Write the expression for electric field intensity at any point outside and inside due to a charged spherical shell.

Explanation

Derivation:
In the above figure
E is the electric field.
R is the radius of a thin spherical shell of centre O.
r is the distance between the point P and O.
P is a point at distance r from O.
`DeltaS` is the area element around the point P.
Let q be the charge uniformly distributed over the surface of the shell. The electric flux through the Gaussian surface is given by
`phi = sum E DeltaS cos theta` ----(1)
The angle between E and `DeltaS` is 0 `:. cos theta=1`
`(1)implies phi=sum E DeltaS`
`phi = E sum DeltaS`----(2)
where `sum DeltaS` = area of the spherical Gaussian surface = `4pi r^2`
`(2)implies phi=E xx 4pi r^2` ----(3)
From Gauss law, `phi =(q)/(epsilon_0)` ----(4)
On comparing the equations (3) and (4) we get,
`E xx 4pi r^2 =(q)/(epsilon_0)`
`E=(q)/(4pi epsilon_0 r^2)`
`E=(1)/(4pi epsilon_0)(q)/(r^2)` Along OP produced
The electric field inside the charged spherical shell is zero. This is because the charge enclosed by the Gaussian surface is zero.