Question

A parallel plate capacitor, with plate area ${}^{\prime }{A}^{\prime }$ and distance of separation ${}^{\prime }{d}^{\prime }$, is filled with a dielectric. What is the capacity of the capacitor when permittivity of the dielectric varies as follows:
$ϵ\left(x\right)={ϵ}_0+kx,$ for $(0<x\le \frac{d}{2})$
$ϵ\left(x\right)={ϵ}_0+k(d-x)$, for $(\frac{d}{2}\le x\le d)$

• A. $\frac{kA}{2ln\left(\frac{2{ϵ}_0+kd}{2{ϵ}_0}\right)}$
• B. $\frac{kA}{2}ln\left(\frac{2{ϵ}_0}{2{ϵ}_0-kd}\right)$
• C. ${\left({ϵ}_0\frac{kd}{2}\right)}^{\frac{2}{ka}}$
• D. $0$

Answer 1

The correct option is A $\frac{kA}{2ln\left(\frac{2{ϵ}_0+kd}{2{ϵ}_0}\right)}$


The net capacity will be the effective capacity of series combination of two capacitors formed by the two halves of the dielectric.
i.e. $\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}$

Taking an element of width $dx$ at a distance
$x$ from left plate $(x<\frac{d}{2})$
$d{C}_1=\frac{({ϵ}_0+kx)A}{dx}$

Capacitance of the first half of the capacitor is,
$\frac{1}{C_1}={\int }_0^{\frac{d}{2}}\frac{1}{dc}=\frac{1}{A}{\int }_0^{\frac{d}{2}}\frac{dx}{{ϵ}_0+kx}$

$\frac{1}{C_1}=\frac{1}{kA}ln\left(\frac{{ϵ}_0+\frac{kd}{2}}{{ϵ}_0}\right)$

Consider another element of width $dx$, at a distance
$x$ from the center $(x>\frac{d}{2})$
$d{C}_2=\frac{A({ϵ}_0+k(d-x\left)\right)}{dx}$

Capacitance of the second half of the capacitor is,
$\frac{1}{C_1}={\int }_{\frac{d}{2}}^d\frac{1}{d{C}_2}=\frac{1}{A}{\int }_{\frac{d}{2}}^d\frac{dx}{{ϵ}_0+kd-kx}$

$\frac{1}{C_2}=\frac{1}{kA}ln\left(\frac{{ϵ}_0+\frac{kd}{2}}{{ϵ}_0}\right)$

As, $\frac{1}{C_1}=\frac{1}{C_2}$

$C_{eq}=\frac{C_1}{2}=\frac{C_2}{2}=\frac{kA}{2ln\left(\frac{2{ϵ}_0+kd}{2{ϵ}_0}\right)}$

Hence, $\left(B\right)$ is the correct answer.