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
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$\begingroup$ Are you sure that you mean the derivative of the electric field - ie $\frac{dE}{dx}$? Where have you seen this written? The electric field $E$ is inversely proportional to the radius of curvature. $\endgroup$– sammy gerbilCommented Aug 29, 2018 at 21:03
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$\begingroup$ I fixed some errors in my question and cited this formula for clarity. Thank you $\endgroup$– EugenSCommented Aug 30, 2018 at 3:25
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$\begingroup$ Where did you get the formula from? What is n? $\endgroup$– sammy gerbilCommented Aug 30, 2018 at 3:33
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$\begingroup$ I took this formula in the book of DV Sivukhin (MIPT). GENERAL COURSE OF PHYSICS. T.III. ELECTRICITY, page 45. $n$ is $r$ in system of cylindrical coordinates I think $\endgroup$– EugenSCommented Aug 30, 2018 at 3:37
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$\begingroup$ Colleagues, why is everyone silent? $\endgroup$– EugenSCommented Aug 30, 2018 at 9:25
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1 Answer
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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 $$EA=E'A'=E'(A+\Delta A)$$ $$E-E'=\frac{\Delta A}{A}E'=(\frac{1}{R_1}+\frac{1}{R_2})zE'$$ $$\lim \limits_{z \to 0}\frac{E-E'}{z}=-\frac{\partial E}{\partial z}=(\frac{1}{R_1}+\frac{1}{R_2})E$$
answered Sep 1, 2018 at 21:29
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$\begingroup$ Thank you, @sammy gerbil! Answer is cool! Figure is very useful. $\endgroup$– EugenSCommented Sep 2, 2018 at 5:14