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Change in Electric Potential from the Surface of a Conducting Sphere to Infinity

  1. Feb 20, 2008 #1
    Let me preface this as this is my first post on this forum. I'm a physics major at Virginia Tech and I've lurked the forum for a while to help understand concepts that may not be intuitive initially. I'm stuck on this one concept, so I decided to give posting a shot.

    Without further ado...

    1. An insulating spherical shell with inner radius 25.0 cm and outer radius 60.0 cm carries a charge of + 150.0 [tex]\mu[/tex]C uniformly distributed over its outer surface. Point a is at the center of the shell, point b is on the inner surface and point c is on the outer surface.

    What will a voltmeter read if it is connected between c and infinity?

    2. Given [tex]\int[/tex]E*dl = V, I'd be integrating over infinity because it's an infinite path to... infinity.

    3. It's more conceptual than anything, so I'm really at a loss. The change in potential from the center of the shell to the inner surface is 0V, and the change between the shell itself is 0V, as it's a conductor.

    I'd rather just get the concept than someone spit out a solution.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Feb 21, 2008 #2
    Stupid me. I was right - I just messed up my units (read it as nano rather than micro).

    Thanks anyways!
  4. Feb 21, 2008 #3
    So then....how did you do this?
  5. Feb 21, 2008 #4
    Well, think about it. If you have a charge, be it a point charge, sphere, or spherical shell, then it has some electric potential. If you bring in a test charge from infinity, you're going to have to do the that amount of work on it to bring it to the charge (assuming the point charge with the potential is a positive charge). It's the summation of all the work from infinity to the surface of the sphere.

    My issue was that I was using a nanocoulomb rather than a microcoulomb, so I was off by a power of 10^3.
  6. Feb 21, 2008 #5
    oh, I see. Makes sense. Thanks.
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