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Q. The capacitance of an air capacitor is $15\, \mu F$ the separation between the parallel plates is $6\, mm$. A copper plate of $3\, mm$ thickness is introduced symmetrically between the plates. The capacitance now becomes

Electrostatic Potential and Capacitance

Solution:

By using $C_{\text {air }}=\frac{\varepsilon_{0} A}{d}, C_{\text {medium }}$
$=\frac{\varepsilon_{0} A}{d-t+\frac{t}{K}}$
For $K=\infty, C_{\text {medium }}=\frac{\varepsilon_{0} A}{d-t}$
$\Rightarrow \frac{C_{\text {medium }}}{C_{\text {air }}}=\frac{d}{d-t}$
$\Rightarrow \frac{C_{\text {medium }}}{15}=\frac{6}{6-3}$
$\Rightarrow C_{\text {medium }}=30\, \mu F$