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Capacitance and electric field

  1. Jul 3, 2006 #1
    I've got a few more I need help with.....I think I might have these ones, but not quite sure....as the concepts are still new to me.


    1. If the voltage between the plates of a parallel plate capacitor is doubled, the capacitance of the capacitor....
    A. is halved
    B. remains the same
    C. is doubled
    D. quadrouples

    ---I think is is A(halved)....because Ceq = Q/V.....so doubling something on the bottom....would be the same as halving C on the other side.


    2. If the voltage applied to a parallel plate capacitor is doubled, the electric field between the plates.....
    A. is halved
    B. remains the same
    C. is doubled
    D. quadrouples

    ----I'm thinking the answer is C(doubled), because E=V/d....
     
  2. jcsd
  3. Jul 4, 2006 #2

    Andrew Mason

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    What has to occur to Q in order for the voltage to double?

    AM
     
  4. Jul 4, 2006 #3
    well if you take Ceq/Q and inverse them. So....then V=Q/Ceq. So....if you double Q, you double V. Likewise.....if you half Ceq, then V will double. Am I right, or not? And, in this question, I'm looking for C....right.
     
  5. Jul 4, 2006 #4

    Andrew Mason

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    Let's assume that plate separation does not change. So if to double V you necessarily must double Q what happens to C? I think that is what the question is asking. Otherwise the question is ambiguous.

    AM
     
    Last edited: Jul 4, 2006
  6. Jul 4, 2006 #5
    If to double V.....and you HAVE to double Q......then I'd say C stays the same.....or am I still off......this is confusing to me for some reason.


    Andrew.....am I correct on the other question then?
     
  7. Jul 4, 2006 #6
    You are right about the second problem.

    To clear up any residual confusion, I will suggest you refer to your problem and then directly to Andrew's post and only if necessary read this:

    The capacitance is given by

    [tex]C = \frac{\epsilon_{0}A}{d}[/tex]

    It is clear that [itex]C[/itex] is not a function of charge [itex]Q[/itex] or potential difference [itex]V[/itex]. Hence, when you write

    [tex]Q = CV[/tex]

    you are actually saying that

    [tex]Q \alpha V[/tex]

    that is the charge is directly proportional to the applied potential difference. It says no more.

    Now when you have a capacitor you have figure out whether it is connected to a source (constant V) or is isolated (constant Q). In your first problem, [itex]V[/itex] changes to [itex]2V[/itex] so obviously the charge must double. However, you aren't changing the geometry so C does not change at all. So you are right.
     
  8. Jul 4, 2006 #7
    Thanks maverick........and you too Andrew. I really appreciate it. That does make a lot more sense now though......I kinda forgot they were perportional.....thanks again!!!
     
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