Capacitors

JEE Main level, single correct, medium

69 of 180

Question 69 of 180

A parallel plate air capacitor has a capacitance CC. When it is half filled as show in figure with a dielectric constant K=5K = 5, the percentage increase in the capacitance is _______.

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Answer: (B) 66.6766.67

Initial capacitance of the air capacitor is given by:

C=ϵ0AdC = \dfrac{\epsilon_0 A}{d}

When the capacitor is half-filled with a dielectric of thickness d/2d/2, it can be considered as two capacitors connected in series.

Capacitance of the dielectric-filled part:

C1=Kϵ0Ad/2=2Kϵ0Ad=2KCC_1 = \dfrac{K \epsilon_0 A}{d/2} = \dfrac{2K \epsilon_0 A}{d} = 2KC

Capacitance of the air-filled part:

C2=ϵ0Ad/2=2ϵ0Ad=2CC_2 = \dfrac{\epsilon_0 A}{d/2} = \dfrac{2 \epsilon_0 A}{d} = 2C

The equivalent capacitance CeqC_{eq} of the series combination is:

Ceq=C1C2C1+C2=(2KC)(2C)2KC+2C=2KK+1CC_{eq} = \dfrac{C_1 C_2}{C_1 + C_2} = \dfrac{(2KC)(2C)}{2KC + 2C} = \dfrac{2K}{K+1} C

Substituting K=5K = 5:

Ceq=2(5)5+1C=106C=53CC_{eq} = \dfrac{2(5)}{5+1} C = \dfrac{10}{6} C = \dfrac{5}{3} C

The increase in capacitance is:

ΔC=Ceq−C=53C−C=23C\Delta C = C_{eq} - C = \dfrac{5}{3} C - C = \dfrac{2}{3} C

The percentage increase in capacitance is:

% increase=ΔCC×100=23×100≈66.67%\% \text{ increase} = \dfrac{\Delta C}{C} \times 100 = \dfrac{2}{3} \times 100 \approx 66.67\%

Answer: 66.6766.67

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