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An air-filled parallel-plate capacitor is charged by a battery, then disconnected. A dielectric slab of constant K=3 and thickness equal to half the…

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An air-filled parallel-plate capacitor is charged by a battery, then disconnected. A dielectric slab of constant

K=3

and thickness equal to half the plate separation, with the same area as the plates, is then slipped in so that it is parallel to the plates. The ratio of the final stored energy to the initial stored energy is

A. $\dfrac12$
B. $\dfrac23$
C. $\dfrac32$
D. $\dfrac13$

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Answer: (B) $\dfrac23$

After disconnection the charge

Q

is fixed, so

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

. Initially

C_0=\dfrac{\varepsilon_0A}{d}

. With the slab, the capacitor is two layers in series: effective separation

\dfrac d2+\dfrac{d/2}{K}=\dfrac d2+\dfrac d6=\dfrac{2d}{3}

, so

C=\dfrac32C_0

. Hence

\dfrac{U_f}{U_i}=\dfrac{C_0}{C}=\dfrac23

. Energy decreases because the slab is pulled in by the field, doing positive work on the agent.

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