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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., $\textbf{E}=4\pi \sigma \,\hat{\textbf{n}}$?

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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.

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  • $\begingroup$ 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. $\endgroup$ – Hilbert Aug 28 '19 at 20:01
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    $\begingroup$ 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! $\endgroup$ – mike stone Aug 28 '19 at 20:18
  • $\begingroup$ Gauss' law tells me that there is an error in your argument. $\endgroup$ – my2cts Aug 28 '19 at 21:13

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