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- Jan 26, 2012

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*Introduction to Electrodynamics, 4th Ed.,*this is Equation 2.55 on page 107. In his derivation, he says that to move a chunk of charge $dq$ against the electric field (necessary to charge a capacitor), you would have to do a chunk of work $dW=V \, dq$. But I'm not sure I buy this. He references Eq. 2.38, which is essentially that $V=W/q$, for the potential difference between two points being equal to the work per unit charge. Rearranging yields $W=qV$. But now, mathematically, if I take the differential of the LHS, I must do so to the RHS: $dW=V\, dq+q\, dV$, via the product rule. $V=q/C$, so it's not constant with respect to $q$. Why can Griffiths (and lots of other physics textbooks doing this derivation) ignore the $dV$ term? With $dV=dq/C$, I get that

$$dW=\frac{q}{C}\, dq+ \frac{q}{C} \, dq= \frac{2q \, dq}{C},$$

in which case $W=CV^{2}$, twice what the book gets.

What's wrong with this picture?