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Potential difference of concentric conducting shells

  1. Sep 27, 2011 #1
    My professor said to in order to solve this, integrate the electric field to find the electric potential...

    [itex]\Delta V = -\int\stackrel{\rightarrow}{E}dl[/itex]

    My question is, using Gauss's Law, ([itex]\oint E_{n}dA = \frac{Q}{\epsilon}[/itex]), how do I go about finding Q?

    Isn't Q just the charge of the shell?
     
  2. jcsd
  3. Sep 27, 2011 #2

    G01

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    In Gauss's law, Q is the charge enclosed by your Gaussian surface. So, first decide where you Gaussian surface will be, then add up all the charge inside of it.

    HINT: You want your surface to be in the region where you want to find the electric field.
     
  4. Sep 27, 2011 #3
    This is what I have drawn (see attachment). Is the basic idea here to integrate the electric field of the outer Gaussian surface from b to a? If so, what about the inner Gaussian surface?
     

    Attached Files:

  5. Sep 28, 2011 #4

    G01

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    You want to find the potential difference between the shells, so you don't need the Gaussian surface outside the larger shell.

    Try this: Take your inner surface and place it at an arbitrary point r. Then, find E using the standard approach when using Gauss's law. You will then have E between the plates as a function of r. Can you use that to find the potential difference between the plates?
     
  6. Sep 28, 2011 #5
    Alright... see attached image.

    Now, how do i determine if i should integrate from a to b or from b to a?
     

    Attached Files:

  7. Sep 28, 2011 #6

    G01

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    That all looks good.

    As far as your integration bounds are concerned, it doesn't matter. The sign of your end result will change, but that's just like hooking up a voltmeter in reverse: You will still get the right pot. difference, just the sign will change.
     
  8. Sep 29, 2011 #7
    thanks for your guidance
     
  9. Sep 29, 2011 #8

    G01

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    No problem! :smile:
     
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