# A hollow charged conductor has a tiny hole cut into its surface . show that the electric field in the hole is $\bigg( \large\frac{\sigma}{2 \in_0}\bigg)$$\hat n, where \hat n is the unit vector in the outward normal direction and \sigma is the surface change density near the hole. ## 1 Answer Comment A) Answer : We can calculate the electric field due to hollow charge sphere by using guass theorem Let the surface charge density near the hole = \sigma Let \hat n be the normal vector which is directed outwards . Let P be the point on the hole. According of guass theorem \int E.ds =\large\frac{q}{\in_o} Where q is the charge near the hole. \sigma= \large\frac{q}{ds} => q= \sigma ds E ds \cos \theta= \large\frac{\sigma ds}{\in_0} But the angle between the electric field and area vector is 0^{\circ} E ds= \large\frac{\sigma ds}{\in_0} E= \large\frac{\sigma }{\in_0} E= \large\frac{\sigma }{\in_0}$$\hat n$
Hence the electric field at P due to each part $\large\frac{1}{2}$$E=\frac{\sigma}{2 \in_0}\hat n$