Using Gauss's Theorem, Derive an expression for electric field intensity due to uniformity charged hollow sphere : (a) At a point outside the shell.

Explanation

Electric field intensity at a point outside the shell : Consider a positive charge q distributed uniformly on the surface of a spherical shell of radius R. Let P be the point outside the shell at distance r from the centre of shell (r>R), where electric field is to be calculated. Draw a Gaussian surface in the from of sphere of radius r with centre O.
According to Gauss.s Theorem
`underset(S)ointvec(E).vec(dS)=q/varepsilon_(0)`
`underset(S)ointEdS=q/varepsilon_(0)`
Since `vecE and vec(dS)` are in same direction
`:." "theta=0`
`underset(S)ointEdScos0^(@)=q/varepsilon_(0)`
`underset(S)ointEdS=q/varepsilon_(0)`
Since E is constant at every point on the Gaussian surface
`:." "Eunderset(S)ointdS= q/varepsilon_(0)`
But`" "underset(S)ointdS= 4 pi r^(2)`
`:." "E xx 4 pi r^(2)=q/varepsilon_(0)`
`E=1/(4 pi varepsilon_(0))q/r^(2)`
In vector from
`vecE=q/(4 pi varepsilon_(0)r^(2))hatr`