$A$ spherical conductor of radius $12 \;cm$ has a charge of $1.6 \times 10^{-7} \;C$ 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) The electric field inside a charged spherical conductor is always zero because the charges reside only on the outer surface,and the net electrostatic force inside the conductor is zero.
$(b)$ The electric field $E$ just outside the conductor is given by $E = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{r^{2}}$.
Given $r = 0.12 \; m$,$q = 1.6 \times 10^{-7} \; C$,and $\frac{1}{4 \pi \varepsilon_{0}} = 9 \times 10^{9} \; N \cdot m^{2} \cdot C^{-2}$.
$E = \frac{9 \times 10^{9} \times 1.6 \times 10^{-7}}{(0.12)^{2}} = 10^{5} \; N \cdot C^{-1}$.
$(c)$ At a distance $d = 18 \; cm = 0.18 \; m$ from the centre,the sphere acts as a point charge.
$E = \frac{9 \times 10^{9} \times 1.6 \times 10^{-7}}{(0.18)^{2}} = \frac{1440}{0.0324} \approx 4.44 \times 10^{4} \; N \cdot C^{-1}$.