Why do we have energy loss same for every length of wire in capacitor charging

Question

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?

Answer 1

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.