• # question_answer The electric field in a hole which is cut into its surface of a hollow charged conductor is $\left( \frac{\sigma }{2{{\varepsilon }_{0}}} \right)\hat{n},$ where $\hat{n}$ is the unit vector in the outward normal direction and $\sigma$ is the surface charge density near the hole.

According to Gauss' theorem, $\oint{E\cdot dS=\frac{{{q}_{in}}}{{{\varepsilon }_{0}}}}$ Since, angle between electric field and area vector is $0{}^\circ .$ $\therefore$      $E\cdot dS=\frac{\sigma dS}{{{\varepsilon }_{0}}}$       $(\because q=\sigma dS)$             $E=\frac{\sigma }{{{\varepsilon }_{0}}}\Rightarrow E=\frac{\sigma }{{{\varepsilon }_{0}}}\hat{n}$ where, $\hat{n}$ is unit vector in normal direction. So, electric field at point P due to each part $=\frac{1}{2}E=\frac{\sigma }{2{{\varepsilon }_{0}}}\hat{n}$