Question

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

• A. $ 3\times 10^{-4}\,C $
• B. $ 3\times 10^{-3}\,C $
• C. $ 3\times 10^{-2}\,C $
• D. $ 3\times 10^{-1}\,C$

Answer 1

Answer: (B)

Using the relation for electric intensity
$F=\left(\frac{1}{4 \pi \varepsilon_{0}}\right) \frac{q}{R^{2}}$ ...(1)
If $q=q_{\max }$ and $F=F_{\max }$
As $E_{\max }=$ dielectric strength of air
$=3 \times 10^{6} V / m$ and $R=3\, m$
So, from equation (1)
$3 \times 10^{6}=9 \times 10^{9} \times \frac{q_{\max}}{(3)^{2}}$
So, $q_{\max }=3 \times 10^{-3}$ coulomb.