Question

Using Gauss’ law deduce the expression for the electric field due to a uniformly charged spherical conducting shell of radius R at a point

(i) outside and (ii) inside the shell.

Plot a graph showing variation of electric field as a function of r > R and r < R.

(r being the distance from the centre of the shell)

Answer 1

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 (r > R) from its centre. Draw the Gaussian surface through point P so as to enclose the charged spherical shell. The Gaussian surface is a spherical shell of radius r and centre O.

Let `vecE`be the electric field at point P. Then, the electric flux through area element vecdsis given by,

`dphi = vecE.vecds`

Since `vecds` s also along normal to the surface,

dΦ = E ds

∴ Total electric flux through the Gaussian surface is given by,

`phi = oint_s Eds = Eoint_s ds`

Now,

`oint ds = 4pir^2`

`therefore phi= E xx 4pir^2   ..... (1)`

Since the charge enclosed by the Gaussian surface is q, according to Gauss theorem,

`phi = q/epsi_0   ......(2)`

From equations (i) and (ii), we obtain

`E xx 4pir^2q/epsi_o`

`E = 1/(4piepsi_0).q/r^2`     (for r>R)

(ii) When point P lies inside the spherical shell:

In such a case, the Gaussian surface encloses no charge.

According to Gauss law,

E × 4πr² = 0

i.e., = E = 0 (r < R)

Graph showing the variation of electric field as a function of r: