question_answer

Four plates of equal area \[A\]are separated by equal distances \[d\] and are arranged as shown in the figure. The equivalent capacity is

A)                     \[\frac{2{{\varepsilon }_{0}}A}{d}\]                      

B)            \[\frac{3{{\varepsilon }_{0}}A}{d}\]

C)                    \[\frac{3{{\varepsilon }_{0}}A}{d}\]

D)                                      \[\frac{{{\varepsilon }_{0}}A}{d}\]

Correct Answer: A

Solution :

           The given circuit is equivalent to a parallel combination two identical capacitors Hence equivalent capacitance between A and B is                    C = \[\frac{{{\varepsilon }_{0}}A}{d}+\frac{{{\varepsilon }_{0}}A}{d}\]                      \[=\frac{2{{\varepsilon }_{0}}A}{d}\]