A capacitor is given a charge `q`. The distance between the plates of the capacitor is `d`. One of the plates is fixed and the other plate is moved away from the other till the distance between them becomes `2d`. Find the work done by the external force.

Answer

A, B, D

Explanation

When one plate is fixed, the other is attracted towards the first with a force
`F=q^2/(2Aepsilon_0)=constant`
Hence, an external force of same magnitude will have to be applied in oppositve direction to increase the separation between the plates.
`:. W=F(2d-d)=(q^2d)/(2Aepsilon_0)`
Alternate solution: `W=/_\U=U_f-U_i=q^2/(2C_f)-q^2/(2C_i)`..........i
Here, `C_f=(epsilon_A)/(2d)` and `C_i=(epsilonA)/d`
Substituting in eqn i we have
`W=q^2/(2((epsilonA)/(2d)))-q^2/(2((epsilon_0A)/d))=(q^2d)/(2epsilon_0A)`