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A parallel plate capacitor with air between its plates is charged to 300 V by a battery, which is then disconnected. A dielectric slab of constant…

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A parallel plate capacitor with air between its plates is charged to

300\ \mathrm{V}

by a battery, which is then disconnected. A dielectric slab of constant

K=3

and thickness equal to half the plate separation is slipped between the plates, parallel to them. The potential difference across the capacitor becomes

A. $100\ \mathrm{V}$
B. $450\ \mathrm{V}$
C. $600\ \mathrm{V}$
D. $200\ \mathrm{V}$

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Answer: (D) $200\ \mathrm{V}$

New capacitance

C=\dfrac{\varepsilon_0A}{d-t+t/K}=\dfrac{\varepsilon_0A}{\dfrac d2+\dfrac d6}=\dfrac{\varepsilon_0A}{2d/3}=1.5\,C_0

. With the battery disconnected,

Q

is constant, so

V=Q/C=300/1.5=200\ \mathrm{V}

. (The energy also falls to

2/3

of its initial value.)

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