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

Answer: (B)

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