Potential difference in uniformly charged cylinder

[Alexander2357]

Homework Statement

Charge is uniformly distributed with charge density ρ inside a very long cylinder of radius R.

Find the potential difference between the surface and the axis of the cylinder.

Express your answer in terms of the variables ρ, R, and appropriate constants.

Homework Equations

$$\int \overrightarrow{E}.d\overrightarrow{A}=\frac{Q }{\epsilon _{0}}$$

$$\Delta V = -\int_{i}^{f}\overrightarrow{E}.d\overrightarrow{s}$$

The Attempt at a Solution

I am struggling with determining which Gaussian surface to use. If I use a cylinder, then the cylinder would have an infinite area, right? How can I deal with that? If I use a sphere (since I am trying to find the potential difference between only two points, one on the surface and one on the axis), what will be the charge inside the sphere?

If I use a sphere as my Gaussian surface, I get:

$$E = \frac{\rho }{4\pi R^{2}\epsilon _{0}}$$

$$\Delta V = \frac{\rho }{4\pi R^{2}\epsilon _{0}} \int_{0}^{R}dR=\frac{\rho }{4\pi R\epsilon _{0}}$$

But this is wrong.

If I use a cylinder as my Gaussian surface instead, I get the following, but it doesn't look right:

$$\Delta V = \frac{\rho }{2\pi \epsilon_{0}}\int_{0}^{R}\frac{1}{R} dR$$

 

[Simon Bridge]

I am struggling with determining which Gaussian surface to use. If I use a cylinder, then the cylinder would have an infinite area, right? How can I deal with that?

It's like when you find the field due to an infinite line of charge. Have you done that one (or had it done for you)?

 

[Alexander2357]

It's like when you find the field due to an infinite line of charge. Have you done that one (or had it done for you)?

Yes, I did that a few weeks ago.

The electric field due to an infinite (very long) line of charge is:

$$\frac{\rho }{2\pi R \epsilon_{0}}$$

If I substitute this into the equation for potential difference, I get:

$$\Delta V = -\frac{\rho }{2\pi \epsilon_{0}}ln(R)$$

Is this correct?

 

[Simon Bridge]

My point is - why not try the same shape gaussian surface as for the long line of charge?

 

[Alexander2357]

My point is - why not try the same shape gaussian surface as for the long line of charge?

Thank you for your help Simon. I managed to get the right answer.

 

[Simon Bridge]

Well done.