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This is a statement from Young and Freedman's University Physics Section 24.1 on Capacitors. Suppose we have any two conductors with charges $-Q$ and $+Q$ on each charged with a battery. Then this gives a fixed potential difference $V_{ab}$ between the conductors (that is, the potential of the positively charged conductor $a$ with respect to the negatively charged conductor $b$) that is just equal to the voltage of the battery.

Now the book says "The electric field at any point in the region between the conductors is proportional to the magnitude $Q$ of charge on each conductor. It follows that the potential difference $V_{ab}$ between the conductors is also proportional to $Q$."

I can't figure out why this statement must hold. In the case of say parallel plates, this is clear, but how do we know in general for any two conductors (which could be of different shapes), with equal in magnitude but opposite charges, the electric field $\vec E$ anywhere between the conductors must be proportional to the magnitude of the charge?

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Imagine that you have two conductors with charges $\pm Q$.
There will be a series of equipotential surfaces around the two conductors each with a value, $V$, as to the potential relative to one of the conductors.

What will increasing the charge stored to $\pm 2Q$ do to the equipotential surfaces?
A surface which had a potential $V$ would now have a potential $2V$.
Given that the electric field at a point is minus the potential gradient at that point this means that the electric at a point will double.

I assume that this is all leading up to the definition of capacitance.
As $Q\propto V \Rightarrow Q=CV$ where $C$, the capacitance, is the constant of proportionality?.

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  • $\begingroup$ Yes this is all leading up to $Q \propto V$. But why does increasing the charge stored to $\pm 2Q$ double the equipotential surfaces? Is this justified from Gauss' Theorem or any properties following from the definition of electric potential? I can't find out why this holds other than specific cases e.g. point charge distributions, parallel plates or conducting spheres where we know the relationship between the source charge and potential explicitly. $\endgroup$
    – nomadicmathematician
    Commented Jun 15 at 9:58
  • $\begingroup$ I am familiar with Poisson's Law for the total potential $V(\vec r)=\frac{1}{4\pi \epsilon_0}\iiint \frac{\rho(\vec r'}{|\vec r-\vec r'|}d\tau$, but unless the surface charge density $\rho$ is uniform, I can't represent this as a function of charge $Q$. $\endgroup$
    – nomadicmathematician
    Commented Jun 15 at 10:05
  • $\begingroup$ If you double the charge then won't the surface charge density follow suit? $\endgroup$
    – Farcher
    Commented Jun 15 at 10:18
  • $\begingroup$ I see so it is the case that the relative charge distribution, i.e. the relative configuration of charges is independent of the total; charge so if we have $\rho \propto Q$ then $k \rho \propto kQ$? $\endgroup$
    – nomadicmathematician
    Commented Jun 15 at 10:39

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