# 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 in figure. Find the capacitance.?

## Answer: C = (ε_0K_1K_2a^2ln[K_1/K_2])/((K_1 - K_2)d)

Mar 14, 2018

Consider an infinitesimal strip of of capacitor of thickness $\mathrm{dx}$ at a distance $x$ as shown. Keeping in view the position of terminlas we see that the infinitesimal capacitors so formed are connected in series.
Plate separation for lower part is $x \tan \alpha$ whereas for the upper part it is $\left(d - x \tan \alpha\right)$.
We know that equivalent capacitance of capacitors connected in series is given by the expression

$\frac{1}{{\mathrm{dC}}_{e q}} = \frac{1}{{\mathrm{dC}}_{{K}_{1}}} + \frac{1}{{\mathrm{dC}}_{{K}_{2}}}$

Inserting value for Parallel-plate capacitor of plate area $A$ and plate separation $t$ as $C = {\epsilon}_{0} \frac{A}{t}$, we get

$\frac{1}{{\mathrm{dC}}_{e q}} = \frac{d - x \tan \alpha}{{K}_{1} {\epsilon}_{0} \left(a \mathrm{dx}\right)} + \frac{x \tan \alpha}{{K}_{2} {\epsilon}_{0} \left(a \mathrm{dx}\right)}$
$\implies \frac{1}{{\mathrm{dC}}_{e q}} = \frac{1}{{\epsilon}_{0} a \mathrm{dx}} \left[\frac{d - x \tan \alpha}{{K}_{1}} + \frac{x \tan \alpha}{{K}_{2}}\right]$
$\implies {\mathrm{dC}}_{e q} = \frac{{\epsilon}_{0} a \mathrm{dx}}{\frac{d - x \tan \alpha}{{K}_{1}} + \frac{x \tan \alpha}{{K}_{2}}}$
$\implies {\mathrm{dC}}_{e q} = \frac{{K}_{1} {K}_{2} {\epsilon}_{0} a \mathrm{dx}}{{K}_{2} \left(d - x \tan \alpha\right) + {K}_{1} \left(x \tan \alpha\right)}$
$\implies {\mathrm{dC}}_{e q} = \frac{{K}_{1} {K}_{2} {\epsilon}_{0} a \mathrm{dx}}{{K}_{2} d + \left({K}_{1} - {K}_{2}\right) \left(\tan \alpha\right) x}$

Integrating both sides with respect to respective variables from $C = 0 \to {C}_{e q} \mathmr{and} x = 0 \to a$ we get

${C}_{e q} = \left({K}_{1} {K}_{2} {\epsilon}_{0} a\right) {\int}_{0}^{a} \frac{\mathrm{dx}}{{K}_{2} d + \left({K}_{1} - {K}_{2}\right) \left(\tan \alpha\right) x}$

Using the following integral we get

$\int \frac{\mathrm{dx}}{b \cdot x + c} = \ln \frac{\left\mid b \cdot x + c \right\mid}{b}$

${C}_{e q} = \frac{{K}_{1} {K}_{2} {\epsilon}_{0} a}{\left({K}_{1} - {K}_{2}\right) \left(\tan \alpha\right)} {\left[\ln \left(| {K}_{2} d + \left({K}_{1} - {K}_{2}\right) \left(\tan \alpha\right) x |\right)\right]}_{0}^{a}$

=>C_(eq)=(K_1K_2epsilon_0a)/((K_1-K_2)(tanalpha))[ln(|K_2d+(K_1-K_2)(tanalpha)a|)-(ln(|K_2d|)]

Inserting $\tan \alpha = \frac{d}{a}$, we get
${C}_{e q} = \frac{{K}_{1} {K}_{2} {\epsilon}_{0} {a}^{2}}{\left({K}_{1} - {K}_{2}\right) d} \left[\ln \left(| {K}_{2} d + \left({K}_{1} - {K}_{2}\right) d |\right) - \ln \left(| {K}_{2} d |\right)\right]$

Simplifying

C_(eq)=(K_1K_2epsilon_0a^2)/((K_1-K_2)d)[ln((K_1d)-(ln(K_2d)]
$\implies {C}_{e q} = \frac{{K}_{1} {K}_{2} {\epsilon}_{0} {a}^{2}}{\left({K}_{1} - {K}_{2}\right) d} \ln \left({K}_{1} / {K}_{2}\right)$