Find the electric potential at any point on the x axis

[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..

 

[vela]

By symmetry, you know the electric field along the x-axis has no vertical component, so all you need to find is E_x. 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.

 

[krtica]

Thank you very much vela.