# Question on the work done while charging up a capacitor.

• I
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##).

Delta2
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Gold Member
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...

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