A) \[\frac{K}{2\left( K+1 \right)}\]
B) \[\frac{2\,K}{K+1}\]
C) \[\frac{2\,K}{K+1}\]
D) \[\frac{4\,K}{3\,K+1}\]
Correct Answer: D
Solution :
\[C\,=\,\frac{{{\varepsilon }_{0}}A}{d}\,\,and\,\,C'=\frac{{{\varepsilon }_{0}}A}{d-\frac{d}{4}\left( 1-\frac{1}{K} \right)}=\frac{4\,{{\varepsilon }_{0}}A}{3d+\frac{d}{K}}\] \[\Rightarrow \,\,\,C'=\frac{4K}{(3K+1)}\left( \frac{{{\varepsilon }_{0}}A}{d} \right)=\frac{4\,KC}{(3K+1)}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{C'}{C}=\frac{4K}{(3K+1)}\]You need to login to perform this action.
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