Electric Field in the Hole of a Charged Conductor
A hollw charged conductor has a tiny hole cut into its surface. Show that the electric in the hole is `((sigma)/(2epsilon_(0)))hat(n)`. Where, `hat(n)` is the unit vector in the outward normal direction and `sigma` is the surface charge density near the hole.

Explanation

Surface charge density near the hole `=sigma` Unit vector `-hat(n)` (normal directed outwards) Let P be the point on the hole. The electric field at point P closed to surface to conductor, according to Gauss. theorem,
`ointE*dS=(q)/(epsilon_(0))`
where, q is the charge near the hole.
`EdScostheta=(sigmadS)/(epsilon_(0))" "(becausesigma=q//dSrArrq=sigmadS," where dS = area")`
`:.` Angle between electric field and area vector is `0^(@)`.
`:." "EdS=(sigmadS)/(epsilon_(0))" "(because cos 0^(@)=1)`
`rArr" "E=(sigma)/(epsilon_(0))`
This electric field is due to the filled up hole and the field due to the rest of the charged conductor. The two fields inside the conductor are equal and opposite. So, there is conductor, the electric fields are equal in the same direction.
So, the electric field at point P due to each part
`=1/2E=(sigma)/(2epsilon_(0))hat(n)`