# Calculating E-field and potential of a charged ring

1. Aug 13, 2013

### earthsandwich

Hello folks!

I've been trying to calculate the E-field of a charged ring. It seems well documented for a symetric point(a line from the center etc.) but what I'm interested in is say if I'm slightly of the center of the ring, how can I make a more general equation?

I've tried calculating the potential and the field from there but I get a dominating 0 somewhere so that must be wrong ( V = ∫(λ(x')*dl'(1/(|x-x'|))) and E = -∇V , substituting dl' and λ(x') I get the product of those vectors to (-ρcos(θ')*ρsin(θ') + ρcos(θ')*ρsin(θ'))λdθ' = 0 ??).

Anybody knows where I'm wrong and what to do?

2. Aug 13, 2013

### Simon Bridge

Welcome to PF;
The reason everyone does the field along the axis is because the symmetry is easy.
Off that axis, things get trickier.

Specify cylindrical polar coords, and specify a vector $\vec{r}$ to an arbitrary point.
The resulting field should look a lot like that in a plane from two like charges - then rotated about the z axis.

It may be easier to solve Poisson's equation or follow the differential form.

Anyway - you could also search for "off axis electric field of a ring of charge"
http://www.mare.ee/indrek/ephi/efield_ring_of_charge.pdf
http://electron.physics.buffalo.edu/sen/documents/field_by_charged_ring.pdf

3. Aug 14, 2013

### earthsandwich

Thank you for the welcome. And thanks, I will check it out, although I found something out that I did miss. Apparently I've treated λ and dl' as vectors when they probably should not have been. Now I went straight to the general integralform of the E-field equation and solved it as λ being a constant over the integration and dl' = ρ*dθ'. For x,y = 0 I get the same equation as for the symetric equations as expected. I'm going to use it in electron path simulation so I solved the integral numericaly, probably going to do it proper later though.