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                 \[=\frac{\frac{5}{3}\frac{{{\varepsilon }_{0}}A}{d}-\frac{{{\varepsilon }_{0}}A}{d}}{\frac{{{\varepsilon }_{0}}A}{d}}\]                 \[=\frac{5}{3}-1=\frac{2}{3}=66.6%\]