How Does Gauss' Law Ensure E-field Perpendicularity on Irregular Conductors?

[iScience]

For an infinite plane sheet of charge it is obvious that the E-field points directly perpendicular to the sheet. but for conductors of irregular shape. say, a wire, or even a sheet with imperfections in it, what guarantees that the E-field will point directly perpendicular from the emanating surfaces?

 

[Nugatory]

For an infinite plane sheet of charge it is obvious that the E-field points directly perpendicular to the sheet. but for conductors of irregular shape. say, a wire, or even a sheet with imperfections in it, what guarantees that the E-field will point directly perpendicular from the emanating surfaces?

Focus down on a small enough region of the surface and it will be flat (if it's not, just go for an even smaller region). Then evaluate the direction of the force at the center of that region, at a distance that is small compared with the size of that region... and you're right back to something that looks like the infinite plane case.

If the object is irregularly shaped, the field may change direction not far from the surface, but as long as the distance from where you're calculating the force to the surface is very small compared with the size of the irregularities, the field will be perpendicular to the surface.

 

[stevendaryl]

Suppose at some point on the surface, the electric field is not perpendicular to the surface. That means that there is a component of the electric field that is parallel to the surface. That means that a freely moving charge at that point would have a force on it causing it to move in a direction that partially cancels the electric field. So if there are plenty of freely moving charges, they would tend to move to cancel the electric field in the direction perpendicular to the surface.

 

[vanhees71]

First of all, this is about electrostatics. Then the electric field is a potential field, i.e., it exists a scalar field such that
$$\vec{E}=-\vec{\nabla} \Phi.$$
Now, inside a conductor the electric field must vanish, because otherwise one had a current due to Ohm's Law, and electrostatics is about the electric field for charges at rest and no currents.

This implies that for electrostatics the $\Phi$ is constant inside a conductor and particularly along its surface. Thus the surface of a conductor is a surface of constant potential. Any curve $\vec{x}(\lambda)$ (where $\lambda$ is an arbitrary parameter for the curve) within the surface is thus an equipotential line, i.e., we have
$$\Phi[\vec{x}(\lambda)]=\text{const}.$$
Taking the derivative with respect to $\lambda$ implies
$$\frac{\mathrm{d} \vec{x}}{\mathrm{d} \lambda} \cdot \vec{\nabla} \Phi=-\frac{\mathrm{d} \vec{x}}{\mathrm{d} \lambda} \cdot \vec{E}=0.$$
This means the electric field at the surface of the conductor is necessarily perpendicular to any tangent vector along the surface, QED.