Physics - Electrostatics Question with Solution | TestHub

Since, we know,

Capacitance of a parallel plate capacitance,

$C=\frac{K{\epsilon }_{0}A}{d}$

Where, symbols have their usual meaning.

In the given figure,

$\frac{1}{C_{\mathrm{n}\mathrm{e}\mathrm{t}}}=\frac{1}{C_1}+\frac{1}{C_2}=\frac{1}{\frac{K_1{\epsilon }_{0}A}{d/2}}+\frac{1}{\frac{K_2{\epsilon }_{0}A}{d/2}}$

$=\frac{1}{\frac{3{\epsilon }_{0}A}{d/2}}+\frac{1}{\frac{5{\epsilon }_{0}A}{d/2}}$

$=\frac{d}{6{\epsilon }_{0}A}+\frac{d}{10{\epsilon }_{0}A}=\frac{d}{2{\epsilon }_{0}A}\left[\frac{1}{3}+\frac{1}{5}\right]$

$=\frac{8d}{15\times 2\times {\epsilon }_{0}A}=\frac{4d}{15{\epsilon }_{0}A}$

Thus,$C_{\mathrm{n}\mathrm{e}\mathrm{t}}=\frac{15{\epsilon }_{0}A}{4d}$