By Euprime B. Regalado
From Gauss theorem, we can show that the surface of a curved charged conductor, the normal derivative of the electric field is given by
where and are the principal radii of curvature of the surface. Gauss’s law in integral form is expressed as
when there are no charges enclosed in the surface S .
Before considering the three dimensional problem, we first tackle the problem in two dimensions. We take a curved Gaussian box next to the surface of a charged conductor at a point where the radius of curvature is R. Application of Gauss’s law yields
where and are the areas of the top and bottom of the box, respectively. We can see that there is no contribution from the sides of the box, because they are taken to be normal to the surface. In polar coordinates the areas are expressed as and . Gauss’s law then yields
and now we get the relation
This allows us to calculate
Noting that the is the same as E when , this maybe written as
which is analogous to two-dimensional expression.
Going back to the 3-dim problem, we use the same method as above. This time however, the areas of the top and bottom of the Gaussian box are
which in turn yields,
Note that this reduces to the two dimensional expression (3) in cylindrical limit,