A) \[2{{\varepsilon }_{0}}A/d\]
B) \[3{{\varepsilon }_{0}}A/d\]
C) \[2{{\varepsilon }_{0}}A/3d\]
D) \[3{{\varepsilon }_{0}}A/2d\]
Correct Answer: D
Solution :
As encircled capacitor is short circuited Hence, \[{{C}_{PQ}}=\frac{{{\varepsilon }_{0}}A}{d}+\frac{{{\varepsilon }_{0}}A}{2d}=\frac{3{{\varepsilon }_{0}}A}{2d}\]You need to login to perform this action.
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