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. $200$ %
• C. $33.3$$ %
• D. $66.6$ %

Answer 1

Answer: (D)

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

When it is half filIed by a dielectric of dielectric constant K, then

$C_1=\frac{K{ϵ}_{o}A}{d/2}=2K\frac{{ϵ}_{o}A}{d}$

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

$\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}=\frac{d}{2{ϵ}_{o}A}(\frac{1}{K}+1)=\frac{d}{2{ϵ}_{o}A}(\frac{1}{5}+1)=\frac{3}{5}\frac{d}{{ϵ}_{o}A}$

$C=\frac{5}{3}\frac{{ϵ}_{o}A}{d}$

Hence % increase in capacitance

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