The electric field strength in air at $$NTP$$  is $$3 \times {10^6}\,V/m.$$   The maximum charge that can be given to a spherical conductor of radius $$3\,m$$  is

Solution :
Given, $${E_{\max }} = 3 \times {10^6}\,V/m$$     and $$R = 3\,m$$
We know that,
$$\eqalign{ & E = \frac{1}{{4\pi {\varepsilon _0}}} \times \frac{Q}{{{R^2}}} \cr & \Rightarrow {Q_{\max }} = 4\pi {\varepsilon _0}{R^2}{E_{\max }} \cr & = \frac{{3 \times 3 \times 3 \times {{10}^6}}}{{9 \times {{10}^9}}} \cr & = 3 \times {10^{ - 3}}C \cr} $$