I don't understand why there is any change in initial and final energy since we have already assumed a perfectly conductive circuit. I mean, theoretically at least, there should be no change in energy.

Now, considering there is a change in energy at all, is it because electrons accelerate in moving from one capacitor to another, so energy gets dissipated in the form of electromagnetic radiation?

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Image source: NCERT Physics Textbook for Class XII Part I, page 82

  • $\begingroup$ Do you mean "total energy" of the entire system, or are you just wondering about the energy stored in the capacitor? $\endgroup$ Mar 11, 2020 at 14:51
  • $\begingroup$ You could consider a simpler situation: take a single charged capacitor and short it so that the voltage becomes $0$. Where does the energy go? $\endgroup$ Mar 11, 2020 at 16:00
  • $\begingroup$ Related: What happens to half of the energy in a circuit with a capacitor? $\endgroup$
    – Vishnu
    May 23, 2020 at 11:39

4 Answers 4


Short Version: When the resistance in the above circuit is zero, the system will never attain equilibrium.

Long Version:

Any time you want to understand what happens with a zero resistance you need to start with a small but non-zero resistance and see what happens in the limit of $R \to 0$.

by John Rennie sir in chat.

When a capacitor of capacitance $C$ is given a charge $Q$, and at time $t=0$, it is connected to an uncharged capacitor of equal capacitance through a resistance $R$, the charge on the second capacitor as a function of time is given by the following equation:

$$Q(t)=\frac Q 2 (1-e^{-2t/RC})$$

When the resistance in the circuit is non-zero, it takes some time before the charges in the two identical capacitors equalise. The charge on the two capacitors after a long time will be equal to $Q/2$. If you calculate the energy stored in the electric field in the capacitors before and after the connections are made, you could clearly see some energy loss (and this is what the example problem in the question is about).

When there is some resistance in the circuit, the energy is lost due as heat developed in the resistors. But your question is how (and why) is the energy lost when there is zero resistance in the circuit? It turns out that when you analyse the system when $R\to0$ using the equation mentioned earlier, the time constant ($RC$) decreases with decrease in $R$. Or in other words, the time required to attain equilibrium reduces.

But when you put $R=0$, we get an invalid equation. Or we can imagine the system to never attain equilibrium. Usually when you have non-zero resistance, the energy lost due to thermal energy damps the oscillations of charges in the capacitors. You can imagine this to be similar to that of a pendulum which stops oscillating after some time due to energy lost to the surroundings. However, when no energy is lost to the surroundings, the pendulum keeps on oscillating. In a similar way, when there is no resistance in the circuit in your example, the system never attains equilibrium. Charges get continuously transferred back and forth between the two capacitors somewhat similar to how energy is interchanged between an inductor and a capacitor in an LC oscillation (as pointed out by Keith McClary).


When charges move due to an electric field (i.e. down the potential gradient) the electric potential energy of the system decreases. When you put in the second capacitor there is a brief moment where an electric field is present that causes excess charges to move to the other plates. Hence you have less energy stored.

Or you can view it as doubling the area of the plates, thus decreasing the field between each plate by half, which decreases the energy by half (two systems each with a quarter of the original energy). Although this is somewhat already covered by the solution you supply, but not explicitly. But this gives a better picture, as it doesn't really matter how we actually make the system, as potential energy only depends on position of the charges, not how they got there.

Any way you look at it, "I mean, theoretically at least, there should be no change in energy" is not a true statement. We are talking about electric potential energy, which can change even if there are no dissipative mechanisms in the system. You don't need to bring in any sort of radiation. You can imagine moving each charge one-by-one very slowly if you want.

  • $\begingroup$ If you move each charge one-by-one very slowly they will be subject to electrostatic forces which will account for the difference in energy. $\endgroup$ Mar 11, 2020 at 5:37
  • $\begingroup$ @KeithMcClary Yes, you got it $\endgroup$ Mar 11, 2020 at 11:40
  • 1
    $\begingroup$ "We are talking about electric potential energy, which can change even if there are no dissipative mechanisms in the system." - In the ideal case of zero resistance, I think, we can't talk about "electrostatics" at all. We'll not see the capacitor system to attain equilibrium. They'd just keep on jumping from one capacitor to the other. Further, when you're moving electrons slowly one by one, an external force does some work which is nothing but absorbing the kinetic energy of the electrons. The situation described in the question is possible only if we consider some non-zero resistance. $\endgroup$
    – Vishnu
    Mar 11, 2020 at 12:38
  • $\begingroup$ @GuruVishnu I'm not concerned about how to actually make the system. That's irrelevant to the question. One capacitor with charge $Q$ will have twice the potential energy as two capacitors with charge $Q/2$ regardless of how you actually make those systems. The OP isn't asking about how one would move from one system to the other. Your answer is correct in explaining one possible way to do this, but it's not really what is being asked about. $\endgroup$ Mar 11, 2020 at 12:45
  • $\begingroup$ @GuruVishnu But if you do look at how we move from one system to the other, then the dissipation doesn't matter in terms of potential energy. Just like how when you drop a ball and it hits the ground. The ground did do work on the ball, but the ground didn't affect the potential energy of the system. The potential energy changed as the ball fell, not when it hit the ground. $\endgroup$ Mar 11, 2020 at 12:50

Yes, even assuming a perfectly conductive circuit, there would be some inductance (you may not have studied this yet) which would create an oscillating LC circuit, which would radiate. Some early spark gap radio transmitters worked like this.

  • $\begingroup$ None of that is taken into account in this problem or the supplied solution though. This can be analyzed purely from an electrostatic point of view. $\endgroup$ Mar 11, 2020 at 4:42

There is a short period of time through which the electrons move from the charged capacitor to the uncharged capacitor. Through this time, assuming the wire has zero resistance, the energy is lost through the magnetic field generated by the accelerating electrons.(We know the electrons accelerate because the potential difference changes, implying the existence of an electric field, which exerts an opposite electric force). If that electric potential is changing, then the force itself is also changing.


You can also think about this in an electrostatic context. Consider two positive charges, when the charges repel and move away from each they lose the potential energy they had. The same is happening to the electrons that were stored in the initially charged capacitor. It is that energy.


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