F on the problem :( I guess $$-\frac{q\pi}{48\epsilon_0a^2}$$ is not the right answer, but I just cant figure out what I did wrong or how to get the right answer...

Thx for you help anyway...

7. Sep 12, 2006

### Galileo

Well, the field points in the +x direction, so the answer should be positive.
$$\frac{q\pi}{48\epsilon_0a^2}$$
Which you get if you take care of all the signs carefully, but I think it's easier to just work with magnitude and determine the direction by inspection.

Anyway, that's the only minor error I can spot. You did mention the field pointed in the +x-direction in your answer, it's definitely not worth an F so I'm curious what comments you got on the answer