Homework Help: Electric field of a charged ring

1. Feb 1, 2009

Seraph404

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

This is a http://session.masteringphysics.com/problemAsset/1003223/45/14677_a.jpg" [Broken] of the problem

-I know that the direction of the electric field at any point on the z-axis is parallel to the z-axis
E = k*q/r^2

http://session.masteringphysics.com/problemAsset/1003223/45/1003223B2.jpg" [Broken]

dE = k* dq/(a^2+z^2)

dEz = (z/(sqrt(a^2+z^2)) *dE

3. The attempt at a solution

So, I need to find the magnitude of the electric field along the positive z axis.

Basically, everything up there was work from previous parts of the problem. Now, I think I just need to put the last two expressions together in order to integrate around the ring (I think with respect to q), but I'm not sure. I need a little help on this last step.

Last edited by a moderator: May 4, 2017
2. Feb 1, 2009

Seraph404

I don't know if there's a better way to explain.

3. Feb 1, 2009

Delphi51

dE = k* dq/(a^2+z^2)
dEz = (z/(sqrt(a^2+z^2)) *dE
Those look good! When you integrate around the ring, a,z and k are constant so you'll have something times Integral of dq from 0 to q, which is just the total charge on the ring, q. The integral just finds the total charge, which you already know anyway!

You don't need calculus for this one because all of the ring charge is equally distant from the point on the z axis where you are calculating and you want only the z component of E.
You just get E = kq/r^2 * (z/(sqrt(a^2+z^2)) where r^2 = (a^2+z^2).

Last edited: Feb 2, 2009