Question

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

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

Answer 1

Answer: (B)

Initial capacitance
$C=\frac{{ϵ}_{0}A}{d}$

When it is half filled by a dielectric of dielectric constant $K$, then
$C_1=\frac{K{ϵ}_{0}A}{d/2}=2K\frac{{ϵ}_{0}A}{d}$
and $C_2=\frac{{ϵ}_{0}A}{d/2}=\frac{2{ϵ}_{0}A}{d}$
$\therefore \frac{1}{C^{\prime \prime }}=\frac{1}{C_1}+\frac{1}{C_2}$
$=\frac{d}{2(ϵ{)}_{0}A}\left(\frac{1}{K}+1\right)=\frac{d}{2(ϵ{)}_{0}A}\left(\frac{1}{5}+1\right)$
$=\frac{6}{10}\frac{d}{{ϵ}_{0}A}$
$\therefore C^{\prime \prime }=\frac{5{ϵ}_{0}A}{3d}$
Hence, increase in capacitance
$=\frac{\frac{5}{3}\frac{{ϵ}_{0}A}{d}-\frac{{ϵ}_{0}A}{d}}{\frac{{ϵ}_{0}A}{d}}=\frac{5}{3}-1=\frac{2}{3}=66.6\%$