A spherical conductor of radius 12 cm has a charge of 1.6 × 10−7C distributed uniformly on its surface. What is the electric field
(a) Inside the sphere
(b) Just outside the sphere
(c) At a point 18 cm from the centre of the sphere?
(a) Radius of the spherical conductor, r = 12 cm = 0.12 m
Charge is uniformly distributed over the conductor, q = 1.6 × 10−7 C
Electric field inside a spherical conductor is zero. This is because if there is field inside the conductor, then charges will move to neutralize it.
(b) Electric field E just outside the conductor is given by the relation,
$E=\frac{q}{4 \pi \epsilon_{0} r^{2}}$
Where,
$\in_{0}=$ Permittivity of free space
$\frac{1}{4 \pi \epsilon_{0}}=9 \times 10^{9} \mathrm{~N} \mathrm{~m}^{2} \mathrm{C}^{-2}$
$\therefore E=\frac{1.6 \times 10^{-7} \times 9 \times 10^{-9}}{(0.12)^{2}}$
$=10^{5} \mathrm{~N} \mathrm{C}^{-1}$
Therefore, the electric field just outside the sphere is $10^{5} \mathrm{~N} \mathrm{C}^{-1}$.
(c) Electric field at a point 18 m from the centre of the sphere = E1
Distance of the point from the centre, d = 18 cm = 0.18 m
$E_{1}=\frac{q}{4 \pi \in_{0} d^{2}}$
$=\frac{9 \times 10^{9} \times 1.6 \times 10^{-7}}{\left(18 \times 10^{-2}\right)^{2}}$
$=4.4 \times 10^{4} \mathrm{~N} / \mathrm{C}$
Therefore, the electric field at a point 18 cm from the centre of the sphere is
$4.4 \times 10^{4} \mathrm{~N} / \mathrm{C}$