Discussion
Uniformly charged solid sphere of radius R has a volume charge density ρ. The strength of electric field, at a distance r, (\(r \gt R\)), from its centre is :
Answer: B
The field is \(\frac{\rho R^{3}}{3\epsilon_{0} r^{2}}\).
Outside a spherically symmetric charge distribution the field is that of a point charge equal to the total charge. The total charge is \(Q = \frac{4}{3}\pi R^{3}\rho\). By Gauss's law on a sphere of radius \(r \gt R\):
\[ \begin{aligned} E \cdot 4\pi r^{2} &= \frac{Q}{\epsilon_{0}} = \frac{4\pi R^{3}\rho}{3\epsilon_{0}} \\ E &= \frac{\rho R^{3}}{3\epsilon_{0} r^{2}} \end{aligned} \]
The field must fall off as \(1/r^{2}\) outside, which rules out the options growing as \(r^{3}\) or falling only as \(1/r\). The expression \(\rho r/3\epsilon_{0}\) (not offered) would be the field inside the sphere.
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