Obtain the expression for electric field due to an uniformly charge spherical shell.

Explanation

Electric field due to a uniformly charged spherical shell: Consider a uniformly charged spherical shell of radius R and total charge Q. The electric field at points outside and inside the sphere is found using Gauss law.
Case (a) At a point outside the shell (r> R): Let us choose a point P outside the shell at a distance r from the center as shown in figure (a). The charge is uniformly distributed on the surface of the sphere (spherical symmetry). Hence the electric field must point radially outward if Q> 0 and point radially inward if Q < 0. So we choose a spherical Gaussian surface of radius r and the total charge enclosed by this Gaussian surface is Q. Applying Gauss law,
`oint vecE.dvecA=Q/epsilon_0` ...(1)

The electric field `vecE` and d`vecA` point in the same direction (outward normal) at all the points on the Gaussian surface. The magnitude of `vecE` is also the same at all points due to the spherical symmetry of the charge distribution
Hence `underset"Gaussian surface"(E oint dA) =Q/epsilon_0`...(2)
But `underset"Gaussian surface"(oint dA)` =total area of Gaussian surface =`4pir^2`.
Substituting this value in equation (2).
`E.4pir^2=Q/epsilon_0`
`E.4pir^2=Q/epsilon_0` (or) `E=1/(4piepsilon_0)Q/r^2`
In vector form `vecE=1/(4piepsilon_0) Q/r^2 hatr` ...(3)
The electric field is radially outward if Q>0 and radially inward if Q<0. From equation (3) we infer that the electric field at a point outside the shell will be same as if the entire charge Q is concentrated at the center of the spherical shell. (A similar result is observed in gravitation, for gravitational force due to a spherical shell with mass M)
Case (b): At a point on the surface of the spherical shell (r= R): The electrical field a points on the spherical shell (r=R) is given by
`vecE=Q/(4piepsilon_0R^2)hatr` ...(4)
Case (c) At a point inside the spherical shell (r< R): Consider a point P inside the shell at a distance r from the center. A Gaussian sphere of radius r is constructed as shown in the figure (b). Applying Gauss law
`underset"Gaussian surface"(oint vecE. dvecA)=Q/epsilon_0`
`E.4pir^2=Q/epsilon_0` ...(5)
Since Gaussian surface encloses no charge , So Q=0 . The equation (5) becomes
E=0 (r < R) ...(6)
The electric field due to the uniformly charged spherical shell is zero at all points inside the shell.