Question

A sphere carrying charge $Q$ and of radius $R$ has its volume charge density proportional to the square of the distance from the centre. Determine the ratio of the magnitude of the electric field at a distance $2 R$ from the centre to that at a distance of $\frac{ R }{2}$ from the centre.

Answer 1

As given,
$\rho= Cr ^{2}$ [where $\rho=$ volume charge density $]$
$q(r) =\int\limits_{0}^{r}\left(4 \pi r^{2} d r\right) \rho$
$=\int\limits_{0}^{r} 4 \pi r^{2} \cdot C r^{2} d r=\frac{4}{5} \pi C r^{5}$
$\left.E\right|_{r=2 R}=\frac{k q_{(2 R)}}{(2 R)^{2}}=\frac{k\left(\frac{4}{5}\right) \pi C R^{5}}{4 R^{2}}$
$\ldots . . \because$ sphere has radius $R$;
so $r \leq R$ for enclosed charge $]$
$=\frac{ k \pi CR ^{3}}{5}$
$\left. E \right|_{ r = R / 2} =\frac{ kq _{( R / 2)}}{\left(\frac{ R }{2}\right)^{2}}$
$=\frac{ k \left(\frac{4}{5}\right) \pi C \left(\frac{ R }{2}\right)^{5}}{\left(\frac{ R }{2}\right)^{2}}=\frac{ k \pi CR ^{3}}{5 \times 2}$
$ \Rightarrow \frac{ E _{(2 R )}}{ E _{( R / 2)}}=2$