Question

A parallel plate air capacitor has a capacitance $C$. When it is half filled with a dielectric of dielectric constant $5$, the percentage increase in the capacitance will be

• A. $400\%$
• B. $66.6\%$
• C. $33.3\%$
• D. $200\%$

Answer 1

The correct option is B $66.6\%$

Initial capacitance$=\frac{{ϵ}_{\circ }A}{d}$

When it is half filled by a dielectric of dielectric constant $k$ then

$C_1=\frac{k{ϵ}_{\circ }A}{\frac{d}{2}}=2k\frac{{ϵ}_{\circ }A}{d}$

and $C_2=\frac{{ϵ}_{\circ }A}{\frac{d}{2}}=\frac{2{ϵ}_{\circ }A}{d}$

$\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}$

$=\frac{d}{2{ϵ}_{\circ }A}(\frac{1}{k}+1)$

$=\frac{d}{2{ϵ}_{\circ }A}(\frac{1}{5}+1)$

$=\frac{d}{2{ϵ}_{\circ }A}\left(\frac{6}{5}\right)$

$=\frac{3d}{5{ϵ}_{\circ }A}$

$\therefore C=\frac{5{ϵ}_{\circ }A}{3d}$

Hence, $\%$ increase in capacitance is

$=\left(\frac{\frac{5{ϵ}_{\circ }A}{d}-\frac{{ϵ}_{\circ }A}{d}}{\frac{{ϵ}_{\circ }A}{d}}\right)\times 100=\frac{2}{3}\times 100=66.6\%$