Homework Help: Physics

Posted by Liz on Sunday, February 3, 2013 at 6:38pm.

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 - Damon, Sunday, February 3, 2013 at 7:13pm
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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