Discussion
Given a spherically symmetric charge density \(\rho(r) = \begin{cases} kr^{2}, & r \lt R \\ 0, & r \gt R \end{cases}\) (k being a constant), the electric field for \(r \lt R\) is (take the total charge as Q)
Answer: A
The field inside is \(\frac{Qr^{3}}{4\pi\varepsilon_{0}R^{5}}\hat{r}\).
First express k through the total charge, then apply Gauss's law to a sphere of radius \(r \lt R\):
\[ \begin{aligned} Q &= \int_{0}^{R} kr^{2}\,4\pi r^{2}dr = \frac{4\pi k R^{5}}{5} \\ q(r) &= \frac{4\pi k r^{5}}{5} = Q\frac{r^{5}}{R^{5}} \\ E &= \frac{q(r)}{4\pi\varepsilon_{0}r^{2}} = \frac{Qr^{3}}{4\pi\varepsilon_{0}R^{5}} \end{aligned} \]
The last option is the field outside the sphere. The other two have the wrong power of r or the wrong numerical factor; at \(r = R\) only the keyed answer matches the outside field \(\frac{Q}{4\pi\varepsilon_{0}R^{2}}\).
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