If we have a battery with voltage $V$ connected in series with a parallel-plate capacitor of capacitance $C$ with dielectric $\epsilon$,and then we pull the dielectric out, the infinitesimal work done on the system while removing the dielectric looks like \begin{equation} dW_{total} = F_{\mathrm{pull}}\,dx + V\,dQ \end{equation}
where $F_{\mathrm{pull}}$ is the force pulling out the dielectric and $x$ depicts the direction which the dielectric is pulled out in. In this scenario the $V$ is held constant by the battery, so the work done is $V\,dQ$ (at least this is what Griffith's says).

Question: Why, physically, is the work done by the battery $V\,dQ$?

Excerpt from Griffith's: "It is, of course, possible to maintain the capacitor at a fixed potential, by connecting up to a battery. But in that case the battery also does work as the dielectric moves, where $V\,dQ$ is the work done by the battery."

Diagram: enter image description here
(source: ju.edu.jo)

Attempt at an explanation:

So, in the diagram we have the system with two states to compare, with and without the dielectric. If we say that the positive terminal of the battery has potential V, and the negative terminal has 0 potential, then from the position of the negative terminal, $t_-$, to the position of the positive terminal, $t_+$, we have

\begin{align} V(t_-) - V(t_+) &= 0 - V = \int_{t_-}^{t_+} \vec{E}\cdot d\vec{l}\\ \therefore -V &= \int_{t_-}^{t_+} \vec{E}\cdot d\vec{l} \;\;\;\;\;(1). \end{align}

Applying the result of Eqn. (1) to the work required to move a charge from $t_-$ to $t_+$ we then obtain \begin{equation} W = \int_{t_-}^{t_+} \vec{F} \cdot d\vec{l} = Q_{test}\int_{t_-}^{t_+}\vec{E}\cdot d\vec{l}=-Q_{test}V \end{equation} for an infinitesimal charge, we then just have \begin{equation} dW = -dQ_{test}V \end{equation} In circuits, it is the electrons who do the moving, so then in order to change the free charge on the plates, we will be moving free electrons on either of the plates "across the battery" from $t_-$ to $t_+$ (or vice-versa). Thus, if the charge on the positively charged plate decreases by $dQ$, then the work done moving $dQ$ electrons across the battery and onto the positive plate to decrease it's net charge is: \begin{equation} dW = V\,dQ \end{equation}

If the charge on the positive plate increases by $dQ$, then this means that $dQ$ electrons move from the positive plate to the negative plate, and the work to traverse the battery is then \begin{equation} dW' = -V\,dQ = -dW \end{equation}

which I suppose makes sense because the work required to move the electron across the battery in the opposite direction should just be the opposite of $dW$.

However the fact that $dW'$ has a minus sign seems like there is a sign error because it is not just $V\,dQ$...

  • $\begingroup$ Does Griffith's text say that the work done by the battery is positive or do you merely assume that it is? $\endgroup$ – CuriousOne Jun 29 '16 at 19:44
  • $\begingroup$ The excerpt describing the work of the battery is: "It is, of course, possible to maintain the capacitor at a fixed potential, by connecting up to a battery. But in that case the battery also does work as the dielectric moves, where $VdQ$ is the work done by the battery. $\endgroup$ – Loonuh Jun 29 '16 at 19:47
  • $\begingroup$ Correct... but it doesn't talk about the sign of $VdQ$, does it? $\endgroup$ – CuriousOne Jun 29 '16 at 19:51
  • $\begingroup$ Not that I can see. Which is part of my confusion. $VdQ$ is given the positive sign in the direct calculation of the work it does. $\endgroup$ – Loonuh Jun 29 '16 at 19:52
  • $\begingroup$ What's ambiguous about it? It's simply a constant voltage. If you use a 1.5V battery, it's $V=1.5V$. Think about it this way: when you pull the dielectric out of the capacitor, its capacitance decreases, so it can hold less charge than before (at the same voltage). This means that the charge difference has to flow into the battery. $\endgroup$ – CuriousOne Jun 29 '16 at 19:53

Your mistake in the above is in what you call the force required to move the charge from across the potential difference $V$. You identify the force as $\vec{F}=Q_{test}\vec{E}$, when it should be $\vec{F}=-Q_{test}\vec{E}$. This makes physical sense -- the reason it takes work to move the charge is because you need to oppose the electric field present in the capacitor.

If you correctly identify the force, you obtain $W=+Q_{test} V$. Then for a constant voltage, the infinitesimal work is just $\mathrm{d}W=V \mathrm{d}Q$.

Thinking about the sign intuitively, I think this makes sense. Moving a positive test charge across a positive potential difference requires you to move against the electric field, and do positive work. As for the magnitude, I think this also makes sense, because (as CuriousOne noted in the comments above) voltage is defined as work per charge.

  • $\begingroup$ I see I see. My only remaining question is why have we chosen to the force against the electric field to be exactly equal to $-q\vec{E}$? I don't think it necessarily has to be. Actually, I think what really happens is that whatever force we apply, the path integral of it from + to - must be equal to the opposite of the work done by the field to move from - to + as such: $W_{applied} = \int_{+}^{-} \vec{F} \cdot d\vec{l} = - q\int_{-}^{+} \vec{E} \cdot d\vec{l} $ = qV $\endgroup$ – Loonuh Jun 30 '16 at 16:31
  • 1
    $\begingroup$ @Loonuh Fair point. We can apply an arbitrary force along the path, but we do need to match the appropriate kinematic boundary conditions. In this case, the kinetic energy needs to vanish at the initial and final point. I think in problems like this, one usually assumes the force applied exactly counters the "ambient" force, because this always results in no change in kinetic energy. $\endgroup$ – jjc385 Jun 30 '16 at 16:45

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