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JEE Main 2019 (10 Jan, Shift 2) · Dielectrics

Dielectrics question from JEE Main 2019 (10 Jan, Shift 2)

Q93JEE Main 2019 · 10 JAN · SHIFT 2Medium

Dielectrics

A parallel plate capacitor having capacitance 12 pF is charged by a battery to a potential difference of 10 V between its plates. The charging battery is now disconnected and a porcelain slab of dielectric constant 6.5 is slipped between the plates. The work done by the capacitor on the slab is:

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Answer: (a)

Battery removed, so $Q$ is constant: $U=\frac{Q^2}{2C}$ .
$U_i=\frac{1}{2}CV^2=\frac{1}{2}(12)(10)^2=600\text{ pJ}$
$U_f=\frac{U_i}{K}=\frac{600}{6.5}\approx 92.3\text{ pJ}$
Work done on slab $=U_i-U_f\approx 508\text{ pJ}$
So the answer is (a) 508 pJ.

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