question_answer

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 %

Correct Answer: B

Solution :

Initial capacitance                 \[C=\frac{{{\varepsilon }_{0}}\,A}{d}\] When it is half filled by dielectric of dielectric constant fc, then                 \[{{C}_{1}}=\frac{K\,{{\varepsilon }_{0}}\,A}{d/2}=2K\frac{{{\varepsilon }_{0}}\,A}{d}\] and        \[{{C}_{2}}=\frac{{{\varepsilon }_{0}}\,A}{d/2}=\frac{2{{\varepsilon }_{0}}\,A}{d}\] \[\therefore \]  \[\frac{1}{C}=\frac{1}{{{C}_{1}}}+\frac{1}{{{C}_{2}}}=\frac{d}{2{{\varepsilon }_{0}}A}\left( \frac{1}{K}+1 \right)\]                                 \[=\frac{d}{2{{\varepsilon }_{0}}\,A}\left( \frac{1}{5}+1 \right)\]                                 \[=\frac{6}{10}\frac{d}{{{\varepsilon }_{0}}A}\] \[\therefore \]  \[C=\frac{5{{\varepsilon }_{0}}A}{3d}\] Hence, increase in capacitance                 \[=\frac{\frac{5}{3}\frac{{{\varepsilon }_{0}}A}{d}-\frac{{{\varepsilon }_{0}}A}{d}}{\frac{{{\varepsilon }_{0}}\,A}{d}}\]                 \[=\frac{5}{3}-1=\frac{2}{3}\]                 \[=\,66\,%\]