Physics - Capacitor Question with Solution | TestHub

The capacitances,$C_1=\frac{k_1{\epsilon }_{0}A/2}{\left[d/2\right]}=\frac{k_1{\epsilon }_{0}A}{d}$

$C_2=\frac{k_2{\epsilon }_{0}A/2}{\left[d/2\right]}=\frac{k_0{\epsilon }_{0}A}{d}$

And$C_3=\frac{k_3{\epsilon }_{0}A/2}{\left[d/2\right]}=\frac{2k_3{\epsilon }_{0}A}{d}$

The equivalent capacitance is given by

$\Rightarrow \frac{1}{C_{eq}}=\frac{1}{C_1+C_2}+\frac{1}{C_3}=\frac{1}{\frac{{\epsilon }_{0}A}{d}\left(k_1+k_2\right)}+\frac{1}{\frac{{\epsilon }_{0}A}{d}\times 2k_3}$

$\Rightarrow \frac{1}{C_{eq}}=\frac{1}{C_1+C_2}+\frac{1}{C_3}=\frac{1}{\frac{{\epsilon }_{0}A}{d}\left(k_1+k_2\right)}+\frac{1}{\frac{{\epsilon }_{0}A}{d}\times 2k_3}$

$\Rightarrow \frac{1}{C_{eq}}=\frac{d}{{\epsilon }_{0}A}\left[\frac{1}{k_1+k_2}+\frac{1}{2k_3}\right]$

$\Rightarrow C_{eq}={\left[\frac{1}{k_1+k_2}+\frac{1}{2k_3}\right]}^{-1}.\frac{\epsilon .A}{d}$

$\therefore \frac{1}{k}=\left[\frac{1}{k_1+k_2}+\frac{1}{2k_3}\right]$$\left(\because C=\frac{K\epsilon A}{d}\right)$

$k=\frac{k_1{k}_3}{k_1+k_3}+\frac{k_2{k}_3}{k_2+k_3}$

$C_{eq}=\frac{{\epsilon }_{o}A/2}{\frac{d}{2k_1}+\frac{d}{2k_3}}+\frac{{\epsilon }_{o}A/2}{\frac{d}{2k_2}+\frac{d}{2k_3}}$

$=\frac{{\epsilon }_{o}A}{d}\left(\frac{k_1.k_{3+}}{k_1+k_3}+\frac{k_2.k_3}{k_2+k_3}\right)$

$\therefore$equivalent

$k=\frac{k_1{k}_3}{k_1+k_3}+\frac{k_2.k_3}{k_2+k_3}$