So, while discussing problems with friends, I came across a capacitors problem which looked something like this:

So, my questions are:

  1. Can this be called a capacitor even though it same polarity of charge on the plates?

  2. Charge – After closing the switch, will there be equal chares on both plates (each ($Q_1 + Q_2$)/2) in order to minimise repulsion? — Update : I’ve figured this out using Guass’s law. You can ignore this part.

  3. How do I find the energy in the first configurations? Can I find the electric field due to one plate, integrate it to get potential and then multiply the charge on the other plate to get the energy? Or is there any other method? — Update : I have found a method a for this. Find the voltage (charge on inner surface)/C and then use $(1/2)(C)(V^2)$. But, I am not sure whether this is correct because it only takes into account the charges on the inner surfaces.

  4. In the second case, will the energy stored be zero because the potential difference is zero? Then again, there are some positive near each other, repelling each other and held together by by the metal plates only. Shouldn’t these have a potential energy just like two point positive charges kept some distance apart? This is why I think that the method in question 3 (i.e, the formula $(1/2)(C)(V^2)$) could be wrong when the plates have unequal charges.

  5. If energies found in part 3 and 4 are not equal, where would the difference go? I think it can't be heat since there is no resistance (assume ideal wires) — Update : I have learned through a comment on my answer to a similar question that this energy is lost in the form of EM radiations.

Your help would be appreciated.

PS: Sorry if there are too many questions. I felt that they are all related to each other, and so I put them in a single post.


3 Answers 3


1) I would not call this a capacitor. Your typical parallel plate capacitor has two charged plates kept at some potential difference (by being hooked up to opposite terminals of a battery, for example). These are just two charged plates that end up being connected and the charges balance out on each side. In otherwords, I would not say you are "storing" charge here like what you would expect a capacitor to do. You could still define a capacitance for the system, but it would not take the general form $C=\frac QV$, since we do not have a single $Q$ to reference.

In general, you can define a capacitance matrix $C_{ij}$ such that $$Q_1=C_{11}V_1+C_{12}V_2$$ $$Q_2=C_{21}V_1+C_{22}V_2$$

Of course, this is more useful when the potentials of the plates are given. However, there is such thing as an "elastance matrix" $P_{ij}$, which is the inverse of the capacitance matrix:

$$V_1=P_{11}Q_1+P_{12}Q_2$$ $$V_2=P_{21}Q_1+P_{22}Q_2$$

These matrices are symmetric so that $C_{12}=C_{21}$ and $P_{12}=P_{21}$. These terms are related to the mutual capacitance between the plates. The diagonal terms deal with the self capacitance.

2) Due to symmetry and the fact that we are dealing with perfect conductors, the charge on each plate must be equal


3) You can still figure out the potential energy difference between the two plates. If the plate separation is small, then between the plates we are looking at distances very close to plates, so we can treat them as infinite planes of charge. Using Gauss's law, we get that $E_1=\frac{\sigma _1}{2\epsilon _0}$ and $E_2=\frac{\sigma _2}{2\epsilon _0}$. Therefore, in between the plates, the field is

$$E=E_1-E_2=\frac{\sigma _1-\sigma _2}{2\epsilon _0}$$

Therefore, the potential difference between the plates is just

$$V=Ed=\frac{\sigma _1-\sigma _2}{2\epsilon _0}d$$.

However, you cannot express this in terms of $U=\frac 12 \epsilon _0 E^2$ for $E$ just inside the plates because the field is not $0$ outside of the plates.

4) There is no energy stored in the system, at least in the sense of energy typically stored in a typical capacitor. There is potential energy since the excess charges on each plate are interacting, but it would take no work to move one charge from one plate to the other since a perfect conductor is an equipotential surface. (Once you move that charge though, then moving another charge would require work, but this would involve some external force keeping the first charge in place, like from a battery, which then makes the system not an ideal conductor). Typically when you talk about energy being stored on a capacitor, you are talking about the energy needed to separate the charges and maintain that separation.

5) The electric field does work to move charges from one plate to the other. This is where the energy goes.

I am not an expert on capacitance, so anyone can correct my reasoning here if something is off. I think something we take for granted in relating the energy stored in the capacitor to the energy in the fields is that in the typical parallel plate capacitor the field is $0$ outside of the system so that the potential difference and the energy and the field are easily to relate. I think in your initial set up you have to be careful in thinking about if you just want to consider the field between the plates or the overall field.

