1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Potential difference in uniformly charged cylinder

  1. Aug 31, 2014 #1
    1. The problem statement, all variables and given/known data

    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.

    2. Relevant equations

    [tex]\int \overrightarrow{E}.d\overrightarrow{A}=\frac{Q }{\epsilon _{0}}[/tex]

    [tex]\Delta V = -\int_{i}^{f}\overrightarrow{E}.d\overrightarrow{s}[/tex]

    3. 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:

    [tex]E = \frac{\rho }{4\pi R^{2}\epsilon _{0}}[/tex]

    [tex]\Delta V = \frac{\rho }{4\pi R^{2}\epsilon _{0}} \int_{0}^{R}dR=\frac{\rho }{4\pi R\epsilon _{0}}[/tex]

    But this is wrong.

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

    [tex] \Delta V = \frac{\rho }{2\pi \epsilon_{0}}\int_{0}^{R}\frac{1}{R} dR [/tex]
     
    Last edited: Aug 31, 2014
  2. jcsd
  3. Aug 31, 2014 #2

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    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)?
     
  4. Aug 31, 2014 #3
    Yes, I did that a few weeks ago.

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

    [tex]\frac{\rho }{2\pi R \epsilon_{0}}[/tex]

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

    [tex]\Delta V = -\frac{\rho }{2\pi \epsilon_{0}}ln(R)[/tex]

    Is this correct?
     
  5. Aug 31, 2014 #4

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    My point is - why not try the same shape gaussian surface as for the long line of charge?
     
  6. Aug 31, 2014 #5
    Thank you for your help Simon. I managed to get the right answer.
     
  7. Aug 31, 2014 #6

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Well done.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Potential difference in uniformly charged cylinder
Loading...