# Doubt regarding work done in separating capacitor plates [duplicate]

Suppose there is a parallel plate capacitor with distance between plates $$d$$ and it is charged to a potential difference $$V$$ and then charging battery is disconnected. Then the plates are further separated by a distance $$d$$. We need to calculate work done in doing this.

Now, I did this problem in two different ways, and they are giving contradictory results.

The first solution involves finding change in electrostatic energy of the capacitor and accordingly find the work done which is $$CV^2/2$$. This gives the right answer.

But, here's another solution which I think is perfectly justified and want your help in finding what is the fault in this approach.

Consider one plate to be at rest and the other one moving. Now the change in potential is from $$V$$ to $$2V$$ as the distance is doubled while field is same between the plates. Therefore, the work done should be the charge on capacitor plate times the potential difference, i.e $$QV$$. Now $$Q=CV$$ which gives the work done as $$CV^2$$.

Now, I know that this answer is incorrect, but even after spending quite a time on this, I cannot figure out the fault.

• @BowlOfRed Thanks for the link. I have one last query: suppose we assume 0 to be voltage of one plate and V be the voltage of second plate initially. Then, is it that there voltages after the operation become -V/2 and 3V/2 so that the net increase of one of them is V/2 giving work as $CV^2/2$ ? Oct 31, 2018 at 7:40