A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is `(sigma/(2in_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.
Solution
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 `in_0` is the permittivity of free space.
Charge `|"q"| = vecsigma xx vec("ds")`
According to Gauss’s law,
Flux, `phi = vec"E".vec("ds") = |"q"|/in_0`
`"E"."ds" = (vecsigma xx vec("ds"))/in_0`
∴ `"E" = sigma/in_0.hat"n"`
Therefore, the electric field just outside the conductor is `sigma/in_0.hat"n"`. This field is a superposition of field due to the cavity (E') and the field due to the rest of the charged conductor (E'). These fields are equal and opposite inside the conductor and equal in magnitude and direction outside the conductor.
∴ E' + E' = E
E' = `"E"/2`
= `sigma/(2in_0).hat"n"`
Therefore, the field due to the rest of the conductor is `sigma/in_0.hat"n"`.
Hence, proved.
