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Capacitance of capacitor with unevenly charged plates

  1. May 27, 2013 #1
    1. The problem statement, all variables and given/known data

    I'm almost embarrassed to post this, but insofar none of my searches online have yielded any results. My question is simply, given a parallel plate capacitor (for the purpose of simplicity let's assume that there are no edge effects) whose constituent plates have charge +aQ and -bQ (a and b are some random number) respectively, what is the capacitance?

    2. Relevant equations


    3. The attempt at a solution

    My main problem lies in determining what the 'Q' term in the equation above is. Is it simply an average of both charges? (Both plates have the same area in this example) All examples I have seen so far assume both plates have the same charge, and I've not been able to find an example where this is not the case. Thank-you in advance.
  2. jcsd
  3. May 27, 2013 #2
    different charges?

    can a PPC have different charges on each plate. I don't think so. a must equal b. if one plate is +aQ then the other has to be -aQ. therefore your capacitance should be C=aQ/V
  4. May 27, 2013 #3
    Are we sure about this? "can a PPC have different charges on each plate. I don't think so"

    Lets assume I have a capacitor and charged it with a battery. Then I touched one lead of the cap to ground. Would the charges on the plates still be equal? Just asking.
  5. May 27, 2013 #4
    don't you have to connect both leads to discharge a capacitor?
  6. May 27, 2013 #5
    I am not sure. I recall a electrostatic problem where there are two spheres with different charges. Isn't this like a capacitor with different charges on the plates? I am just asking. I don't know the answer.
  7. May 28, 2013 #6
    Hello PedroB

    When charged conducting plates are placed parallel to each other ,the two outermost surfaces get equal charges and the facing surfaces get equal and opposite charges i.e let plate A has charge Q1 and plate B has charge Q2 then the charge on outermost surfaces of both the plates will be (Q1+Q2)/2 .The innermost surfaces will have charge (Q1-Q2)/2 and -(Q1-Q2)/2 .

    To get this result use Gauss Law and the fact that the electric field inside a conductor in electrostatic condition is zero.

    Hope this helps .
  8. May 30, 2013 #7


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    i agree with this.
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