derivative of the electric field along the normal to the surface near the conductor how to derive the formula approves the derivative of the electric field along the normal to the surface near the conductor is inversely proportional to the principal radii of curvature?
$\frac{\partial E}{\partial n}=-E \left(\frac{1}{R_1}+\frac{1}{R_2}\right)$.
I tried to derive it by Gauss's theorem for cylindrical coordinates, but it didn't help
 A: This formula can be deduced by applying the Conservation of Flux : the flux through any volume above the conductor is zero because it encloses no charge.
Imagine such a volume close to the surface of the conductor. In profile this volume (vastly exaggerated in size) looks like the shaded region ABCD in the diagram below in each of the two principal planes of curvature. The conductor has local radius of curvature $R$ in this plane, and the volume subtends angle $\theta$ from the local centre of curvature. The red arrows represent electric field lines, which are perpendicular to the surface of the conductor and parallel to the sides of the shaded volume.

The lower and upper faces of this volume are approximately rectangular and have areas $$A=R_1\theta_1 R_2 \theta_2$$ $$A'=(R_1+z)\theta_1 (R_2+z)\theta_2=(R_1R_2+[R_1+R_2]z+z^2)\theta_1\theta_2$$ In the limit that $z \to 0$ so that terms in $z^2$ can be neglected, the increase in area is $$\Delta A=(R_1+R_2)z \theta_1\theta_2=(R_1+R_2)z \frac{A}{R_1R_2}=(\frac{1}{R_1}+\frac{1}{R_2})zA$$
The flux through these two faces are the same so $$EA=E'A'=E'(A+\Delta A)$$ $$E-E'=\frac{\Delta A}{A}E'=(\frac{1}{R_1}+\frac{1}{R_2})zE'$$ $$\lim \limits_{z \to 0}\frac{E-E'}{z}=-\frac{\partial E}{\partial z}=(\frac{1}{R_1}+\frac{1}{R_2})E$$
