As far as I know the voltage between the plates of a capacitor will be reduced due to discharging. I wonder how much the voltage would be changed if the capacitor lost %25 of its stored energy.

Basically in order to calculate the energy I use these three well-known formulas including $U=\frac12CV^2$, $U=\frac12QV$, $U=\frac{1}{2c}Q^2$ .

In the current circumstance which the capacitor has been already charged and also has been disconneced from the battery, I'm not pretty sure which one should be used. That's what I can't figure it out well.

I went ahead and yield these two possible solutions. First one has been offered by the book which the question is coming from and the second one is my own challenging solution! (I can provide some shots)

Solution a) We know that $Q$ (charge) will be constant. Using the formula $U=\frac12QV$ we can say $U'=\frac34U$ so then $V'= \frac34V$. It means $V$ will be reduced $25$ percent.

Solution b) We also know that $C$ (capacitance) will not be changed due to discharging. So we can say $C$ is constant and we can use the formula $U=\frac12CV^2$. As $U'=\frac34U$ then $V'=\frac{\sqrt3}2V$. So $V$ has been reduced about $13$ percent.

Please let me know which formula should be used in this particular circumstance.

Any help would be really appreciated.


closed as off-topic by Aaron Stevens, GiorgioP, Yashas, John Rennie, Jon Custer May 2 at 14:46

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  • 1
    $\begingroup$ Why do you think (a) is correct? In particular, what do you think discharging means? $\endgroup$ – J. Murray May 2 at 2:14
  • $\begingroup$ @J.Murray The reason is somehow funny! Last couple questions which I was solving were all about a charged capacitor (as an isolated system). So for all of them I should consider Q as a constant and find other parameters like E, V & etc. I just forgot it's not an isolated system anymore! Indeed the book also has offered solution a! Quite strange! So you may agree it's not my fault too much..! I liked second part of your amazing comment! lmao $\endgroup$ – Marelbiker May 2 at 6:05
  • $\begingroup$ The question has been edited. $\endgroup$ – Marelbiker May 3 at 1:01
  • $\begingroup$ I've edited my question but still "on hold". I'm not sure Mods here are okay! $\endgroup$ – Marelbiker May 8 at 12:37

(b) is correct. Because in discharging, charge decreases. So use, $(1/2)CV^2$ or $Q^2/(2C)$. C obviously remains constant. But Q and V change (decrease) by same amount.

  • $\begingroup$ Yes. You're completely right. Thanks for participating. $\endgroup$ – Marelbiker May 2 at 5:57

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