# Electric field at edge of circular plate capacitor

I know that the Electric field between 2 parallel oppositely charged plates is $E = \frac{\sigma}{\epsilon}$ and that this can be calculated using Gauss’ law, but this only applies under the assumption you’re not near the edges of the plates. So I was wondering how one would go about calculating an expression for the electric field near the edges of the plates. For simplicity I will assume the plates are circular, but even then I’m not sure what shape of Gaussian surface I should choose to make this problem simpler. Any help would be much appreciated.

• Due to a lack of necessary symmetry, the use of Gauss's law will not be helpful to find the electric field in this problem. – freecharly Feb 23 '18 at 22:45
• There are a few references that may help you, for example G. T. Carlson and B. L. Illman, “The circular disk parallel plate capacitor,” Am. J. Phys. 62 (12), 1099–1105 (1994). For others, check the link aapt.scitation.org/doi/abs/10.1119/1.1463738 – jim Feb 23 '18 at 22:45

What is usually easier is to calculate the potential $\phi$, whish is a scalar, and from it find the electric field $\mathbf{E} = -\nabla \phi$.

This is for a time-independet situation, for a charging capacitor then you'd have to also find the vector potential $\mathbf{A}$ which would give yuou a magnetic field $\mathbf{B} = \nabla \times \mathbf{A}$ and an electric field $\mathbf{E} = -\nabla \phi - \frac{\partial \mathbf{A}}{\partial t}$.

Formulae for $\mathbf{A}$ and $\phi$ can be found here, where the integration region $\Omega$ would be the area of the plate of the capacitor in your case.

Another way is to solve Laplace's equation $\nabla^2 \phi = 0$ in vacuo and apply the boundary condition that the potential needs to be constant across the plate, like it's done here.

• This isn't quite there - the formulae you've linked to are for known charges and currents, whereas OP deals with a known constant potential at the boundary instead. However, the core message (the hard truth that there is no gaussian-surface analysis that will work in this situation) is on target. – Emilio Pisanty Feb 23 '18 at 20:04
• You can still use the $\phi$ formula for a general charge density $\rho$ no? Just need to find a suitable area element $\mathrm{d}^2\mathbf{r}$ and integrate on the plate. – SuperCiocia Feb 23 '18 at 20:07
• You can, but you don't know $\rho$ - you just know that the potential needs to be constant. It is a substantial amount of work to find out $\rho$ from the potentials, and by the time you're in a position to calculate it you already know the fields you wanted in the first place. – Emilio Pisanty Feb 23 '18 at 20:09
• Oh yeah I guess the density at the edge won't be constant anymore... – SuperCiocia Feb 23 '18 at 20:11
• Well, it depends on what you're doing. OP lucked out by specifying that the plates are charged, presumably with constant density - but having a constant charge density at the edge is incompatible with having conducting plates. – Emilio Pisanty Feb 23 '18 at 20:19

When you have circular plates you can reduce the problem to a two-dimensional one at the edge of the capacitor, by using cylindrical coordinates r,z and then consider a large r so that the problem at the edges is approximately two-dimensional. Then you can use conformal mapping to obtain the general potential solution in this fringe region of the capacitor. You have to find the complex function which describes the boundary conditions of the charged plates. The real and imaginary part of all complex functions, in the domain where they are analytic, are solutions of the Laplace equation for the electrostatic potential. Thus if you find an analytic function describing the boundary conditions of the plates, you have the potential solutions and from this by differentiation you get the correct electric field solutions between and around the plates at the fringes.

• This is inaccurate as stated. Conformal mapping works as a technique for electrostatic analysis when there is translational (instead of rotational) symmetry, i.e. when the electrostatic potential obeys the Laplace equation in cartesian (not polar) two-dimensional coordinates. The result thus obtained can only be made to work for a circular-plate capacitor in an approximate way when one 'zooms in' to the edges so much that the curvature of the edge is negligible. (Fix it or don't, up to you. This is my single comment here.) – Emilio Pisanty Feb 23 '18 at 20:00
• @EmilioPisanty - Typical nitpicking! It is obvious that the edges are considered. But I will improve the wording to avoid ambiguities. – freecharly Feb 23 '18 at 22:10

Given two infinite planes, under your symbols for field inside planes $E=\frac{\sigma}{\epsilon}$. At the edge of planes, in the middle of edge-connecting line $E=\frac{\sigma}{2 \epsilon}$. To obtain that result, you consider the mirroring of the planes along that line.

If you move away from middle-point, you will get less $E$, but your result will have the member depending from dimensions. If your dimensions are ideal, you will obtain same result on all dots of line connecting the edges.