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Force between cylindrical capacitors

  1. Apr 10, 2013 #1
    1. The problem statement, all variables and given/known data

    6sz82b.png

    2. Relevant equations



    3. The attempt at a solution

    rsg41i.png

    I'm not sure what is being varied here, radial separation between capacitors x = (b-a) or whether the capacitors can slide up and down so as to change the length L of the capacitor..
     
  2. jcsd
  3. Apr 12, 2013 #2
  4. Apr 12, 2013 #3

    TSny

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    Since the question states that the inner cylinder is partially withdrawn along the common axis, I think it's L that is varied.
     
  5. Apr 15, 2013 #4
    OK, so for constant Q, the force is (1/2)Qo(∂V/∂L), and constant V, the force is (0.5)V^2 (∂C/∂L):

    v45tag.png
    2a91v1z.png
     
    Last edited: Apr 15, 2013
  6. Apr 15, 2013 #5

    TSny

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    How are you defining x? Is it the amount by which the inner conductor is withdrawn or the amount by which the inner conductor is still inside the outer conductor? Or does x represent something else?

    How would you write C in terms of x?
     
  7. Apr 15, 2013 #6
    Won't it be much better to use the equation ##U=\frac{Q^2}{2C}## and calculate the force when the charge is constant?
     
  8. Apr 15, 2013 #7
    Sorry i meant L instead of x.
     
  9. Apr 15, 2013 #8
    The C is in the denominator, which gives a C-2(∂C/∂L) term
     
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