2

The usual derivation of energy stored in a capacitor is as follows $$dU=Vdq\\dU=\frac QCdq$$$$U=\frac12\frac {Q^2}C\equiv\frac12QV\tag1$$ Where $V$ is the final potential. Explicitly $$V=-\int\vec E\cdot d\vec l\tag2$$ Where $\vec E$ is the net electric field (that is, this field has contributions from both plates) between the plates.

From a mathematical point of view, it makes perfect sense why a factor of $1/2$ is coming.

From a physical point of view, we can justify this factor for the parallel plate capacitor as follows.

Firstly we will distribute our system of capacitor into two sub-systems, one being the positive plate and another being the negative plate. It is important to realize that we are only interested in the interaction energy of these two systems, usually called "energy stored by the capacitor" which is given in $(1)$. Hence we will not give any regard to the way these plates were themselves created (which does not matter anyways due to the conservative nature of electrostatics) and to the energy needed to create these plates (which is infinite anyways due to infinite charge on each plate in our approximation used). Thus taking this into account we will create our capacitor as follows: Take two plates with opposite polarity very close to each other so that the net electric field is zero everywhere. Now hold the negative plate and pull the positive plate away to some distance $l$. This procedure is used specifically to simplify the calculations done below.

Consider a charge of $dq$ on the positive plate. The energy needed to bring this to $l$ is $$\Delta\phi dq\tag3$$ where $\phi$ is the potential due to the negative plate. The difference in $\phi$ is calculated at $0$ and $l$. Since we are only talking about interaction energy, the net energy needed to bring the plates close is simply $$\int\Delta\phi dq=Q\Delta\phi\tag4$$ since $\Delta\phi$ is a constant and $Q$ is $\int dq$.

Calculation of $\Delta\phi$:

The electric field due to the negative plate is $$E_n=-\frac\sigma2\tag5$$

here $\epsilon_0=1$ is used. The potential due to this, $\Delta\phi$ is $$\Delta\phi=-\int E_ndl=\frac\sigma2 l\tag6$$ Thus the energy becomes $$U=\frac\sigma2 Ql=\frac {Q^2}2\frac Al =\frac12QV\tag7$$ Where the definition of capacitance is used and $V$ is defined as in $(2)$.

How to generalize this kind of "see-through" reasoning in terms of charge distributions and electric fields as opposed to direct definitions of capacitance, to any type of capacitor?

