# Find the electric potential at any point on the x axis

1. Feb 16, 2010

### krtica

Two positive point charges each have a charge of +q and are fixed on the y axis at y = +a and y = -a. (Use k, q, a, and x as necessary.)

(a) Find the electric potential at any point on the x axis.
V = 2kq/[sqrt(x^2+a^2)]

(b) Use your result in part (a) to find the electric field at any point on the x axis.

For part B, would I differentiate the potential with respect to x? If so, my answer would be (-2xkq)/[(x^2+a^2)^(3/2)]

If you can please, I'm also having trouble understanding why the electric field is the derivative of the potential with respect to its distance..

2. Feb 16, 2010

### vela

Staff Emeritus
By symmetry, you know the electric field along the x-axis has no vertical component, so all you need to find is Ex. Your answer looks correct except for a sign.

The electric potential is the integral of the electric field, so the electric field is the gradient of the potential. It's essentially the fundamental theorem of calculus.

3. Feb 16, 2010

### krtica

Thank you very much vela.