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I have a simple doubt that energy $(U)$ stored between two parallel plates of capacitance $C$, having charge $Q$ and applied potential is $V$ is given by three ways by formulas

● $U= \frac{C×V^2}{2}$

● $U= \frac{Q×V}{2}$

● $U= \frac{Q^2}{2C}$

And yes I know how to prove them. But If I am asked that how does energy stored in such capacitor varies with capacitance $C$? OR how it varies with $Q$?. Then should I say that it is inversely proportional to $C$ or it is directly proportional to $C$? Or it is directly proportional to $Q$ or $Q^2$? And also is it directly proportional to $V$ or is independent of it?

I have asked this question because I have come through a question given below:

The plate seperation in a parallel plate capacitor is $d$ and plate area is $A$. If it is charged to $V$ volt and battery is disconnected then work done in increasing the plate separation to $2d$ will be?

I used $W=U2-U1$ and if my calculations are correct then calculating value of $U$ from $U= \frac{C×V^2}{2}$ and $U= \frac{Q^2}{2C}$ gave me two different answers for $W$.

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The energy stored is a function of two independent variables $(Q=CV)$, so if asked about how the energy stored depends on the capacitance you need also to be told or state yourself which of the independent variables is kept constant, either voltage or charge in this case.

So the energy stored is proportional to the capacitance if the voltage is kept constant or the energy stored is inversely proportional to the capacitance if the charge is kept constant.

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  • $\begingroup$ So suppose I am changing capacitance along with Voltage then obviously Charge will also keep changing i.e everything has changed, then if I am asked how U varies with C. Then does this question has any answer or is it a useless question? $\endgroup$ – Avi Mar 14 '17 at 10:17
  • $\begingroup$ I have updated my question, please have a look on it. $\endgroup$ – Avi Mar 14 '17 at 10:30
  • $\begingroup$ So in my updated question which term among Q, C and V is being varied by varying seperation d. $\endgroup$ – Avi Mar 14 '17 at 10:38
  • $\begingroup$ @Avi With your supplementary question as soon as you mention that the supply is is disconnected this immediately means that the charge stored on the capacitor will be constant. $\endgroup$ – Farcher Mar 14 '17 at 10:48
  • $\begingroup$ So should I use the third equation or the second one? $\endgroup$ – Avi Mar 14 '17 at 11:22

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