why does in chargingcapacitor half of energysupplied islost.supplied by battery isQV,stored in capacitor QV/2?

Question

how does that energy get converted to mechanical energy?

Answer 1

Resistive circuit: I (current) is proportional to V (Voltage)
or I = (1/R) * V (current can not be stored, heat loss)

In capacitor Q is proportional to V
or Q = C* V (charge can be stored, loss due reactance)

Compare: resistive reactance R equivalent to capacitive reactance (1/C)

In capacitor circuit, charging will take energy + parasitic reactance will (1/C) will result in heat loss 0.5*C*V^2 or 0.5*Q*V.

Energy stored = supplied - loss = QV - 0.5 QV = 0.5 QV

Answer 2

Capacitor is one that stores the excess energy. energy supplied is QV but capacitor stores exess half of ithence QV/2

Answer 3

Actually in charging of capacitor the energy supplied is not lost.Battery supplies energy equal to QV/2 only.You have done mistake in calculating the work done by the battery.It is right that for constant p.d.,we have the equationW=VQ.BUT,in the work done by battery in charging of capacitor,the potential difference(p.d.)is not constant.So for evaluating the work done by battery ,you will have to take integration simply or take avarage potential difference.p.d. is not constant because p.d across capacitor gradually builds up and in the opposite direction.
Hope you will make sense of it.

Answer 4

It's not "lost". It's expended and converted to heat/mechanical energy.