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.

Answer:

            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 \[=\frac{1}{2}E=\frac{\sigma }{2{{\varepsilon }_{0}}}\hat{n}\]