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. $\frac{Q}{4\pi {\epsilon }_0{r}_1^2}$
• B. $\frac{Q{r}_1^2}{4\pi {\epsilon }_0{R}^4}$
• C. $\frac{Q{r}_1^2}{3\pi {\epsilon }_0{R}^4}$
• D. zero

Answer 1

Answer: (B)

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

$\oint \vec{E}\cdot d\vec{s}=\frac{q_{inside}}{{\epsilon }_0}$
$\oint Edscos{0}^{\circ }=\frac{1}{{\epsilon }_0}\oint \rho \left(r\right)dV$
$E{\left(4\pi r_1\right)}^2=\frac{Q}{{\epsilon }_0\pi R^4}$$\underset{0}{\overset{r_1}{\int }}r(4\pi r^2)dr$
$=\frac{4Q}{{\epsilon }_0{R}^4}\left(\frac{r_1^4}{4}\right)=\frac{Q}{{\epsilon }_0}{\left(\frac{r_1}{R}\right)}^4$.
$\therefore E=\frac{Q}{4\pi {\epsilon }_0}\frac{r_1^2}{R^4}$