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