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A parallel plate air capacitor of capacitance 5 muF is charged to 100 V and then disconnected from the battery. A dielectric slab of constant K = 4…

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A parallel plate air capacitor of capacitance

5\ \mu\mathrm{F}

is charged to

100\ \mathrm{V}

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

K = 4

and thickness equal to half the plate separation, with the same area as the plates, is slowly inserted so that it lies flat against one plate. The change in the energy stored in the capacitor is

A. decrease of $9.375\ \mathrm{mJ}$
B. increase of $15\ \mathrm{mJ}$
C. decrease of $15.6\ \mathrm{mJ}$
D. increase of $9.375\ \mathrm{mJ}$

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Answer: (A) decrease of $9.375\ \mathrm{mJ}$

With the slab, the gap behaves as two layers in series:

C = \dfrac{\varepsilon_0 A}{d/2 + (d/2)/4} = \dfrac{8}{5}C_0 = 8\ \mu\mathrm{F}

. The battery is disconnected so the charge stays

Q = C_0V = 500\ \mu\mathrm{C}

. Energy

U = Q^2/2C

U_i = \dfrac{(5\times10^{-4})^2}{2\times5\times10^{-6}} = 25\ \mathrm{mJ}

and

U_f = \dfrac{(5\times10^{-4})^2}{2\times 8\times10^{-6}} = 15.625\ \mathrm{mJ}

. The energy decreases by

9.375\ \mathrm{mJ}

(the slab is pulled in).

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