Question on the work done while charging up a capacitor.

[bubblewrap]

So the work done when charging up a capacitor is $dW=VdQ$
However, when we add a charge $dQ$ to the capacitor, $V$ also changes accordingly, so I was wondering why the work done wasn't written as $dW=VdQ+QdV$ (one that also takes into account t he change in $V$).

Thanks in advance.

 

[Delta2]

I think the correct equation for dW is $dW=(V+dV)dQ=VdQ+dVdQ$. This is because the voltage changes to V+dV but remains constant and equal to V+dV as the charge dQ moves from one plate of the capacitor to the other.

But when we do a single (not double) integration on dVdQ the result is 0. That is $\int dVdQ=0$ I can expand for this if you want.

EDIT: Why the equation you wrote is not correct for the case of a charging capacitor:

It all starts from the equation $V=W/Q$. If this equation was correct in all cases then we could simply handle it as follows $W=VQ$ and differentiating using the product rule of differentiation (and omitting the denominators $dt$) we ll end up with $dW=VdQ+QdV$. So at first place it seems that the equation you wrote is correct.

However the mistake lies at our starting point: The equation $V=W/Q$ does not hold for the case of a charging capacitor. This equation is true only if the voltage V is constant and independent of time t or the charge Q (as to why this is true we ll have to go back to electrostatics and studying the motion of a charged point particle which moves between two points of an electrostatic field that have potential difference V). In the case of a charging capacitor V is not constant in time and it depends on the charge Q in the capacitor plates, so this equation does not hold.

It holds though if we apply it for an infinitesimal time interval $(t,t+dt)$ for which we consider the voltage constant and equal to $V$ (or $V+dV$ as I wrote above). Then the infinitesimal work $dW$ for this infinitesimal time interval $dt$ during which the charge dQ moves from one plate of the capacitor to the other, is $dW=(V+dV)dQ$. And from here we go back at the start of my message...