A) 1
B) 2
C) 3
D) 4
Correct Answer: D
Solution :
\[C=\frac{{{\varepsilon }_{0}}A}{d}\] ......(i) \[C'=\frac{{{\varepsilon }_{0}}KA}{2d}\] ......(ii) From equation (i) and (ii) \[\frac{C'}{C}=\frac{K}{2}\] Þ \[2=\frac{K}{2}\,\,\Rightarrow K=4\]You need to login to perform this action.
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