When we charge a capacitor via a battery then why does exactly $1/2$ of the work done by the battery get converted to heat?

Question

I know the $\frac{QV}{2}$ & $QV$ stuff but I want the logic. You would say it's due to the resistance that the work gets converted to heat. But, why is it that exactly half the value of work done by the battery no matter what the $emf$ of the battery or the capacitance of the capacitor be?
Why is it so precisely half the work done by the battery?

Answer 1

If we have a capacitor $C$ with voltage $V$ applied across it; we will have to do $W$ work to get $Q$ inside the capacitor, the small amount of work done per small amount of charger is given by:

$$dW=VdQ$$

So the total work done on the capacitor is given by:

$$W=\int_0^Q dW$$ $$=\int_0^Q VdQ$$ $$=\int_0^Q \frac{Q}{C}dQ$$ $$=\frac{1}{2C}(Q^2-0^2)=\frac{Q^2}{2C}=\frac{QV}{2}=\frac{V^2C}{2}$$

It has nothing to do with a resistance or anything like that, it is the amount of energy stored in a capacitor and it is amount of energy taken from the battery. Remember that as a Capacitor saturates it becomes an insulator, namely $I \rightarrow 0$, so there is no more power consumed after a certain point, unlike a resistor that consumes power indefinitely.