# Does charge distribute itself uniformly on a conductor?

An excerpt from a beginning E&M book

[...] In other words, the surface of a conductor is an equipotential surface under static conditions. [...] Summarizing the boundary conditions at the conductor surface, we have $$\Tiny \color{maroon}{\boxed{\begin{array}{c} \color{black}{\text{Boundary Conditions}} \\ \color{black}{\text{at a Conductor/Free Space Interface}} \\[5pt] \hline \color{black}{E_t=0} \\ \displaystyle \color{black}{E_n = \frac{\rho_s}{\epsilon_0}} \\ \end{array}}}$$

This challenges my intuition because this would mean a charged gourd at equilibrium would be in stress. (Presumably the skinny side would be pushed away from the base.) Is this another approximation for the sake of computational simplicity? The charge is distributed uniformly on a spherical surface, but that is a function of the high degree of symmetry on the situation.

In general the charge tends to accumulate most strongly near pointy bits and most weakly in depressions in the surface.

There are several way to understand this. My favorite is not necessarily the most helpful for a beginner, but I think it is illuminating.

We begin by rigging a small Gaussian surface that encloses a bit of the surface and lies parallel to it a small ways outside, perpendicular to the surface to penetrate and closes inside. The flux is separately proportional to both the enclosed charge and to the area of the parallel surface outside the conductor.

So $$A_c \rho_s \propto \Phi \propto A_G \,,$$ where $A_c$ is the area of conductor surface enclosed and $A_G$ is the area of the parallel part of the Gaussian surface.

From that it is clear that $$\rho_s \propto \frac{A_G}{A_s} \,.$$

In regions where the shape is convex this ratio is greater than 1, where concave it is less than 1. The tighter the radius of curvature the more pronounced the effect.

• That's not what the book says. – Christian Chapman Jul 19 '19 at 0:06
• The portion of the book you quoted talks about potential and about electric field not about charge density. The bulk of my post is about how to use the electric field tile given in the text to deduce the surface charge density. – dmckee --- ex-moderator kitten Jul 19 '19 at 3:39