A parallel plate capacitor with plate area A and plate separation d is filled with a dielectric material of dielectric constant $\mathrm{K}=4$. The thickness of the dielectric material is x, where $\mathrm{x}<\mathrm{d}$. Let $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ be the capacitance of the system for $\mathrm{x}=\frac{1}{3} \mathrm{~d}$ and $\mathrm{x}=\frac{2 \mathrm{~d}}{3}$, respectively. If $\mathrm{C}_{1}=2 \mu \mathrm{~F}$ the value of $\mathrm{C}_{2}$ is $\_\_\_\_$ $\mu \mathrm{~F}$

JEE Main · 2023 · 06 Apr 2023 · Shift 1 · Capacitance

A parallel plate capacitor with plate area A and plate separation d is filled with a dielectric material of dielectric constant

K = 4

. The thickness of the dielectric material is x, where

x < d

.

Let

C_{1}

and

C_{2}

be the capacitance of the system for

x = \frac{1}{3} d

and

x = \frac{2 d}{3}

, respectively. If

C_{1} = 2 μ F

the value of

C_{2}

____

μ F

Answer: 3

The capacitor can be treated as two capacitors in series: one with dielectric (thickness

x

) and one without (thickness

d - x

). The effective capacitance is

C = \frac{ϵ _{0} A}{\frac{x}{K} + ( d - x )}

. For

x = \frac{d}{3}

C_{1} = \frac{ϵ _{0} A}{\frac{d /3}{4} + \frac{2 d}{3}} = \frac{ϵ _{0} A}{d} \times \frac{12}{9} = \frac{4}{3} \frac{ϵ _{0} A}{d} = 2 μ

F. For

x = \frac{2 d}{3}

C_{2} = \frac{ϵ _{0} A}{\frac{2 d /3}{4} + \frac{d}{3}} = \frac{ϵ _{0} A}{d} \times 2

. From

C_{1}

, we get

\frac{ϵ _{0} A}{d} = \frac{3}{2}

, so

C_{2} = 2 \times \frac{3}{2} = 3 μ

JEE Main · 2023 · 06 Apr 2023 · Shift 1 · Integer answer