Capacitors

JEE Main level, single correct, medium

117 of 180

Question 117 of 180

A parallel plate capacitor having a separation between the plates d\mathrm{d}, plate area A\mathrm{A} and material with dielectric constant K\mathrm{K} has capacitance C0\mathrm{C}_0. Now one-third of the material is replaced by another material with dielectric constant 2 K2 \mathrm{~K}, so that effectively there are two capacitors one with area 13 A\frac{1}{3} \mathrm{~A}, dielectric constant 2 K2 \mathrm{~K} and another with area 23 A\frac{2}{3} \mathrm{~A} and dielectric constant K\mathrm{K}. If the capacitance of this new capacitor is C\mathrm{C} then CC0\frac{\mathrm{C}}{\mathrm{C}_0} is

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Answer: (B) 43\frac{4}{3}

C0=k∈0 A d\mathrm{C}_0=\frac{\mathrm{k} \in_0 \mathrm{~A}}{\mathrm{~d}}
C=k∈023 d+2k∈0 A3 d=43k∈0 A d\mathrm{C}=\frac{\mathrm{k} \in_0 2}{3 \mathrm{~d}}+\frac{2 \mathrm{k} \in_0 \mathrm{~A}}{3 \mathrm{~d}}=\frac{4}{3} \frac{\mathrm{k} \in_0 \mathrm{~A}}{\mathrm{~d}}
∴CC0=43k∈0 A dk∈0 A d=43\therefore \frac{\mathrm{C}}{\mathrm{C}_0}=\frac{\frac{4}{3} \frac{\mathrm{k} \in_0 \mathrm{~A}}{\mathrm{~d}}}{\frac{\mathrm{k} \in_0 \mathrm{~A}}{\mathrm{~d}}}=\frac{4}{3}

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