A capacitor, once charged up with charge $Q$ will have a given voltage $V$. I tried to work out this voltage by simply plugging into coulomb's equation. The potential difference between two static point charges is


where $q$ is a unit of charge located on the capacitor, $d_1$ and $d_2$ are the distances from the center of the other electron, and $d=d_2-d_1$; the distance through which pd is measured. If we assume that $q$ are distributed smoothly across the surface, then the size of the unit $\displaystyle{q = \frac{Q}{A}}$, if the area is a unit defined by the area one coulomb takes up on a capacitor. Therefore, the voltage should be

$$V = \frac{Qdk}{4\pi\epsilon_0d_1d_2A}\qquad(1)$$

and $k$ is some constant.

I found a much simpler derivation (in a textbook) that assumes Gauss's law. It says that the electric field $\displaystyle{E=\frac{cQ}{A}}$ by Gauss's law, and that it is also $\displaystyle{E=\frac{V}{d}}$, therefore


and $c$ is $\displaystyle{\frac{1}{\epsilon_0}}$.

I don't understand this. According to the first equation, the electric field does not depend on the distance. Since area and charge don't change with distance, how can that be? Could someone explain how this works, and why my equation is incorrect?


The derivation that you found is for a parallel-plate capacitor (in which the electric field is indeed constant, assuming that the plates are large relative to the separation between them). It won't apply to a spherical capacitor, though Gauss's law would.

The real issue here, though, isn't Gauss's law, it's the definition of potential: $$\Delta V = -\int_C \vec{E}\cdot d\vec{s}$$

If the field is constant along the path $C$, then this reduces to: $$\Delta V = -\vec{E}\cdot\vec{d}$$ That's fine for the field of a parallel-plate capacitor ($E = \dfrac{\sigma}{\epsilon_0}=\dfrac{Q}{\epsilon_0A}$), but the electric field of a point charge or a spherically-symmetric distribution varies as the distance from the center $r$ changes ($E=\dfrac{Q}{4\pi\epsilon_0r^2}$), so (unless the path stays at the same $r$, which is the trivial case), you'd have to actually perform the integral above to determine the potential change.

The potential change is vital to find, since the definition of capacitance relies on it: $$\dfrac{1}{C}=\dfrac{dV}{dQ}$$

If you want to see how Gauss's law is used to determine spherical (http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elesph.html) and parallel-plate (http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elesht.html) fields, there are plenty of good resources out there.

  • $\begingroup$ apparently not, the equation in the textbook was for parallel plate capacitors. fairly sure... $\endgroup$ – lucky-guess Oct 6 '16 at 13:45
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    $\begingroup$ The first? You'd need to integrate in order to determine the capacitance point-by-point, since the distances between charges are all different. $\endgroup$ – DeltaG Oct 6 '16 at 14:29
  • $\begingroup$ I suppose my main issue in this was that i didn't stop to think about the sum of the effects of all the point charges. I now understand why V/d is constant, $\endgroup$ – lucky-guess Oct 10 '16 at 14:57

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