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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.

So, please help.

Thanks.

Abhinav Dhawan
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  • @BowlOfRed Q is constant as the battery is removed and though C is not constant, I have only used a relation Q=CV which is valid – Abhinav Dhawan Oct 31 '18 at 06:25
  • https://physics.stackexchange.com/questions/265149/energy-changes-due-increasing-plate-separation-of-unconnected-charged-capacitor – BowlOfRed Oct 31 '18 at 07:07
  • I think that I see the problem: The E-field between the plates is due to the charges on BOTH plates. But to calculate the energy required to move one charged object with respect to another charged object, you have to use the E-field for only ONE of the objects while you move the other object with respect to it. That explains the missing factor of 1/2. Unfortunately, it's getting late here so I don't have time to write a detailed answer. –  Oct 31 '18 at 07:10
  • @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 $CV^2/2$ ? – Abhinav Dhawan Oct 31 '18 at 07:40

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