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Question about capacitors with dielectrics

  1. Sep 20, 2009 #1
    Say i have a capacitor and i supply it with a voltage V and after that it has charge density sigmainitial on each plate.

    Now i remove the voltage source and then introduce a dilectric with area A that it equal to the area of one of the plates in the capacitor to the capacitor.

    But this dielectric is only touching one plate.

    My question is: is the net charge density on each plate(sigmainitial - sigmainduced) the same in both plates or different?

    please explain...
     
  2. jcsd
  3. Sep 20, 2009 #2

    sophiecentaur

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    I think the answer is that the same charge is there but the molecules in the dielectric have all polarised a bit. This has increased the capacity ov the capacitor. Equivalent to moving the plates closer together. Now
    Q=CV
    so the voltage will have dropped. As for the energy difference, some work has been done in moving the dielectric into place- nothing violated either way, I think.
     
  4. Sep 20, 2009 #3
    Hi Anon-
    Sophie is correct. Charge is conserved, and the charge distribution is unchanged. The stored energy is

    E = Q2/2C

    So when the dielectric is pulled (not pushed) in, the stored energy drops, and work is done.

    Bob S
     
  5. Sep 20, 2009 #4

    sophiecentaur

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    Yes- the same work that the plates would do if they were allowed to come together and have the equivalent capacitance.
     
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