Question

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

• A. $\frac{A{ϵ}_{0}K}{d}\left(1+{\left(\frac{\alpha d}{2}\right)}^2\right)$
• B. $\frac{AK{ϵ}_0}{d}\left(1+\frac{\alpha d}{2}\right)$
• C. $\frac{A{ϵ}_{0}K}{d}\left(1+\frac{{\alpha }^2{d}^2}{2}\right)$
• D. $\frac{AK{ϵ}_0}{d}\left(1+\alpha d\right)$

Answer 1

Answer: (B)

Capacitance of element $=\frac{k{\epsilon }_{0}A}{dx}$
Capacitance of element, $C^{\prime }=\frac{k_0\left(1+\alpha x\right){\epsilon }_{0}A}{dx}$
$\sum \frac{1}{C^{\prime }}=$$\underset{0}{\overset{d}{\int }}$$\frac{dx}{k_0{\epsilon }_{0}A\left(1+\alpha x\right)}$
$\frac{1}{C}=\frac{1}{k_0{\epsilon }_{0}A\alpha }\ell n\left(1+\alpha d\right)$
Given $\alpha d<<1$
$\frac{1}{C}=\frac{1}{k_0{\epsilon }_{0}A\alpha }\left(\alpha d-\frac{{\alpha }^2{d}^2}{2}\right)$
$\frac{1}{C}=\frac{d}{k_0{\epsilon }_{0}A}\left(1-\frac{\alpha d}{2}\right)$
$C=\frac{k_0{\epsilon }_{0}A}{d}\left(1+\frac{\alpha d}{2}\right)$