Potential difference in uniformly charged cylinder

AI Thread Summary
The discussion focuses on calculating the potential difference between the surface and the axis of a uniformly charged cylinder. Participants explore the appropriate Gaussian surface to use for determining the electric field, debating between a cylinder and a sphere. The electric field due to a long line of charge is noted as ρ/(2πRε₀), which leads to a potential difference expressed as ΔV = -ρ/(2πε₀)ln(R). Ultimately, the correct approach involves using the same Gaussian surface as for the infinite line of charge, confirming the solution. The conversation concludes with a participant successfully arriving at the correct answer.
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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
 
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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)?
 
Simon Bridge said:
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?
 
My point is - why not try the same shape gaussian surface as for the long line of charge?
 
Simon Bridge said:
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.
 
Well done.
 
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