Let, R be the radius of the shell and Q be the charge uniformly distributed on the surface.
i) For a point outside the shell :
By Gauss's law, E.4πr2=Qen/ϵ0
here r be the distance from centre of shell (r>R) and charge enclosed by surface Qeq=Q.
So, the electic field outside the shell is E=Qen/(4πr2ϵ0) , the electric potential outside the shell is V(r)=−∫∞r4πr12ϵ0Qendr1=4πrϵ0Qen.
Comments
Dear Awesh, thank you for correcting us.
Totally wrong... According to question find 'Electric Potential' ("V")... But in the solution the expert has found 'Electric Field' ("E").