Work done in moving the plates of the capacitors

[bs vasanth]

A parallel plate capacitor of circular cross section r>>d, and separation d. It is charged to potential V then disconnected from the circuit . What will be the work done in moving the capacitor from d to d1?

Answer:
Here the initial energy is 1/2C$V^{2}$
Where C=Aε/d
While moving the capacitor both the capacitance and potential change:
Cf= Aε/d1
Vf=∫E.dl (with the limits 0 to d1)
Vf= E*d1
We know that the E doesn't change ( Under the approx. as r>>d )
Therefore initial potential and final potential can be related as V= E*d
Vf=V*d1/d
Work done = 1/2( Cf*$Vf^{2}$ - C*$V^{2}$ )
= Aε/2 ($(V*d1/d)^{2}/d1$ - C*$V^{2}/d$ )
= $AεV^{2}( d1/d -1)/2d$
I want to know if this answer is correct.

 

[Meir Achuz]

It's easier to use U=Q^2/2c because Q doesn't change for a disconnected capacitor, but your answer is correct.