The dielectric strength of air at NTP is 3 ti | vedclass

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(B) The electric field $E$ at the surface of a spherical conductor of radius $R$ carrying charge $Q$ is given by $E = \frac{1}{4\pi\epsilon_0} \frac{Q

Vedclass · Accepted Answer

$3 	imes 10^{-3} \, C$

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(B) The electric field $E$ at the surface of a spherical conductor of radius $R$ carrying charge $Q$ is given by $E = \frac{1}{4\pi\epsilon_0} \frac{Q}{R^2} = k \frac{Q}{R^2}$.Given $E_{max} = 3 	imes 10^6 \, V/m$,$R = 3 \, m$,and $k = 9 	imes 10^9 \, N \cdot m^2/C^2$.Substituting the values into the formula: $3 	imes 10^6 = 9 	imes 10^9 	imes \frac{Q}{3^2}$.$3 	imes 10^6 = 9 	imes 10^9 	imes \frac{Q}{9}$.$3 	imes 10^6 = 10^9 	imes Q$.$Q = \frac{3 	imes 10^6}{10^9} = 3 	imes 10^{-3} \, C$.

The dielectric strength of 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:

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