Electrical Power and The Product Rule

Question

As we know, the formula for Instantaneous Electrical Power is:

$$P(t) = I(t)V(t) $$

But, if work is expressed as:

$$W(t) = q(t)V(t)$$

And Power is the time derivative of Work:

$$P(t) = W'(t)$$

Then by the Product Rule, since Voltage varies with time:

$$P(t) = V(t)I(t) + q(t)V'(t)$$

But, as we know, the V'(t) term does not exist in the formula. The common explanations for this is: Voltage being the charge derivative of Work. This of course would result in:

$$P(t) = (dW/dq)(dq/dt) = (dW/dt)$$

But this assumes Voltage is independent of time. Fine for a constant voltage, but this proves problematic otherwise.

If Voltage is time variant, then some charge(s), q(t), in a time-varying voltage should provide power generation. We can imagine some charge distribution, whose charges are pushed closer together, without adding or subtracting any charges (let's imagine something like Maxwell's Demon pushing each charge simultaneously). These charges would then each be at a higher potential. Thus, work has been done on the system, and so Power has been supplied to it.

Another explanation stems from the formula: $$P = F \cdot v$$ Where a variable force over some curved integral can, using the Fundamental Theorem of Calculus, be expressed as the product of force and velocity. We can then sub $qE$ for $F$ and $d/(q/I)$ (where d is distance traversed in time t) for $v$ (velocity). But this explanation feels off to me, as it doesn't conclusively disprove that second, $dV/dt$ term.

Ultimately, my question is: Why do we neglect this second term when taking the derivative of Work, if we acknowledge that voltage varies with time? There should be a physical explanation for this, but right now I can't think of one rigorous enough to put my mind at ease.

Answer 1

Work is expressed as $W=qV$ only in the case of electrostatics, not when $V$ is time dependent. When the EM field is time dependent you have both electric AND magnetic fields and Poynting's theorem, see https://en.wikipedia.org/wiki/Poynting%27s_theorem.

Answer 2

work is expressed as: $$W(t) = q(t)V(t)$$

The expression W=qV applies when a charge is moved from one place (with a starting potential) to another place (with a different potential). But particles can't be moved instantaneously and work can't be done instantaneously so we can't extend this equation to your version where all three variables are taken as functions of time.

We can make a differential form,

$$dW = dq\ V$$

meaning that the increment of work done is the product of the increment of charge moved through a potential difference V. And if these increments are done over a time increment $dt$ we can recover the usual expression $P = IV$.