# Is the $E$-field near and on the outer surface of a conductor null?

Let us say we have a conductor the outermost surface $$S$$ of which is given by $$S: \big(\,x(u,v),\, y(u,v),\, z(u,v)\,\big )$$

where $$u$$ and $$v$$ are parameters.

Since it is a conductor, the potential $$\phi (x,y,z)$$ is constant over the surface $$S$$: $$\forall {P(x,y,z)} \in S,~~~~ \phi( P) =cst$$

We also have:$$\textbf{E}\,(x,y,z)=-\nabla\phi \,(x,y,z)$$

applying this equality to the points belonging to the surface on which the potential is constant, we get:

$$\textbf{E}\,(x,y,z)=-\nabla\phi \,(x,y,z)=\textbf{0}$$

Also, by taking into account the fact that the potential is continuous, we find the $$E$$-field must also be very small outside, in the vicinity of the surface. Isn't this in contradiction with the formula for the $$E$$-field near the surface of a conductor, i.e., $$\textbf{E}=4\pi \sigma \,\hat{\textbf{n}}$$?

No. The $${\bf E}$$ field is not zero immediately outside the surface of a charged conductor. All your argument has shown is that $${\bf E}\cdot {\bf t}=0$$ for any tangent $${\bf t}$$ to the surface. In other words $${\bf E}$$ is perpendicular to the surface of the conductor.
• I didn't say the $E$- field is zero just outside the surface, but very close to zero, which I think is in contradiction with the known formula $E=4\pi \sigma$ on the surface of a conductor. – Hilbert Aug 28 '19 at 20:01
• It is not close to zero either. The surface being an equipotential does not prevent there being a (possibly large) potential gradient imediately adjacent to a surface. It just has to be perpendicular to the surface. Such a gradient is indeed what $|E|=\sigma/\epsilon_0$ necessitates. A slope on a topographical map is not inconsistent with there being contour lines! – mike stone Aug 28 '19 at 20:18