The loss of energy of a capacitor is independent of length of wire connecting it to voltage source $$(\frac{1}{2}×CV^2)$$So where and when is the coverted into heat?
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$\begingroup$ See questions in Related column. $\endgroup$– sammy gerbilCommented Jun 18, 2018 at 17:06
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$\begingroup$ As long as the charge stays on the capacitor no energy is lost. $\endgroup$– my2ctsCommented Jun 18, 2018 at 18:07
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2$\begingroup$ I find this an odd question because, on one hand, in the first sentence you seem concerned about the length of the wire(s) which, to me, implies that you're assuming the wire(s) have non-zero resistance per unit length. On the other hand, you ask where and when it is converted into heat which, to me, implies that you're assuming the wire(s) are ideal, i.e., have zero resistance per unit length. Would you mind editing your question to clarify if you're considering the wires ideal or not (or something else entirely)? $\endgroup$– Alfred CentauriCommented Jun 19, 2018 at 1:02
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I'm not sure if this addresses your question or not but, it's easy to show that, for any non-zero value of resistance (including, e.g., radiation resistance), the energy supplied by a voltage source charging a capacitor is twice the energy stored in the capacitor.
For a series RC circuit with a constant voltage source of voltage $V_{DC}$ connected at $t = 0$, the charging current is straightforwardly found to be
$$i(t) = \frac{V_{DC}}{R}e^{-t/RC},\quad t \ge 0$$
and so it must be that the voltage across the resistor is
$$v_R(t) = V_{DC}e^{-t/RC}$$
Then, the work done by the source is
$$W_S = V_{DC}\int_0^\infty\mathrm{d}t\, i(t) = CV^2_{DC}$$
while the work done on (heating) the resistor is
$$W_R = \int_0^\infty\mathrm{d}t\, v_R(t)i(t) = \frac{1}{2}CV^2_{DC}$$
Remarkably, the work done on the resistor, for non-zero $R$, is independent of the value of R!
So where and when is the coverted into heat?
Assuming the wires have non-zero resistance, I've answer that question above. If you're assuming ideal wire (zero resistance), you must consider other sources for loss such as radiation resistance which accounts for the fact that the energy, rather than being converted to heat, can be lost to electromagnetic radiation.
answered Jun 19, 2018 at 1:23