question_answer

Two parallel conducting plates each of area A, are placed 3d apart and are both earthed. A third plate, identical with the first two, is placed at a distance d from one of the earthed plates and is given a charge of Q coulomb. The potential of the central plate is

A)  \[\frac{3}{2}\frac{Qd}{{{\varepsilon }_{0}}A}\]                 

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

C)  \[\frac{2}{3}\frac{Qd}{{{\varepsilon }_{0}}A}\]                 

D)  \[\frac{3Qd}{{{\varepsilon }_{0}}A}\]

Correct Answer: C

Solution :

 The arrangement of the plates can be rearranged as two capacitors connected in parallel. Equivalent capacitance of the system \[{{C}_{eq}}={{C}_{1}}+{{C}_{2}}=\frac{{{\varepsilon }_{0}}A}{d}+\frac{{{\varepsilon }_{0}}A}{2d}=\frac{3{{\varepsilon }_{0}}A}{2d}\] Hence\[Q={{C}_{eq}}(V-0)=\left( \frac{3{{\varepsilon }_{0}}A}{2d} \right)V\Rightarrow V=\frac{2Qd}{3{{\varepsilon }_{0}}A}\]