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 $3m$ 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
$F=\left(\frac{1}{4\pi {\epsilon }_0}\right)\frac{q}{R^2}$ ...(1)
If $q=q_{\max }$ and $F=F_{\max }$
As $E_{\max }=$ dielectric strength of air
$=3\times 10^{6}V/m$ and $R=3m$
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.