A) \[400%\]
B) \[66.6%\]
C) \[33.3%\]
D) \[200%\]
Correct Answer: D
Solution :
\[C=\frac{{{\varepsilon }_{0}}A}{d}\] and \[C'=\frac{{{\varepsilon }_{0}}A}{2d}+\frac{{{\varepsilon }_{0}}\left( 5A \right)}{2d}\] \[=\frac{{{\varepsilon }_{0}}A}{2d}(1+5)=\frac{6{{\varepsilon }_{0}}A}{2d}=\frac{3{{\varepsilon }_{0}}A}{d}\] \[\Rightarrow \,\,\Delta C=C'-C=\frac{3{{\varepsilon }_{0}}A}{d}-\frac{{{\varepsilon }_{0}}A}{d}=\frac{2{{\varepsilon }_{0}}A}{d}\] Percentage change in capacitance \[\frac{\Delta C}{C}=\frac{\frac{2{{\varepsilon }_{0}}A}{d}}{\frac{{{\varepsilon }_{0}}A}{d}}\times 100%=200%\]
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