When we find the electric field between the plates of a parallel plate capacitor we assume that the electric field from both plates is $${\bf E}=\frac{\sigma}{2\epsilon_0}\hat{n.}$$ The factor of two in the denominator comes from the fact that there is a surface charge density on both sides of the (very thin) plates. This result can be obtained easily for each plate. Therefore when we put them together the net field between the plates is $${\bf E}=\frac{\sigma}{\epsilon_0}\hat{n}$$ and zero everywhere else. Here, $\sigma$ is the surface charge density on a single side of the plate, or $Q/2A$, since half the charge will be on each side.

But in a real capacitor the plates are conducting, and the surface charge density will change on each plate when the other plate is brought closer to it. That is, in the limit that the two plates get brought closer together, all of the charge of each plate must be on a single side. If we let $d$ denote the distance between the plates, then we must have $$\lim_{d \rightarrow 0}{\bf E}=\frac{2\sigma}{\epsilon_0}\hat{n}$$ which disagrees with the above equation. Where is the mistake in this reasoning?

Or more likely, do our textbook authors commonly assume that we are in this limit, and that this is why the conductor behaves like a perfectly thin charged sheet?


3 Answers 3


When discussing an ideal parallel-plate capacitor, $\sigma$ usually denotes the area charge density of the plate as a whole - that is, the total charge on the plate divided by the area of the plate. There is not one $\sigma$ for the inside surface and a separate $\sigma$ for the outside surface. Or rather, there is, but the $\sigma$ used in textbooks takes into account all the charge on both these surfaces, so it is the sum of the two charge densities.

$$\sigma = \frac{Q}{A} = \sigma_\text{inside} + \sigma_\text{outside}$$

With this definition, the equation we get from Gauss's law is

$$E_\text{inside} + E_\text{outside} = \frac{\sigma}{\epsilon_0}$$

where "inside" and "outside" designate the regions on opposite sides of the plate. For an isolated plate, $E_\text{inside} = E_\text{outside}$ and thus the electric field is everywhere $\frac{\sigma}{2\epsilon_0}$.

Now, if another, oppositely charge plate is brought nearby to form a parallel plate capacitor, the electric field in the outside region (A in the images below) will fall to essentially zero, and that means

$$E_\text{inside} = \frac{\sigma}{\epsilon_0}$$

There are two ways to explain this:

  • The simple explanation is that in the outside region, the electric fields from the two plates cancel out. This explanation, which is often presented in introductory textbooks, assumes that the internal structure of the plates can be ignored (i.e. infinitely thin plates) and exploits the principle of superposition.

    electric fields in superposition

  • The more realistic explanation is that essentially all of the charge on each plate migrates to the inside surface. This charge, of area density $\sigma$, is producing an electric field in only one direction, which will accordingly have strength $\frac{\sigma}{\epsilon_0}$. But when using this explanation, you do not also superpose the electric field produced by charge on the inside surface of the other plate. Those other charges are the terminators for the same electric field lines produced by the charges on this plate; they're not producing a separate contribution to the electric field of their own.

    electric field from one plate to the other

Either way, it's not true that $\lim_{d\to 0} E = \frac{2\sigma}{\epsilon_0}$.

  • 2
    $\begingroup$ Why does all the charge migrate inside? $\endgroup$
    – drake01
    Commented Dec 24, 2018 at 5:22
  • 2
    $\begingroup$ @drake01 they are attracted to the opposite charge on the other side. $\endgroup$
    – KingLogic
    Commented Jan 19, 2021 at 3:33
  • $\begingroup$ I don't quite understand why we can't use superposition in the second case. If I have a dipole and I want to find the electric field in between, don't I have to use superposition even though it is true that the negative charge is the terminator for the same electric field line of the positive charge? $\endgroup$ Commented Mar 18, 2021 at 20:15
  • $\begingroup$ "The more realistic explanation is that essentially all of the charge on each plate migrates to the inside surface. " While it doesn't affect the conclusion, technically don't the mobile charges (electrons) on the left plate migrate to the outside of the plate leaving the immobile (protons) on the inside? $\endgroup$
    – Bob D
    Commented Apr 8 at 21:10
  • $\begingroup$ @BobD Yes, that's exactly how migration of positive charge is "implemented" in a real conductor: by mobile electrons moving away from the part of the conductor that the positive charge is migrating to. (Don't forget the distinction between movement of charge and movement of the particles that carry that charge.) $\endgroup$
    – David Z
    Commented Apr 10 at 3:21

The very short, but perhaps terse answer is that it does not matter on which side of the plate the charge resides. The field outside a charged plate, conducting or not, is $E = \sigma/2\epsilon_0$ if the surface density of both sides combined is $\sigma$. The plate does not even have to be thin.


Let us assume a uniform field $E$ . Applying $\nabla \cdot D = Q$ , and noting that all components of $E$ vanish inside a perfect conductor, gives $\sigma = \epsilon_0 E$ at one surface and $E\sigma = -\epsilon_0 E$ at the other. The conductors are not infinitesimal sheets.


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