Posted by Liz on .
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
C)1 (no change).
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)
Uf/Ui = .5 Q^2 / (K Ci) /.5 Q^2 / (Ci)
= 1/K or B)