Question

A capacitor of capacity C is connected with a battery of potential V in parallel. The distance between its plates is reduced to half at once, assuming that the charge remains the same. Then to charge the capacitance up to the potential V again, the energy given by the battery will be
A) $ C{V^2}/4 $
B) $ C{V^2}/2 $
C) $ 3C{V^2}/4 $
D) $ C{V^2} $

Answer 1

Hint: In this question, we will first calculate the charge induced across the plates of the capacitor. When the distance between the plates is halved, the capacitance will increase by two times too.

Complete step by step answer
We know that the capacitance of a capacitor is given as
 $ C= \dfrac {\epsilon A}{d} $
where $ A $ is the area of the capacitor and $ d $ is the distance between the plates of the capacitor.
So, we can see that when the distance between the plates of the capacitor is halved, the capacitance will increase to twice as $ C \propto \dfrac{1}{d} $ . If we indicate the new capacitance by $ C' $ ,
 $ \dfrac{{C'}}{C} = \dfrac{d}{{d/2}} = \dfrac{2}{1} $
So, we can write
 $ C' = 2C $
Now, the charge of a capacitor is given as
 $ Q = CV $
Since the capacitance doubles up, the charge across the plates $ Q' $ in the new situation will also double up. So $ Q' = 2Q = 2CV $ .
The work done by the battery when the charge across the capacitor will double will be
 $ W = \Delta QV $
Since $ \Delta Q = Q' - Q $
 $ \Rightarrow \Delta Q = 2CV - CV = CV $ ,
The work done will be
 $ W = \left( {CV} \right)V $
 $ \Rightarrow W = C{V^2} $
Hence the correct choice will be option (D).

Note
Since the battery will remain connected to the capacitor plates, the potential across the capacitor will also remain constant. Hence the change in charge stored by the capacitor will only be due to the change in the capacitance due to a change in the distance between the plates. In the process, the battery will do work on the capacitor to provide more energy to the capacitor.