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A parallel plate capacitor of plate separation d is charged and then disconnected from the battery. A dielectric slab of constant K=3 and thickness…

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A parallel plate capacitor of plate separation

d

is charged and then disconnected from the battery. A dielectric slab of constant

K=3

and thickness

d/2

, with the same area as the plates, is now inserted so that it fills half the gap (one face touching a plate). The ratio of the final stored energy to the initial stored energy is

A. $\dfrac32$
B. $\dfrac34$
C. $\dfrac13$
D. $\dfrac23$

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Answer: (D) $\dfrac23$

The filled and empty halves of the gap act as two capacitors in series, so

\dfrac{1}{C}=\dfrac{d/2}{\varepsilon_0A}+\dfrac{d/2}{3\varepsilon_0A}=\dfrac{2d}{3\varepsilon_0A}

, giving

C=\dfrac32\cdot\dfrac{\varepsilon_0A}{d}=1.5\,C_0

. Since the battery is disconnected, the charge

Q

stays constant, and

U=\dfrac{Q^2}{2C}

. Hence

\dfrac{U}{U_0}=\dfrac{C_0}{C}=\dfrac{1}{1.5}=\dfrac23

. The energy decreases because the slab is pulled in by the field. If the battery had stayed connected, the ratio would be

\dfrac32

instead.

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