Question

Let $\rho\left(r\right)=\frac{Qr}{\pi R^{4}}$ be the charge density distribution for a solid sphere of radius R and total charge $Q$. For a point $P$ inside the sphere at a distance $r_1$ from the centre of the sphere, the magnitude of electric field is

• A. $\frac{Q}{4\pi\varepsilon_{0}r^{2}_{1}}$
• B. $\frac{Qr^{2}_{1}}{4\pi\varepsilon_{0}R^{4}}$
• C. $\frac{Qr^{2}_{1}}{3\pi\varepsilon_{0}R^{4}}$
• D. zero

Answer 1

Answer: (B)

In figure dotted sphere of radius $r_1$ is the Gaussian surface.
According to Gauss’s theorem

$\oint\,\vec{E}\cdot d\vec{s}=\frac{q_{inside}}{\varepsilon_{0}}$
$\oint\,E\,ds \,cos \,0^{\circ}=\frac{1}{\varepsilon_{0}}\oint \rho\left(r\right)dV$
$E\left(4\pi r_{1}\right)^{2}=\frac{Q}{\varepsilon_{0}\pi R^{4}}$$\int\limits^{r_1}_{{0}}r(4\pi r^2)dr$
$=\frac{4Q}{\varepsilon_{0}R^{4}}\left(\frac{r^{4}_{1}}{4}\right)=\frac{Q}{\varepsilon_{0}}\left(\frac{r_{1}}{R}\right)^{4}$.
$\therefore E=\frac{Q}{4\pi\varepsilon_{0}} \frac{r^{2}_{1}}{R^{4}}$