Question

A parallel plate capacitor $C$ carries a charge $Q$. The distance between the plates is doubled by application of force, the work done by force is

• A. $zero$
• B. $\frac{Q^2}{2C}$
• C. $\frac{Q^2}{C}$
• D. $\frac{Q^2}{4C}$

Answer 1

The correct option is B $\frac{Q^2}{2C}$

Energy before separation,

$E=\frac{Q^2}{2C}$

Energy after separation,

$E=\frac{Q^2}{C}$

Work done is difference in energy

$W=\frac{Q^2}{2C}$