A) \[\frac{{{\varepsilon }_{0}}A}{d+\frac{b}{2}}\]
B) \[\frac{{{\varepsilon }_{0}}A}{2d}\]
C) \[\frac{{{\varepsilon }_{0}}A}{d-b}\]
D) \[\frac{2{{\varepsilon }_{0}}A}{d+\frac{b}{2}}\]
Correct Answer: C
Solution :
(c) \[\frac{{{\varepsilon }_{0}}A}{d-b}\] As capacitance, \[{{C}_{0}}=\frac{{{\varepsilon }_{0}}A}{d}\] \[\therefore \]After inserting copper plate, \[C=\frac{{{\varepsilon }_{0}}A}{d-b}\]You need to login to perform this action.
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