A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is `( sigma //2 epsilon_(0)) hat n` , where `hatn` is the unit vector in the outward normal direction, and `sigma` is the surface charge density near the hole.

Answer

B

Explanation

Surface charge density near the hole `=sigma` . Unit vector n (normal directed outwards).
Let P be the point on the hole. The electric field at Point P closed to the surface of conductor, according to Gauss's theorem,
`int vec E.overline(ds)=q/e_0`
`(because =q/(ds) implies q=0` ds, where ds= area) where, q is the charge near the hole. `Eds cos theta=(ads)/e_0`
`because` Angle between electric field and area vector is `0^@` `therefore Eds=(ads)/e_0=E=sigma/e_0 implies E=0/e_0 n`
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 no electric field inside the conductor. Outside the conductor, the electric field at P due to each part `=1/2 E=sigma/(2e_0) overline n`