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I’ve been told that coulomb repulsion pushes excess electrons to the surface of a conductor (i.e. sphere) electrostatic equilibrium, and this symmetry causes the net electric field inside to be zero. However, why can’t excess charges just be evenly distributed throughout the sphere? Would that symmetry not cancel the fields from each charge?
It would be helpful if anyone can provide a diagrammatic answer.
If the "excess charges" are in a conductor then they are, by definition, free to move. If there are excess charges distributed throughout the conductor then they will be compelled to move by any electric field within the conductor.
Gauss's law tells us that if there is any net free charge within the conductor then this also produces an electric field within the conductor. The electric field will cause the charges to move, and they will do so until this electric field is zero. Gauss's law then also tells us that this will be accomplished once there is no net free charge within the conductor, so the free charge must accumulate on the surface.
Suppose in an arbitrary shaped conductor, there are some unbalanced positive charges. Draw a Gaussian surface within the conductor surrounding the charge. From Gauss Law, we would find there is a net flux which is not possible inside a conductor.
It wouldn't.
It is a well-known fact from gravity, which also has a law $\nabla\cdot\vec F_g \propto \rho$ expressing a $1/r^2$ force, that the force of a spherical shell of mass is zero inside that shell, but outside it is $-\hat r G M m / r^2$ as if all of that mass were at the shell's center.
The same mathematics will mean that the $\vec E$ field within a spherical insulator with uniform charge density $\rho$ will be $$\vec E = \hat r \frac{Q_{\text{encl.}}}{4\pi\epsilon_0 r^2} = \hat r \frac{1}{4\pi\epsilon_0 r^2} \rho \cdot \frac{4\pi}{3}r^3 = \hat r \frac{\rho r}{3\epsilon_0}$$which is plainly not zero unless $\rho$ is.
Interestingly, you can kill these sorts of arguments by extending the radius of the sphere out to all space, at which point Gaussian arguments get a little hazier and there are symmetry reasons to think that there is no net force on you. However, Gauss still wins in this way: any little noise in that background distribution of charge gets amplified by the laws as time goes on, eventually leading to a very uneven distribution of charge, so it's like balancing a pen on its tip: the physics says it's possible but unstable.
This is only in case of a charged conductor in which no currents are flowing through it. Now, consider a gaussian surface inside the conductor everywhere on which electric field is zero (since there is no current). From Gauss' s law, we get q(inside) = 0 since E=0. Thus, the sum of all the charges inside the Gaussian surface is zero. This surface can be taken just inside the surface of the conductor, hence, any charge on the conductor must be on the surface of the conductor.