GedankenExperimentalist
  • 607
  • 3
    Does this answer your question? Why does a $1/2$ appear in this simple capacitor equation? – The Photon Jan 09 '23 at 00:03
  • @ThePhoton No, Here I am talking specifically in terms of fields and charge distributions rather than the process of charging the capacitor, thus involvement of current is not necessary. One can imagine building a capacitor by bringing two charged plates from infinity to some distance $l$, and it won't affect the problem due to the conservative nature of electrostatics. – GedankenExperimentalist Jan 09 '23 at 00:19
  • RE: "One can imagine building a capacitor by bringing two charged plates from infinity to some distance l, and it won't affect the problem due to the conservative nature of electrostatics. " Exactly. So the answer based on driving current onto a capacitor with fixed geometry is just as good as one based on moving the plates around with a fixed charge. – The Photon Jan 09 '23 at 00:22
  • You could certainly calculate the work required to move the two plates together from an infinite distance, but you'll get the same answer and the same factor of 1/2. – The Photon Jan 09 '23 at 00:23
  • @ThePhoton Yeah, I agree with that but I want a more "see-through" approach (like the one I provided for the parallel plate capacitor) if possible. – GedankenExperimentalist Jan 09 '23 at 00:26
  • I'm finding your approach thought-provoking. Here are two of my worries about it... (a) Why can we ignore the energy needed to put the charge $Q$ on the second plate? You seem to be taking this charge as a 'given'. (b) I agree with your $dU=EL\ dq$ for the first $dq$ of charge on the first plate, but as you put more and more increments $dq$ on to that first plate it will start to set up its own electric field, and that will change the $E$ in your $dU=EL\ dq$. – Philip Wood Jan 09 '23 at 13:03
  • You can avoid these problems, but still maintain the spirit of your approach, by starting with uncharged plates and carrying increments $dQ$ across the gap from one to the other. When charge ±Q has built up, the field in the gap will be $E=Q/\epsilon_0 A$, so the next increment $dQ$ will, as it goes across the gap, acquire energy $dU=El\ dQ=\frac{Ql\ dQ}{\epsilon_0 A}=\frac {Q\ dQ} C.$ And so on. – Philip Wood Jan 09 '23 at 14:07
  • This is an interesting approach but I don't see how it could possibly generalize to an arbitrary pair of conductors, since the charge distribution on finite conductors will change depending on their distance from each other. Heck, it won't even work for finite parallel plates once their separation exceeds their size, since you can't make the approximation $|\vec{E}| \approx \sigma/2$ beyond that point. – Michael Seifert Jan 09 '23 at 15:27
  • @Michael Seifert You generalise simply by replacing $El$ by $\int_{r=0}^l \mathbf E.d \mathbf{r}=V$ in which $V$ is the pd between the plates. So the energy acquired by $dQ$ as it goes from one plate to the other is $dU=V \ dQ=V \ CdV$. So the energy stored when enough charge has been transferred for the pd to have risen to $V_0$ is $\int_0^{V_0} CV\ dV=\frac 12 CV_0^2$. It doesn't matter, of course, whether the charge increments are transferred by some mechanical means straight across the gap between the plates, or through a battery! – Philip Wood Jan 09 '23 at 17:32
  • 1
    @William Martens Thank you. Hope it's put right. I find the time limit for editing comments frustrating, though I try to remember to keep copying the raw text on to my cyberclipboard, so as to be able to delete and re-instate when barred from publishing the edited comment. – Philip Wood Jan 09 '23 at 17:36
  • @PhilipWood No worries! sorry now it all seems great; thanks for the thank you as well; and yes I have major trouble with comments as well. really; +1 – William Martens Jan 09 '23 at 17:37
  • @PhilipWood: I'm aware (of course) that the standard method generalizes. What I was trying to point out was that the OP's method relies on being able to explicitly calculate the potential energy of the charges on one conductor due to the other conductor. For two infinite plates, this can be done, but for most other configurations it's much much harder. – Michael Seifert Jan 10 '23 at 01:35

1 Answers1

3

Remember that charging a capacitor means shifting charge from one of its plates (initially neutral) to the other (initially neutral). Thus one plate becomes more and more positively charged while the other becomes equally and oppositely charged. The shifting is most easily done by connecting a battery across the plates. The battery pumps charge through itself from one plate to the other. The charge shifting will take a finite time because of the battery's internal resistance.

Consider the time during the charging when the capacitor has a pd $V$ between its plates and a charge $Q=CV$ (in other words charges $±Q$ on its plates). When a further charge $dQ$ passes from one plate to the other, it gains potential energy $V\ dQ$. So the extra energy gained by the capacitor is $$dU=V\ dQ=V\ CdV$$ Suppose that $V_0$ is the 'final' pd between the plates. [To all intents and purposes $V_0$ will be the battery emf.] The 'final' energy stored in the capacitor will be $$U=\int_0 ^{V_0} V\ CdV=\tfrac 12 CV_0^2$$ Intuitive justification: you are transferring a total charge $Q_0=CV_0$ from one plate to the other through a pd that was $0$ initially and $V_0$ finally, that is through a mean pd of $\tfrac 12 (V_0+0)$, so the gain in PE is $\tfrac 12 (V_0+0)\times CV_0 =\tfrac 12 CV_0^2$.

Philip Wood
  • 35,641
  • I apologise for not commenting directly on your argument, but it does seem complicated and, in places, challenging (e.g. " here is the field due to one plate whereas is the voltage due to both plates in the region bounded by .") I have presented instead what I hope is a simple and easy argument. – Philip Wood Jan 08 '23 at 21:18
  • I have eased the language a bit and taken your objections made below the post into account. Please refer to the original post. – GedankenExperimentalist Jan 09 '23 at 13:54
  • I think that your method is now essentially correct. Because you're now starting with two equally and oppositely charged plates very close together, there's no "single charged plate energy" to worry me. [I still prefer the method I've given in my last comment ("You can avoid these problems...") on your question, but I would, wouldn't I ?] – Philip Wood Jan 09 '23 at 15:02