A parallel plate capacitor has plates of area A separated by distance 'd' between them. It is filled with a dielectric which has a dielectric constant that varies as $k(x)=K(1+ \alpha x)$ where 'x' is the distance measured from one of the plates. If $(\alpha d) \ll 1$, the total capacitance of the system is best given by the expression:

(A) $\frac{AK\varepsilon_{0}}{d}\left(1+\frac{\alpha d}{2}\right)$

(B) $\frac{A \varepsilon_{0} K}{d}\left(1+\frac{\alpha^{2} d^{2}}{2}\right)$

(C) $\frac{AK \varepsilon_{0}}{d}(1+\alpha d)$

(D) $\frac{A \varepsilon_{0} K}{d}\left[1+\left(\frac{\alpha d}{2}\right)^{2}\right]$

JEE Main · 2020 · 07 Jan 2020 · Shift 1 · Capacitance

A parallel plate capacitor has plates of area A separated by distance 'd' between them. It is filled with a dielectric which has a dielectric constant that varies as

k (x) = K (1 + α x)

where 'x' is the distance measured from one of the plates.

If

(α d) ≪ 1

, the total capacitance of the system is best given by the expression:

(A) $\frac{A K ε _{0}}{d} (1 + \frac{α d}{2})$ ✓ Correct
(B) $\frac{A ε _{0} K}{d} (1 + \frac{α ^{2} d ^{2}}{2})$
(C) $\frac{A K ε _{0}}{d} (1 + α d)$
(D) $\frac{A ε _{0} K}{d} [1 + (\frac{α d}{2})^{2}]$

Answer: (A)

Solution

Treat the slab as a stack of thin layers of thickness

d x

at distance

x

from a plate; each behaves like a capacitor

d C = \frac{K ( 1 + α x ) ε _{0} A}{d x}

, and these add in series, so

\frac{1}{C} = \int_{0}^{d} \frac{d x}{K ε _{0} A ( 1 + α x )} = \frac{l n ( 1 + α d )}{K ε _{0} A α}

. For

α d ≪ 1

use

ln (1 + α d) \approx α d - \frac{( α d ) ^{2}}{2}

, giving

\frac{1}{C} \approx \frac{d}{K ε _{0} A} (1 - \frac{α d}{2})

. Inverting with

(1 - x)^{- 1} \approx 1 + x

yields

C \approx \frac{K ε _{0} A}{d} (1 + \frac{α d}{2})

JEE Main · 2020 · 07 Jan 2020 · Shift 1 · Single correct option

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JEE Main · 2020 · 07 Jan 2020 · Shift 1 · Capacitance

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