Proving Gauss Theorem for Surface of Curved Charged Conductor

[fluidistic]

Homework Statement

Use Gauss theorem to prove that at the surface of a curved charged conductor, the normal derivative of the electric field is given by \frac{1}{E} \frac{\partial E}{\partial n}=-\left ( \frac{1}{R_1}+ \frac{1}{R_2} \right ) where R_1 and R_2 are the principal radii of curvature of the surface.

Homework Equations

\int \vec \nabla \cdot \vec E d V= \int _{\partial V} \vec E \cdot \hat n dS.

The Attempt at a Solution

The first thing that crosses my mind is to sketch the situation and this is where I'm stuck. Why has the conductor 2 principal radii of curvature? What if it's a sphere?!
Should I sketch the conductor as a "lenslike" solid like we see in geometrical optics sketch?

And by the way isn't the normal derivative \frac{\partial E}{\partial n} instead of the same expression but with a factor of \frac{1}{E}?

 

[tiny-tim]

hi fluidistic!

… Why has the conductor 2 principal radii of curvature? What if it's a sphere?!

if it's a sphere, the two principal radii are the same (and can be in any direction).

start with the simplest case, a cylinder, so that R₂ = ∞ …

then you have the same calculation as for finding E in a cylindrical capacitor …

you'll see that essentially you're finding the formula for ∂/∂n(1/A), where A is surface area within a fixed solid angle

(and yes, the normal derivative is ∂E/∂n … the phrasing is a little unusual, but they're saying that ∂E/∂n is given by the equation (1/E)∂E/∂n = … )

 

[fluidistic]

Hi tiny-tim,
I still don't understand. By radii of curvature I assue the conductor must have a spherical shape. So a cylinder as you said would do the job but I don't know the "centre" where the radii start. In other words to me a cylinder has only 1 radius of curvature.
But you said its second radius of curvature is infinite, thus I believe it has a plane part? You mean a base of the cylinder? And where does the centre of the radii is?Edit: I just found 2 solutions on the internet of this problem: http://www-personal.umich.edu/~pran/jackson/P505/F07_hw01a.pdf.
It's still not clear to me as what are the radii of curvature.
I prefer the differential approach of page 5 of the PDF. However I do not understand why he takes \vec \nabla \cdot \vec E =0, in other words why does he consider a chargeless part of the space when doing the algebra since all the charge of a conductor must lie within its surface when it's in steady state.
I'm also trying to derive the expression \vec \nabla \cdot \hat n =\frac{1}{R_1}+\frac{1}{R_2} without success. Should I get a parametric expression of a sphere and then use some facts/formulae to derive it or there's a simpler way to show it?

 

[tiny-tim]

hi fluidistic!

I assue the conductor must have a spherical shape.

no, it can have any smooth shape

(unfortunately, wikipeida is pretty useless on radius of curvature)

every curve in a 2D plane has a radius of curvature, that you're familiar with

on a 2D surface in 3D space, at each point you can cut the surface with planes going through the normal: these give curves along the surface in each 2D plane (ie in each direction at that point) …

the radius of curvature in each such direction is defined as the radius of curvature of that curve

half-way up an egg, for example, the "horizontal" radius of curvature is small, but the "vertical" radius of curvature is large

and for a cylinder, one radius of curvature is infinite

you can solve the given problem by considering the divergence of ñ, the unit normal vector …

do it first for a sphere, then do the same thing for any two perpendicular directions on a general surface ​

 

[tiny-tim]

i've just seen your edit:

E = 0 inside the conductor, it jumps suddenly at the surface, and then it follows the differential equation given

the .pdf uses a volume just above the surface (as in the picture)