Gauss's Law states that the net outward flux through any closed surface is equal to `1/epsi_0` times the charge enclosed by the closed surface.
(i) When the point `P` is inside the shell.
In this case, the Gaussian surface lies inside the spherical shell and hence no charge is enclosed by it.
`∮ vecE . vec(ds) = ∮ E. ds cos0 = E∮ds = E xx 4pi r_1^2` ................(i)
and by Gauss's law
`∮ vecE . vec(ds) = 1/epsi_0 xx 0 = 0` ................(ii) (since no charge is enclosed)
`therefore` From (i) and (ii), we have
`E xx 4 pir_1^2 = 0`
or `E = 0`, i.e. there is no electric field inside a charged spherical shell.
(ii) When the point P lies outside the shell
Consider a spherical shell of radius `r_1` charged to a potential `V` by a charge `Q.` .To find the electric intensity at a point `P` at a distance `r_2` from the centre of the spherical shell imagine a spherical Gaussian surface to be drawn around the charged shell. At every point of this shell, the `vecE` vector and `vec(dS)` vector are directed outwards
in the same direction, i.e.` theta = 0`.
`therefore ∮ vecE . vec(ds) = ∮ E .ds = E ∮ds = E xx 4 pir_2^2` ............(i)
Also, by Gauss's law
`∮vecE . vec(ds) = 1/epsi_0 Q` ................(ii)
From (i) and (ii), we get
`E xx 4 pi r^2 = 1/epsi_0 Q => E = 1/(4pi epsi_0) Q/r^2 \ \ \ \ \ \ \ [ because r = r_2]`