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(C) The number of electric field lines per unit area is equivalent to the electric field intensity $E$ at that surface.For a uniformly charged sphere

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(C) The number of electric field lines per unit area is equivalent to the electric field intensity $E$ at that surface.For a uniformly charged sphere of radius $R$ and volume charge density $ho$,the electric field at the surface $(r = R)$ is given by Gauss's Law as $E = \frac{q}{4 \pi \epsilon_{0} R^{2}}$.Since $q = ho 	imes 	ext{Volume} = ho 	imes \frac{4}{3} \pi R^{3}$,we substitute this into the expression for $E$:$E = \frac{ho 	imes \frac{4}{3} \pi R^{3}}{4 \pi \epsilon_{0} R^{2}} = \frac{ho R}{3 \epsilon_{0}}$.Given $ho = 2 \, \mu C \, m^{-3} = 2 	imes 10^{-6} \, C \, m^{-3}$,$R = 6 \, m$,and $\epsilon_{0} = 8.85 	imes 10^{-12} \, C^{2} N^{-1} m^{-2}$.$E = \frac{2 	imes 10^{-6} 	imes 6}{3 	imes 8.85 	imes 10^{-12}} = \frac{12 	imes 10^{-6}}{26.55 	imes 10^{-12}} \approx 0.4519 	imes 10^{6} 	imes 10^{6} = 0.4519 	imes 10^{12} \, N C^{-1}$.Rounding to the nearest integer as per the options,we get $45 	imes 10^{10} \, N C^{-1}$.

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