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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1 Answers
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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