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March 29, 2017

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A parallel-plate capacitor is charged by a battery and then the battery is removed and a dielectric of constant K is used to fill the gap between the plates. Inserting the dielectric changes the energy stored by a factor of
A)2.
B)1/K.
C)1 (no change).
D)K.
E)K-1.

  • Physics - ,

    Q could not change, no flow of current
    but the initial capacitance Ci may change
    Ui = .5 Q^2 / Ci

    so what happens to C ?
    K = C/Ci
    so now C = K Ci
    Uf = .5 Q^2 / (K Ci)
    so
    Uf/Ui = .5 Q^2 / (K Ci) /.5 Q^2 / (Ci)
    = 1/K or B)

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