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Capacitor and dielectriscs

  1. Jan 10, 2010 #1
    1. The problem statement, all variables and given/known data

    A parallel plate capacitor has a capacitance C when there is a vacuum between teh plates. How does this change if the gap between the plates is half filled with a dielectric with dielectri constant [tex]\epsilon[/tex]r

    Calculate the change in energy stored in the capacitor if the dielectric is inserted while the capacitor is connected to a battery of fixed voltage V.

    2. Relevant equations

    C=[tex]\stackrel{\epsilon_0 A}{d}[/tex]
    3. The attempt at a solution

    For the first part, i assumed it was equivalent to 2 capacitors (one with the dielectric and one without in parallel)

    C=[tex]\stackrel{\epsilon_0 0.5A}{d}[/tex] =0.5C

    C2= [tex]\epsilon_r[/tex]*[tex]\stackrel{\epsilon_0 0.5A}{d}[/tex] - 0.5[tex]\epsilon_[/tex]C

    C(total) = 0.5C+0.5[tex]\epsilon_r[/tex]C.

    Second part, not really sure how to start...

    V is fixed, and U=0.5CV2.. So if C increases by (1+[tex]\epsilon[/tex]r) U increases by the same factor??
  2. jcsd
  3. Jan 10, 2010 #2


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    Homework Helper

    Why in parallel? The battery's voltage is being applied across the 2 capacitors, not across each capacitor individually.

    Yes, except C doesn't increase by (1+[tex]\epsilon[/tex]r).
  4. Jan 10, 2010 #3

    Not entirely sure, just seen other examples where they said this

    C(total)=C(0.5+0.5[tex]\epsilon[/tex]r) so it increases by a factor of (0.5+0.5[tex]\epsilon[/tex]r)?
    Last edited: Jan 10, 2010
  5. Jan 10, 2010 #4
    tried it again...

    [itex] U_0 =1/2 C_0 V^2 [/itex] and [itex] U_1 =1/2 C_1 V^2 [/itex]

    [itex] C_1 = 1/2 \epsilon[/tex]r so

    [itex] U_1 =1/2 *1/2 C_0 \epsilon[/tex]r V^2 [itex] =1/2 \epsilon[/tex]r *U_0 ?
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