I just learnt the formula for calculating Electric Potential Energy

$W=\frac{1}{C}\int_0^Qqdq = \frac{1}{C}[\frac{1}{2}q^2]_0^Q=\frac{Q^2}{2C}$

I understand the methodology, but what I do not understand is how one can extract the $\frac{1}{C}$ as if it were a constant, when C itself actually depends on q and Q (as $C=\frac{Q}{V}$) and Q grows as a result of adding more q.

Can somebody help out?

Thank you!


This is an important misconception. When considering a capacitor with capacitance C, the capacitance is a geometric quantity. It only depends on the structure of the capacitor, so it does not depend on the charge or potential. Your formula just tells you how much charge Q you can store given a potential V and a capacitance C.

  • $\begingroup$ Thank you! One more question, though: I learnt that the Capacitance of a cable depends on the area of the positive plate and the distance between the positive and the negative plates, mathematically: C = εo·(A/D). We arrived at that conclusion by plugging V=Ed=(Qd)/(εοA) into C=Q/V. If Capacitance depends on area and distance, /through/ the definition of V, does that not imply that Capacitance depends at least on Potential? (thanks to your answer I now understand that it does not depend on Q, but I am having difficulty seeing how it does not depend on V, given the above). $\endgroup$ – Pregunto Aug 29 '18 at 17:09
  • $\begingroup$ You almost gave the answer yourself. Look at the formula $C = \varepsilon_0\frac{A}{d}$. All these values are fundamental/geometrical units and hence the capacitance does not depend on the potential. You merely used the formula for the potential as way to find the capacitance. $\endgroup$ – NDewolf Aug 29 '18 at 17:17
  • $\begingroup$ Thanks once more! One last thing: we use the formula C=Q/V in many exercises to determine what the Q is. If it only tells me how much charge /can/ be stored, and not how much /is/ actually stored, then how can I use this formula? Do I assume that the capacitance is "being used to the max", storing the full charge that it can store? $\endgroup$ – Pregunto Aug 29 '18 at 17:19
  • $\begingroup$ The capacitor will charge until it is fully charged. This time is important when considering circuits with resistors and inductors. But for these one needs to solve differential equations. $\endgroup$ – NDewolf Aug 29 '18 at 17:22

For adding more q it will consequently increase the V so the ratio of q/V remains constant if I thought your way.


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