  • $\begingroup$ First off, it's a rather difficult question, so thanks for posting an answer. The system here is such that one can just ignore the question instead of putting thought into it. Ok, so now, from what I gather, the system has a capacitance (at least in the first case). This should be ${\epsilon_0}A/d$, since capacitance depends only on the geometry, and not in the charge, voltage etc, right? $\endgroup$ Aug 25, 2018 at 17:34
  • $\begingroup$ And with regard to the energy stored, I gather that it would be difficult to find the total energy stored, i.e., in the electric field both inside/between and outside. However, we could still calculate the energy stored in the electric field between the plates using $U = \frac{CV^2}{2}$, where$V$ is the p.d. b/w the plates, or using $U = \frac{{\epsilon_0}E^2}{2}$, where$E$ is the electric field b/w the plates. Is that correct? $\endgroup$ Aug 25, 2018 at 17:42
  • $\begingroup$ ^^What was I thinking! Can't blame the system ... It's probably just that being a capacitor problem, my question has a small audience, making it a little harder to get answers. Not really the system's fault ... Also, even though I think that your answer is correct, I'll wait for a few days before accepting the answer just to see what others have to say, if anything. $\endgroup$ Aug 25, 2018 at 17:59
  • $\begingroup$ @AnuragBaundwal I'll have more time to edit my answer later, but you have to consider the mutual capacitance in this situation. Although it might work out to the normal relations anyway, since the plates are the same geometry. $\endgroup$ Aug 25, 2018 at 19:31

I will answer the questions in order 2-3-4-1.

  1. Charge – After closing the switch, will there be equal charges on both plates (each $(Q_1+Q_2)/2$) in order to minimise repulsion?

Ans– Yes, both plates will gain equal charge which can be seen through symmetry arguements.

  1. How to find energy– The simplest way to find energy is to use the fact that energy density is given by the formula : $U = \frac{1}{2}\epsilon_0E^2$

Now simply multiply the volume to find the stored energy in the electric field.

EDIT-I realize it after writing this answer that a different approach will be needed to calculate energy as electric field is present in unbounded condition(ie on the side away from the other plate). In this case Finding energy will be quite complex since formula for infinite sheet wont work here as region electric field is present till infinity.In the conventional capacitor case due to different polarity of charges the electric field cancelled out in the unbounded region and we were able to use the infinite sheet formula

  1. I recommend you to solve for energy before and after and find out for yourself.

  1. A capacitor is defined as a device which can hold charge and has energy stored in electric field. Technically speaking, you could say that this device is a capacitor as it fulfills both the criteria.
  • $\begingroup$ The formula for energy stored in an electric field! How could I forget?! 😅 ... Thanks for your answer! $\endgroup$ May 17, 2018 at 17:59
  • $\begingroup$ @Anurag._. you can do it the HC verma way too-Find the work done in seperating the two plates by distance d $\endgroup$
    – Polario
    May 17, 2018 at 18:00
  • 1
    $\begingroup$ Yeah.. but one doubt – After the charges are equally distributed, the electric fields of the plates would cancel out and the formula for energy/volume would suggest that the energy is now zero when it is clearly not so. Thoughts? $\endgroup$ May 17, 2018 at 18:03
  • 1
    $\begingroup$ @Anurag._. mistake is plain enough-The approximation formula for infinite sheet wont work in this case since region is unbounded.Calculating energy will be quite complex since in a conventional capacitor electric field in unbounded region cancels anyways $\endgroup$
    – Polario
    May 17, 2018 at 18:19
  • 1
    $\begingroup$ I'm attempting the JEE Adv. as well! All the best!! $\endgroup$ May 17, 2018 at 18:25

Case 1

Think, why the charges will remain at the plates.

In capacitors, there are opposite charges on each plate so they attract each other. But here the condition is opposite, the both charges will repel each other, so the charges would also reside on wires not only on plates.

The configuration would depend closely on how the wires are kept, at what distance the plates are and maybe many more. Eventually they will achieve lowest energy state, which is beyond calculation in Case I.

Not confirm about Case 2, but I think it would be like this -

The charges will distribute themselves over the wire and the plates, though not uniformly. But it would act as a single charge and energy would be zero as there are no other forces on the charge.

Remember, potential energy due to a conservative force depends on configuration not on how it is achieved.

  • $\begingroup$ You also have to keep in mind that in a normal capacitor there is a battery keeping the charges separate. Also, if the plates hold much more charge than the wires then you can ignore the charge in the wires. Technically the normal capacitor will also have excess charge in it's wires. $\endgroup$ Aug 27, 2018 at 1:30
  • $\begingroup$ @AaronStevens Yes, you are right that there would be some excess charge in wires in normal capacitors also. But, in normal capacitors forces are working to keep the charges on plates, while here forces are working to keep the charges away from plates. This is because in normal capacitor forces are attractive, while here forces are repulsive. So, result is- in normal capacitors excess charges in wire can be ignored, but here it can't be ignored. $\endgroup$
    – TontyTon
    Aug 27, 2018 at 8:48
  • $\begingroup$ @AaronStevens Also, it is not necessary that the two charges are separated by a battery. You can first connect the battery then plates can get charged, now you remove the battery. The condition you get is similar to Case 1. Here, you charge up the plates (yes, not by 'a battery') and remove the charging source. $\endgroup$
    – TontyTon
    Aug 27, 2018 at 8:52
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    $\begingroup$ Ok then let's imagine a system where the wires are very small and go far away from the plates. I think the OP is primarily focused on the plates. $\endgroup$ Aug 27, 2018 at 10:10

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