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

Explanation

Electric field due to a uniformly charged shell : Consider a uniformly charged spherical shell of radius R and total charge Q . The electric field at point outside and inside the sphere is found using Gauss law .

Case (a) At a point outside the shell `(r gt 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 surface of the sphere ( spherical symmetry) . Hence the electric field must point radially outward if `Q gt0` and point radially in ward if `Q lt 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))`
The electric field `vecE` and `dvecA` point in the same direction ( outward normal ) at all the point 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 `E underset("Gaussian surface ")(int) dA = (Q)/(epsilon_(0))`
But `underset("Gaussian surface ")(int)` 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`
The electric field is radially outward if `Q gt0` and radially inward if `Q lt 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 centre 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 fiedl at points on the spherical shell ( r= R) is given by
`vecE= (Q)/(4piepsilon_(0)R^(2)) hatr `
Case (c ) At a point inside the spherical shell `(r lt 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 teh figure (b) Applying law
`underset("Gaussian furface ")(int) vecE .d vecA= (Q)/(epsilon_(0))`
`E.4pir^(2)=(Q)/(epsilon_(0))`
Since Gaussian surface encloses no charge So Q = 0 . The equation ( 5) becomes
E=0 `(r lt R)`
The electric field due to the uniformly charged spherical shell is zero at all points inside the shell .