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What is the dielectric constant of the slab on a parallel plate capacitor ?

  1. Jun 19, 2009 #1
    1. The problem statement, all variables and given/known data
    When a certain air-filled parallel-plate ca-
    pacitor is connected across a battery, it ac-
    quires a charge (on each plate) of magnitude
    214 μC. While the battery connection is
    maintained, a dielectric slab is inserted into
    the space between the capacitor plates and
    completely fills this region. This results in
    the accumulation of an additional charge of
    272 μC on each plate.

    http://images.upload2world.com/get-6-2009-upload2world_com_ohteyhd.jpg [Broken]

    What is the dielectric constant of the slab?

    2. Relevant equations

    i used
    k= Uf/Ui

    3. The attempt at a solution

    what i did is that i k=[Q2^2/2C]/[Q1^2/2C] .. in this case the 2C will cancel each other .. we will be left out with k=[Q2^2/Q1^2] .. which as a result gave me 1.61551

    i really dont know were im going wroing .. so please help mee !!

    thanx in advance
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jun 19, 2009 #2


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    Gold Member

    I would take a step by step approach.

    1. By how much did the capacitance increase in the second case from the first case?

    2. How does the capcitance depend upon the dielectric constant in the case of a parallel plate capacitor?

    3. Therefore, what must the increase in dielectric constant of the gap have been?
  4. Jun 20, 2009 #3
    i didnt undcerstand :confused:
  5. Jun 20, 2009 #4


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    shouldn't the dielectric constant be like; k=C/Co?
    where C is the capacitance with the dielectric material in between the plats and Co the same but in vaccum
  6. Jun 20, 2009 #5

    so it will be k= Q/Qo.. were the volt cancel each other
  7. Jun 20, 2009 #6


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    no, I don't think V is the same, there is Vo and V, where V is less than Vo [due to the inverse electric field induced by putting the dielectric material in the capacitor, which reduces the main E, thus the sum of E is less than before and as a result the applied voltage decreases as well]
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