# Electric Potential @ center of an equilateral triangle

1. Oct 25, 2009

### Nyquilt

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

What is the electric potential at point P, the center of an equilateral triangle with side length of 2a and total charge of Q? Use infinity as the reference point.

2. Relevant equations

∆v = -∫E∙ds
v(r)= kq/r

3. The attempt at a solution

I already have the answer which is k(Q/a) ln(2+sqrt(3)) but I don't know how to get it. This is what I tried:

v = kq/r = ∫kdq/r = ∫k lamda dx/r = k lamda ∫dx/sqrt(x^2 + (a/sqrt(3))^2) = k lamda ln(x + sqrt(x^2 + (a/sqrt(3))^2)).

I evaluate that integral from 0 to 2a

k lamda ln(2a - sqrt(4a^2 + (a^2)/3) - k lamda ln(a/sqrt(3))
= k lamda ln(2sqrt(3) - sqrt(13)) = kQ/(2a) ln(2sqrt(3) - sqrt(13))

2. Oct 25, 2009

### flatmaster

Your substitution for r = sqrt(x^2 + (a/sqrt(3))^2 Fixes the origin for your x axis somewhere. Where is that point? How does that change the limits for your integral?

3. Oct 25, 2009

### Nyquilt

Hmm I guess the origin for the x axis is at the axis of symmetry. If that's the case then I integrate from -a to a?

4. Oct 25, 2009

### flatmaster

Yes. Either that, or you can further exploit symetry by only integrating from 0 to a and multiplying by 2. Also, your integral was the potential from only one side of the triangle. I'm not sure if you multiplied by 3.

5. Oct 25, 2009

### Nyquilt

That worked great. Thanks a lot for the help. I was stuck on this for a while.

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