From these questions:
Energy loss in Capacitors
What happens to half of the energy in a circuit with a capacitor?
I come to the conclusion that energy loss in a capacitor is explained by the fact that the the potential applied accelerates charges and since the end configuration is stationary, there needs to be some damping to get a steady state.
Now assuming a hypothetical situation where we somehow bring $dq$ charge infinitely slowly. We repeat this a number of times and thus charge the capacitor to charge $Q$. Thus we lose zero energy as heat as we have avoided any acceleration of charges and the work done and the energy stored would both be $\frac{Q^2}{2C}$. Ignoring the practical limitations of this experiment, is this theoretically correct?
The charges are brought from infinity are used to charge a single spherical capacitor of capacitance C.
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$\begingroup$ What mechanism do you actually use to transfer the charge? If you bring your microscopic tweezers in and start picking up electrons from one capacitor and moving them, then the electrons will do work on the tweezers to account for the change in energy between the initial and final states. $\endgroup$– The PhotonCommented Nov 19, 2018 at 18:02
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$\begingroup$ Yes, the charge we pick up by our "microscopic tweezers" would be repelled by the charge already present on capacitor. That is why work needs to be done on the system to change the configuration. Though, i cannot see the relevance of an actual mechanism here. $\endgroup$– newuserCommented Nov 19, 2018 at 18:12
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$\begingroup$ The energy loss doesn't really doesn't have anything to do with the speed of the charge transfer. I answered a similar question to yours awhile back. See physics.stackexchange.com/questions/209215/… and the "two water tanks" analogy in my answer. Even if you trickle the water transfer extremely slowly between the water tanks, there will still be a loss of energy. $\endgroup$– user93237Commented Nov 19, 2018 at 18:26
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$\begingroup$ @SamuelWeir Well, i read your analogy and have come up with a different conclusion. Suppose we redo that experiment (assuming 0 atmospheric pressure), then by Bernoulli theorem, we must have the same energy. Since potential energy decreases then water must acquire some velocity. Assuming elastic collision of water molecules with walls, the molecules must acquire some velocity in the equilibrium state. We cannot use this energy to regain initial configuration and entropy is increased and it is thermodynamic limitation. Bringing water molecules from infinity slowly has no such limitation. $\endgroup$– newuserCommented Nov 19, 2018 at 18:55
1 Answer
I come to the conclusion that energy loss in a capacitor is explained by the fact that the the potential applied accelerates charges and since the end configuration is stationary, there needs to be some damping to get a steady state.
First, there is no energy loss in a (ideal) capacitor$^*$.
The first question linked regards the case of 'missing' energy when two capacitors, one charged and one not, are connected together at some instant. In that case, energy is lost from the circuit to the environment in the form of heat and/or radiation. This case has been covered in several Q & A, e.g., here.
The second linked question regards the case of 'missing' energy when charging a capacitor with a constant voltage source. Again, energy is lost to the environment in the form of heat and/or radiation. And again, this case has been covered in several Q & A, e.g., here.
Regarding your idea of slow charge transfer, note that the results in the answers linked above are independent of the rate at which the capacitor charges. The essence of the energy loss in these types of problems is in the fact that voltage across the capacitor cannot change instantly.
Note that if, instead of a voltage source, a current source is employed to charge the capacitor, there is no 'missing' energy. If the current through is given by $i_S(t)$, then the voltage across the capacitor is given by (assume $i_S(t)$ is zero for $t<0$ and the capacitor is initially uncharged)
$$v_S(t) = v_C(t) = \frac{1}{C}\int_0^t\,i_S(\tau)\,\mathrm{d}\tau = \frac{Q(t)}{C}$$
and the work done by the source equals the energy stored in the capacitor for any time $t$.
$^*$ there's a subtlety here that's not worth bringing up in this answer.
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$\begingroup$ Thanks. I now realize that my earlier understanding was not complete. Can you just mention that "subtlety" very briefly, possibly just name ? $\endgroup$– newuserCommented Nov 19, 2018 at 19:18
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$\begingroup$ @newuser, it's been quite a while since I looked into this but an ideal parallel plate capacitor can radiate. See, for example, this arXiv article: Capacitors can radiate - some consequences of the two-capacitor problem with radiation $\endgroup$ Commented Nov 19, 2018 at 19:31