A) 10
B) 20
C) 15
D) 13.33
Correct Answer: D
Solution :
\[C=\frac{A{{\varepsilon }_{0}}}{d}=10\mu F\] \[{{C}_{1}}=\frac{A{{\varepsilon }_{0}}}{d-t+\frac{t}{k}}=\frac{A{{\varepsilon }_{0}}}{d-\frac{d}{2}+\frac{d}{2k}}=\frac{A{{\varepsilon }_{0}}}{\frac{d}{2}\left( 1+\frac{1}{2} \right)}=\frac{4}{3}.\frac{A{{\varepsilon }_{0}}}{d}\] \ \[{{C}_{1}}=\frac{4}{3}\times 10=13.33\mu F\]
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