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Voltage of an infinite cylinder with nontrivial reference point

  1. Feb 12, 2014 #1
    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. jcsd
  3. Feb 12, 2014 #2
    Assuming your field calculations are correct, these look right.
     
  4. Feb 12, 2014 #3

    haruspex

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    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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