# Difference of electric potential between the axis of the cylinder and its surface?

1. ### ashe540

1
We have a piece of a cylinder of infinite length and radius R, charged with a uniform and positive volumetric density of charge p.
Compute:

a) Difference of electric potential between the axis of the cylinder and its surface.

b) Work to carry a charge q from the axis to the surface of the cylinder. Is this work done by the forces of the electric field, or is it external to the field forces?

c) Compute the electric potential of a point on the surface of the cylinder (this potential could be computed as the difference of potential between such point and the infinite). But this potential can not be computed. Why is it not possible to compute the potential on a point, but it´s possible to compute differences of potential in this problem?

a) What I did was using the difference of electric potential formula
*********A******
V(A)-V(B)= ∫ E*dr **
********* B *****

I set the A and B points as 0 (for the axis which I took as the origin) and R as the surface. Then I pulled E out of the integral because it´s constant and integrated dr. I´m not sure if this is the right way about doing it.

b) As for the work I wasn´t sure what to do here. I never have clear which formula to use for which cases. Whether W=qEd or W=q* [V(A)-V(B)]

c) I started thinking about it and I was thinking that maybe it had something to do with the fact that given the formula V=W/Q , that W is not given. Does this have anything to do with it?

Help please, I'm really lost

Last edited: Oct 10, 2011
2. ### lightgrav

1,247
Re: Difference of electric potential between the axis of the cylinder and its surface

where did you get the idea that E was uniform at different radii?
E.A = E 2πrL = 4πk (rho)πr^2L , so the E increases with radius.