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

Explanation

Let us consider a conductor with a cavity or a hole. Electric field inside the cavity is zero.
Let E is the electric field just outside the conductor, q is the electric charge, `sigma` is the charge density, and `epsilon_(0)` is the permittivity of free space.
Charge `"|"q"|"=barsigmaxxbards`
According to Gauss's law,
`Flux, phi=barE.bards=(|q|)/(epsilon_(0))`
`Eds=(bar sigma xx bar ds)/(epsilon_(0)) `
herefore, the electric field just outside the conductor is `(sigma )/(epsilon_(0))hatn` This field is a superposition of field due to the cavity `(E)` and the field due to the rest of the charged conductor (E') hese fields are equal and opposite inside the conductor, and equal in magnitude and direction outside the conductor.
`therefore E'+E'=E`
`E'=(E)/(2)`
`=(sigma)/(2epsilon_(0))hatn` Therefore, the field due to the rest of the conductor is `(sigma)/(2epsilon_(0))hatn` ltbrrgt Hence, proved.