Show that the electric field at the surface of a charged conductor is given by ` oversetto E = ( sigma )/( 60 ) hat n ` , where ` sigma ` is the surface charge density and ` hatn ` is a unit vector normal to the surface in the outward direction.

Explanation

Consider a charged conductor whose surface charge density is ` sigma . ` To derive electric field at its surface consider a short cylinder (pill box) as the Gaussian surface about a given point . The cylinder is partly inside and party outside the surface of the conductor. It has a small area of cross section ` delta S ` and negligible height.
Just inside the surface , the electric field is zero but just outside the field has a magnitude E and is directed normal to the surface. Thus , the contribution to the electric flux comes only from the outside circular cross-section of the cylinder.
` therefore phi_in =oversetto E . oversetto (delta S) = E.delta S `
As per Gauss theorem.
` phi _in =(1)/( in_0) ` (charged close) `=(1)/( in_0) .(delta sigma S) " "...(ii) `
Comparing (i) and (ii) we get
`E =( sigma)/( in_0) rArr " "oversetto E =(sigma)/( in_0) hatn `