Question

A capacitor is formed by two square metal-plates of edge $a$ separated by a distance $d$. Dielectrics of dielectric constants $K_1$ and $K_2$ are filled in the gap as shown figure . Find the capacitance.

Answer 1

Let a small strip of length $dx$ at distance $x$ from one end making an angle $\varphi$

Now both capacitors are in series hence

$\frac{1}{d{C}_{net}}=\frac{1}{d{C}_1}+\frac{1}{d{C}_2}$

$=\frac{d_1}{k_1{\epsilon }_0{A}_1}+\frac{d_2}{k_2{\epsilon }_0{A}_2}$

$=\frac{d-x\tan \varphi }{k_1{\epsilon }_0{A}_1}+\frac{x\tan \varphi }{k_2{\epsilon }_0{A}_2}$

$d{C}_{net}=\frac{{\epsilon }_{0}adx}{\frac{d-x\tan \varphi }{k_1}+\frac{x\tan \varphi }{k_2}}$

$={\epsilon }_{0}a{k}_1{k}_2\int \frac{dx}{k_{2}d+(k_1-k_2)x\tan \varphi }$

Now integration we get

$=\frac{{\epsilon }_{0}a{k}_1{k}_2}{(k_1-k_2)\tan \varphi }(\log K_1-\log K_2)$

From figure $\tan \varphi =\frac{d}{a}$

$=\frac{{\epsilon }_{0}a{k}_1{k}_2}{(k_1-k_2)\frac{d}{a}}(\log K_1-\log K_2)$

$d{C}_{net}=\frac{{\epsilon }_0{a}^2{k}_1{k}_2}{(k_1-k_2)d}\left(\log \frac{k_1}{k_2}\right)$