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