I have two very related questions:
In the derivation of power in a circuit we can do the following:
\$P=\frac{dw}{dt}=\frac{d(QV)}{dt}=Q\frac{dV}{dt}+V\frac{dQ}{dt}\$
and then assuming that \$V\$ is constant:
\$P=V\frac{dQ}{dt}=VI\$
This formula holds in every case (that I can think of) in electronics, but why can we assume V is constant, even if we have an ac signal? Furthermore if we are assuming that V is constant why do we not assume that Q is constant also?
Secondly:
When deriving the equation for energy stored in a capacitor you can work out the work done to move charge from one side plate to the other. But in the act of removing charge from one plate, you will change the potential between the plates, so why can we assume that the potential is constant when moving this charge from one plate to another. (the charge is usually infinitesimal (\$dQ\$) and the energy \$dU=VdQ\$ is then integrated over)
Best Answer
One of your equations is incorrect: the energy in a capacitor is \$ \frac{1}{2}QV\$. Then, the power is
\$P = \frac{dw}{dt} = \frac{d}{dt}(\frac{1}{2}QV) = \frac{1}{2}[Q\frac{dV}{dt} + V\frac{dQ}{dt}]\$
Then, since V = \$\frac{Q}{C}\$, \$\frac{dV}{dt} = \frac{1}{C}\frac{dQ}{dt}\$, so
\$ P = \frac{1}{2}[\frac{Q}{C}\frac{dQ}{dt} + V\frac{dQ}{dt}] = \frac{1}{2}[V\frac{dQ}{dt} + V\frac{dQ}{dt}] = IV \$
with no assumptions about constant voltage.
I haven't thought about your second question, but I think the same idea will help out there too.