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Suppose we have a plate capacitor, placed in a uniform background electric field (in a way that the electric field is perpendicular to the capacitors plates.

Without the electric field, the relationship of "voltage" and charge in the electric field would be \begin{align} U = \frac{Q}{C} \end{align} Here, U denotes the line-integral of the electric field. There is no rotational electric field, the situation is completetely static.

If I place the capacitor in the electric field, at first the voltage will be bigger (because there is an additional electric field present). But will charges subsequently move in a way that restores above mentioned equation? What will happen? More importantly: What is the steady state?

Or is it, in this situation, "wrong" to use the definition of voltage being a line integral, and one has to use the definition of voltage being the difference in coulomb-potential instead, to still be able to use the formular mentioned?

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  • $\begingroup$ Just to be very clear - you charge a capacitor C with a voltage source U which yields charge Q. Now you place this setup (assuming stability has been reached) in an electric field. You want to know if the expression still remains U=Q/C for the same values of U, Q and C right? $\endgroup$ – Paddy Jun 18 at 3:11
  • $\begingroup$ If that is the case, I think you can consider your parallel plate capacitor setup to be placed between another parallel plate capacitor (with the outer capacitor as your source of external field). If this is sounds right to you, take a look at arxiv.org/pdf/1712.04442.pdf $\endgroup$ – Paddy Jun 18 at 3:16
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    $\begingroup$ @Paddy yes, that's what I want to know about. $\endgroup$ – Quantumwhisp Jun 18 at 6:08
  • $\begingroup$ Did my second comment help? $\endgroup$ – Paddy Jun 18 at 8:14
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    $\begingroup$ @Paddy yes, the answer is somewhere in there. I will write an answer myself with the solution the paper didn't explicitly state, but implicitly show. $\endgroup$ – Quantumwhisp Jun 19 at 18:11
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In my answer, "voltage" means difference in the scalar potential (in either coulomb or lorentz-gauge, which is the same, because there are no fields with time dependence present). This definition of voltage is also equivalent to the negative line integral of the electric field, because the field is conservative. However, applying this definition of "voltage" means that the electric field in the capacitor (and the voltage across the capacitor plates) isn't produced solely by charges on the capacitor plates.

For the answer of the question, it doesn't matter what the steady states are. The law in question, \begin{align} U = \frac{Q}{C} \end{align} should hold for every possible $Q$ in the capacitor, because in every imaginable situation there is an amount $Q$ on the capacitor, and voltage $U$ measurable across the capacitor plates.

We will however look at the steady state of a simple circuit, and show, that even in that state the law in question does not hold anymore: Let's look at the most simple situation: The two plates of the capacitor are connected to an ideal voltage source that provides $U_0$. In steady state, the electric field intensity is zero in the conductors, and it follows that the two plates of the capacitor have the same voltage between them. However, this potential is generated by the charge on the plates AND the background field. While the first contribution is proportional to $Q$, the 2nd is independent of it.

You can see this when you set $U_0$ to $0$ (effectively short-circuiting the capacitor). The plates are at the same potential, which means that charges, following the background electric field, have travelled from one plate to the other, until the field strength between the plates (and thus the voltage) was zero. However, in this situation, the charge stored in the plates is not zero.

In general, for a plate-capacitor in a uniform electric field perpendicular to the plates of field strength $E$, the relationship between charge and voltage will be:

\begin{align} U = \frac{Q}{C} - E*d \end{align} Where $d$ is the distance between the plates.

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