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Electric potential at the center of a cylinder enclosed in a cylindrical shell

  1. Oct 28, 2012 #1
    1. The problem statement, all variables and given/known data

    A solid metal cylinder of radius R and length L, carrying a charge Q, is surrounded by a thick coaxial metal shell of inner radius a and outer radius b. The shell carries no net charge. Find the potential at the center, using r=b as the reference point.

    2. Relevant equations

    V= (1/4πεo)q/r

    V = -∫E dot dl

    3. The attempt at a solution

    The cylinder is not infinite but I assume I'd have to ignore end effects.

    I want to approach this problem using gauss's law to find the field and then integrating. I don't even think I can use the first equation since my reference point is not at infinity.

    This is easy enough but I'm having problems because I don't know the electric field at the boundaries. I can find the field easily when r<R when R < a < b and when a < r < b

    I don't know what the field is exactly at the surfaces?
     
  2. jcsd
  3. Oct 28, 2012 #2

    TSny

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    Hello, aftershock.

    The fact that E is discontinuous at the surfaces will not cause any problem. You do not need to worry about the value of E at a surface. E will be a piecewise continuous function. When you integrate a piecewise continuous function, you do not need to know the value of the function at the discontinuities.
     
  4. Oct 28, 2012 #3
    Thank you, but I'm still confused. I actually have the solution but I can't make sense of it.

    the solution tells me V(b)-V(a) = ∫Q/(2πoLr) integrated from a to b

    Why am I integrating the electric field in the region R < r < a from a to b. Isn't the electric field zero from a to b since it's inside a conductor?

    Then it goes on to claim V(0) = V(a)

    How does that happen?
     
  5. Oct 28, 2012 #4

    TSny

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    You're right, the integration should be from R to a. The integration form a to b would be zero since E is zero in that region.
     
  6. Oct 28, 2012 #5
    So the answer is the integral of the field from R to a?

    And b does not appear in the solution even though it's the reference point?
     
  7. Oct 28, 2012 #6

    TSny

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    Yes. Technically you should be integrating from R to b, but the part of the integral form a to b is zero. So, you just need to integrate from R to a. :smile:

    [Edit: Actually, technically, you should be integrating from r = 0 to r = b. But the integral from 0 to R is also zero.]
     
    Last edited: Oct 28, 2012
  8. Oct 28, 2012 #7
    Thank you!

    It's actually integrating from a to R though, right? Or are you reversing the order and throwing away the negative sign?
     
  9. Oct 28, 2012 #8

    TSny

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    Yes, that's right. I was reversing the limits to get rid of the negative sign. Good.
     
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