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

(A) Consider a point $P$ just outside the surface of a charged conductor and a point $Q$ just inside the surface.
Let $E$ be the electric field just outside the conductor,which is given by $E = (\sigma / \varepsilon_{0}) \hat{n}$.
This field $E$ is the sum of the electric field $E_1$ due to the small patch of charge at the hole and the electric field $E_2$ due to the rest of the conductor.
At point $P$ (outside),$E_1$ and $E_2$ are in the same direction,so $E_1 + E_2 = E = \sigma / \varepsilon_{0}$.
At point $Q$ (inside),$E_1$ is directed inward (opposite to $\hat{n}$) and $E_2$ is directed outward,so $-E_1 + E_2 = 0$ (since the field inside a conductor is zero).
From the second equation,$E_1 = E_2$.
Substituting this into the first equation,$2E_2 = \sigma / \varepsilon_{0}$,which gives $E_2 = \sigma / (2 \varepsilon_{0}) \hat{n}$.
Thus,the electric field in the hole is due to the rest of the conductor,which is $(\sigma / 2 \varepsilon_{0}) \hat{n}$.