Question

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 which varies as $k\left(x\right)=k(1+\alpha x)$, 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. $\frac{A{\epsilon }_{0}k}{d}[1+{\left(\frac{\alpha d}{2}\right)}^2]$
• B. $\frac{Ak{\epsilon }_0}{d}[1+\left(\frac{\alpha d}{2}\right)]$
• C. $\frac{A{\epsilon }_{0}k}{d}[1+\left(\frac{{\alpha }^{2}d}{2}\right)]$
• D. $\frac{Ak{\epsilon }_0}{d}[1+\alpha d]$

Answer 1

The correct option is B $\frac{Ak{\epsilon }_0}{d}[1+\left(\frac{\alpha d}{2}\right)]$

Given, $k\left(x\right)=k(1+\alpha x)$
$dC=$ $\frac{A{\epsilon }_{0}k}{dx}$
Since all capacitance are in series, we can apply
$$\frac{1}{Ceq}=\int \frac{1}{dC}={\int }_0^d\frac{dx}{k(1+\alpha x){ϵ}_{0}A}$$
$$\frac{1}{Ceq}={\left[\frac{\ln (1+\alpha x)}{k{ϵ}_{0}A\alpha }\right]}_0^d$$
On putting the limits from 0 to d
$$=\frac{\ln (1+\alpha d)}{k{ϵ}_{0}A\alpha }$$
Using expression $\ln (1+x)=x-\frac{x^2}{2}+...$
And putting $x=\alpha d$ where, x approaches to 0.
$$\frac{1}{C}=\frac{d}{k{ϵ}_{0}Ad\alpha }[\alpha d-\frac{{\left(\alpha d\right)}^2}{2}]$$
$$\frac{1}{C}=\frac{d}{k{ϵ}_{0}A}[1-\frac{\alpha d}{2}]$$
$$C=\frac{k{ϵ}_{0}A}{d}[1+\frac{\alpha d}{2}]$$