A parallel-plate capacitor of capacitance 6…

A parallel-plate capacitor of capacitance 6 muF (air-filled) is connected permanently across a 10 V battery. A dielectric slab of constant K=3 and…

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A parallel-plate capacitor of capacitance

6\ \mu\mathrm{F}

(air-filled) is connected permanently across a

10\ \mathrm{V}

battery. 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 lies flat against one plate. The increase in the energy stored in the capacitor is

A. $450\ \mu\mathrm{J}$
B. $300\ \mu\mathrm{J}$
C. $75\ \mu\mathrm{J}$
D. $150\ \mu\mathrm{J}$

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Answer: (D) $150\ \mu\mathrm{J}$

The slab and the remaining air gap act as two capacitors in series, with total effective separation

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

. Hence

C'=\dfrac{3}{2}C=9\ \mu\mathrm{F}

. The battery holds the voltage fixed at

10\ \mathrm{V}

, so

U_i=\tfrac12(6)(100)=300\ \mu\mathrm{J}

and

U_f=\tfrac12(9)(100)=450\ \mu\mathrm{J}

. The increase is

150\ \mu\mathrm{J}

. (The battery supplies

\Delta Q\,V=300\ \mu\mathrm{J}

, half of which is stored and half is done as work on the slab.)

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Original question written in the JEE Main 2018 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake