Write the expression for electric field intensity at any point outside and inside due to a charged spherical shell.

Explanation

Electric field due to a uniformly charged thin spherical shell :

(i) When point P lies outside the spherical shell Suppose that we have to calculate electric field at the point P at a distance `r(rgtR)` from its centre. Draw the Gaussian surface through point P so as to enclose the charged spherical shell . The Gaussian surface is spherical shell of radius r and centre O.
Let `vecE` be the electric field at point P, then the electric flux through area element is `vecds` given by ,
`dphi=vecE*vecds`
Since `vecds` is also along normal to the surface,
`dphi=Eds`
`therefore` Total electric flux through the Gaussian surface is given by , `phi=oint_(s)Eds=Eoint_(s)ds`
Now , `oint_(s)ds=4pir^(2)`
`thereforephi=Exx4pir^(2)` ... (i)
Since the charge enclosed by the Gaussian surface is q , according to Gauss theorem,
`phi=(q)/(epsilon_(0))` ...(ii)
From equations (i) and (ii) , we obtain
`Exx4pir^(2)=(q)/(epsilon_(0))`
`E=(1)/(4piepsilon_(0)).(q)/(r^(2))("for "rgtR)`
(ii) When point P lies inside the spherical shell : In such a case , the Gaussian surface encloses no charge . According to Gauss law,
`Exx4pir^(2)=0`
i.e., E=0 `("for" r lt R)`
Graph showing the variation of electric field as a function of r :