# Homework Help: Voltage of an infinite cylinder with nontrivial reference point

1. Feb 12, 2014

### programmer1813

1. The problem statement, all variables and given/known data
Find V(r), the electric potential due to an infinitely long cylinder with uniform charge density ρ and radius R.
Use V(r = 2R) = 0 as your reference point.

2. Relevant equations
E at r < R = $\frac{(ρr)}{2ε_{0}}$
E at r > R = $\frac{(ρR^{2})}{(2ε_{0}r)}$

3. The attempt at a solution
This is basically simple. Just integrate the electric fields to get voltage. The only problem is figuring out the limits of integration. I BELIEVE that if you have a nontrivial reference point outside of a cylinder, you will need to integrate from r to 2R for both inside and outside. So I did this:

$$V(r < R) = \int_r^R \frac{\rho r}{2ε_{0}}dr + \int_R^{2R} \frac{\rho r}{2\epsilon_{0}}dr = \frac{\rho}{4\epsilon_{0}}(4R^{2} - r^{2})$$

$$V(r > R) = \int_r^{2R} \frac{\rho R^{2}}{2\epsilon_{0}r}dr = \frac{\rho R^{2}}{2\epsilon_{0}}ln(\frac{2R}{r})$$

Does this look right?

2. Feb 12, 2014

### jackarms

Assuming your field calculations are correct, these look right.

3. Feb 12, 2014

### haruspex

I'm not following the logic behind your equations.
For r > R, the field is $E = \frac{\rho r}{2\pi\epsilon_0 R^2}$, yes?
So for r > R, isn't the potential $\int_{s=r}^{2R}\frac{\rho s}{2\pi\epsilon_0 R^2}.ds$?

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