# Homework Help: Help with electric potential!

1. Oct 21, 2004

### stunner5000pt

The electric field inside a NONCONDUCTING sphere of radius R containing uniform charge densiy is radially directed and has magnitude of

E = qr / 4 pi epsilon0 R^3

where r is the distance from the centre

q is the charge on the sphere

a) Find potential V inside the sphere taking V=0 @ r = 0

Since V = Integral E dr
Then V = qr^2 / 8 pi epsilon0 R^3

then if r = 0 then V = 0

am i correct here?

Show that the potential at distance r from the centre where r <R is given by V = q (3R^2 - r^2) / 8pi epsilon R^3

It looks like it has been integrated from r to (root 3) R. But i dont understand why??

Another problem is

A geiger counter has a metal cylinder of 2.10 dimater with a wire stretched along it's axis whose diamtere is 1.34 x 10^-4 cm in dimater. If 855 V is applied between these two what is the electric field at the surface of the wire and the cylinder??

lets say lambda = Q / L

then flux = EA = E 2 pi r L = 4 pi k Qenc = 4 pi k lambda L

so 2 k lambda / r = E

then i integrate because V = integrate E dr

so that V = 2k lambda Log r

thanks a lot

2. Oct 26, 2004

### artybear

account for the integration constant

This is only a response to your first question about the potential withing a non-conducting sphere. You are right that you can choose the potential at the center of the sphere to be zero. You can set it to whatever you want, since you are generally only concerned with the *change* in potential.

But, what about that sqrt(3) in the equation that some people derive?

Consider the other integral you take to determine the potential of a point *outside* the non-conducting sphere. In that case, it is common to choose the potential at r=infinity to be 0. Now you have two equations that describe the potential near the sphere, based on whether or not it is inside, and presumably, the transition of the potential going from inside the sphere to the outside of the spehre should be continuous. In fact, the slope of the potential at r=R using the equation from inside the sphere equals the slope at r=R from outside the sphere. To make the curves of the two potential equations meet up, we add a constant to the equation from inside the sphere. I hope that helps. I don' t have time right now to write out equations in detail.