Electric field due to a uniformly charged spherical shell

Case 1. At a point outside the shell :

Consider a charged shell of radius R (Fig.). Let P be a point outside the shell, at a distance r from the centre O. Let us construct a Gaussian surface with r as radius. The electric field E is normal to the surface. 

Fig. Field At a point outside the shell

The flux crossing the Gaussian sphere normally in an outward direction is,  

φ = ∫ E * dS = ∫ E dS = E (4π r^2)         

            (since angle between E and ds is zero)

By Gauss’s law, E (4π r^2)  = q / εo

                          E = q / 4π r^2 εo

It can be seen from the equation that, the electric field at a point outside the shell will be the same as if the total charge on the shell is concentrated at its centre.

Case 2. At a point on the surface :

The electric field E for the points on the surface of charged spherical shell is,        

      E = q / 4π R^2 εo    (∵ r=R)

Case 3. At a point inside the shell :

Consider a point P′ inside the shell at a distance r′ from the centre of the shell. Let us construct a Gaussian surface with radius r′.

The total flux crossing the Gaussian sphere normally in an outward direction is 

Fig.  At a point inside the shell

φ = ∫ E * dS = ∫ E dS = E (4π r′^2)      

since there is no charge enclosed by the gaussian surface, according to Gauss’s Law

   E (4π r′^2) = q / εo 

∴ E = 0

(i.e) the field due to a uniformly charged thin shell is zero at all points inside the shell.