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Electric potential help

  1. Jan 30, 2008 #1
    1. The problem statement, all variables and given/known data
    An empty capacitor is connected to a 11.7-V battery and charged up. The capacitor is then disconnected from the battery, and a slab of dielectric material (κ = 3.1) is inserted between the plates.
    Find the magnitude of the amount by which the potential difference across the plates changes.


    2. Relevant equations
    I did the following steps but did not get the right answer. Is there another step to the problem?

    3. The attempt at a solution
    q=CV, therefore C=q/V
    C is affected by the k... therefore C(dielectric)= k*C
    then the equation reads: q/V = k (q/11.7)... and the Q's cancel out.

    The final equation I worked with is (1/V)=3.1 (1/11.7). The answer I got however is not correct. help.
     
  2. jcsd
  3. Jan 30, 2008 #2

    berkeman

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    Staff: Mentor

    Looks like you are on the right track. Since Q is constant (no place for the charge to go), CV must be constant. Since C goes up by 3.1x, V must do the inverse. What answer exactly did you try?
     
  4. Jan 30, 2008 #3
    the answer i tried was 3.77... originally i had tried 36.27 which was wrong. But it turns out that both are wrong. What do you think?
     
  5. Jan 30, 2008 #4

    berkeman

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    Staff: Mentor

    Well, that's pretty much what I get for 11.7/3.1. And unless I'm missing something, that seems like the right way to get it -- even talks about using the dielectric insertion method for measuring k here:

    http://en.wikipedia.org/wiki/High-k

    How many sig figs does this homework program expect? Maybe they just want 3.8V?
     
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