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Capacitors

JEE Main level, numerical answer, medium

72 of 180

Question 72 of 180

The space between the plates of a parallel plate capacitor of capacitance C (without any dielectric) is now filled with three dielectric slabs of dielectric constants $K_{1}=2, K_{2}=3$ and $K_{3}=5$ (as shown in figure). If new capacitance is $\frac{n}{3} C$ then the value of $n$ is $\_\_\_\_$ .

Show answer and solution

Answer: 8

The initial capacitance is

C = \frac{\epsilon_0 A}{d}

.
From the figure, the capacitor is divided into three parts.
The bottom half is divided into two capacitors

C_2

and

C_3

in parallel, each with area

A/2

and thickness

d/2

C_2 = \frac{K_2 \epsilon_0 (A/2)}{d/2} = K_2 \frac{\epsilon_0 A}{d} = 3C

C_3 = \frac{K_3 \epsilon_0 (A/2)}{d/2} = K_3 \frac{\epsilon_0 A}{d} = 5C

The equivalent capacitance of the bottom part is

C_{eq, bottom} = C_2 + C_3 = 3C + 5C = 8C

.
The top half is a single capacitor

C_1

with area

A

and thickness

d/2

C_1 = \frac{K_1 \epsilon_0 A}{d/2} = 2K_1 \frac{\epsilon_0 A}{d} = 2(2)C = 4C

.
The total capacitance

C'

is the series combination of

C_1

and

C_{eq, bottom}

\frac{1}{C'} = \frac{1}{C_1} + \frac{1}{C_{eq, bottom}} = \frac{1}{4C} + \frac{1}{8C} = \frac{2+1}{8C} = \frac{3}{8C}

C' = \frac{8}{3}C

Comparing with

C' = \frac{n}{3}C

, we get

n = 8

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