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