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 a dielectric of dielectric constant K, then \[{{C}_{1}}=\frac{K{{\varepsilon }_{0}}A}{d/2}=2K\frac{{{\varepsilon }_{0}}A}{d}\] and \[{{C}_{2}}=\frac{{{\varepsilon }_{0}}A}{d/2}=\frac{2{{\varepsilon }_{0}}A}{d}\] \[\frac{1}{C}=\frac{1}{{{C}_{1}}}+\frac{1}{{{C}_{2}}}=\frac{d}{2{{\varepsilon }_{0}}A}\left( \frac{1}{K}+1 \right)\] \[=\frac{d}{2{{\varepsilon }_{0}}A}\left( \frac{1}{5}+1 \right)\] \[=\frac{6}{10}\frac{d}{{{\varepsilon }_{0}}A}\] \[C=\frac{5{{\varepsilon }_{0}}A}{3d}\] Hence, increase in capacitance \[=\frac{\frac{5}{3}\frac{{{\varepsilon }_{0}}A}{d}-\frac{{{\varepsilon }_{0}}A}{d}}{\frac{{{\varepsilon }_{0}}A}{d}}\] \[=\frac{5}{3}-1=\frac{2}{3}=66.6%\]