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

Explanation

Electric Field due to a uniform charged spherical shell :
Consider a uniformly charged spherical shell.
Radius - R
Total charge - Q
(a) At a point outside the shell (`r gt R`) :
P is a point ourside the shell at a distance r from the centre. The charge is uniformly distributed on the surface of the sphere. If `Q gt 0`, field point radially outward.
If `Q lt 0,` field point readially inward.
Applying Gauss law
`ointvecE.vec(dA) = Q/epsi_(0)" ...(1)"`
`vecE` and `vec(dA)` are in the same direction.
Hence E`oint "dA"=Q/epsi_(0)`
But `oint`dA = total area of Gaussian suface `= 4 pi r^2`
Substituting in (1)
`E.4 pir^2 =Q/epsi_0 (or)E=1/(4 piepsi_0)Q/r^2`
In vector from `vecE= 1/(4 pi epsi_0) Q/r^2 hatr`
(b) At a point on the suface of the spherical shell ( r = R). Electric field at point on the spherical shell, is r = R
`vec E = Q /(4 pi epsi_(0) R^2) hatr`
(c) At a point inside the shell ( ` r gt R`):
Consider a point P inside the shell at a distance r from the center.
`ointvecE.vec(dA) = Q/epsi_(0)" ...(1)"`
`E. 4 pi r^2 = Q/epsi_0`
Since Gaussian surface encloses no charge,
so Q = 0
`:. " E =0"`