JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Capacitance

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A parallel plate capacitor with plate area 'A' and distance of separation 'd' is filled with a dielectric. What is the capacity of the capacitor when permittivity of the dielectric varies as:

ε (x) = ε_{0} + k x

, for

(0 < x \leq \frac{d}{2})

ε (x) = ε_{0} + k (d - x)

, for

(\frac{d}{2} \leq x \leq d)

(A)0
(B) $\frac{kA}{2} ln (\frac{2 ε _{0}}{2 ε _{0} - kd})$
(C) $\frac{kA}{2 ℓ n ( \frac{2 ε _{0} + kd}{2 ε _{0}} )}$ ✓ Correct
(D) $(ε_{0} + \frac{kd}{2})^{2/ kA}$

Answer: (C)

Solution

For a thin slice of thickness

d x

at position

x

, the capacitance is

d C = \frac{ε ( x ) A}{d x}

. For the first half (

0 < x \leq d /2

\frac{1}{C _{1}} = \int_{0}^{d /2} \frac{d x}{ε ( x ) A} = \frac{1}{A} \int_{0}^{d /2} \frac{d x}{ε _{0} + k x} = \frac{1}{k A} [ln (ε_{0} + k x)]_{0}^{d /2} = \frac{1}{k A} ln (1 + \frac{k d}{2 ε _{0}})

. By symmetry,

C_{2} = C_{1}

. The two halves are in series, so

C_{t o t a l} = \frac{C _{1} C _{2}}{C _{1} + C _{2}} = \frac{C _{1}}{2} = \frac{k A}{2 l n ( 1 + \frac{k d}{2 ε _{0}} )} = \frac{k A}{2 l n ( \frac{2 ε _{0} + k d}{2 ε _{0}} )}

JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Single correct option

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JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Capacitance

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