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Capacitance and charge

  1. Jan 31, 2012 #1
    1. The problem statement, all variables and given/known data

    If the separation between the plates of a parallel plate capacitor is doubled and the potential difference is held constant by a battery, the magnitude of the charge on the plates will:

    *a. double

    b. quadruple

    c. be cut in half

    d. not change

    2. Relevant equations

    CV=Q
    C=εoA/2d

    3. The attempt at a solution

    CV=Q
    C proportional to Q
    εoA/2d proportional to Q
    1/d proportional to Q

    now if the separation d is doubled the charge Q should be cut in half instead of doubled (as the answer key says) right? so have I made a mistake? Is it supposed to be done differently?
     
  2. jcsd
  3. Jan 31, 2012 #2

    I like Serena

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    Homework Helper

    Hi physgrl! :smile:

    You are right and the answer key is wrong.

    Btw, I have a different formula for the capacitance of 2 parallel plates.
    It's ##C={\epsilon A \over d}##, as you can see on the wiki page: http://en.wikipedia.org/wiki/Capacitance

    But that does not change your line of reasoning, which is correct.
     
  4. Jan 31, 2012 #3
    ohh yeah the 1/2 was a mistake...thanks
     
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