Question

Let there be a spherically symmetric charge distribution with charge density varying as $\rho \left(r\right)={\rho }_0(\frac{5}{4}-\frac{r}{R})$ upto r = R and $\rho \left(r\right)=0$ for r > R, where r is the distance from the origin. The electric field at a distance r(r < R) from the origin is given by:

• A. $\frac{{\rho }_{0}r}{3{\epsilon }_0}(\frac{5}{4}-\frac{r}{R})$
• B. $\frac{4\pi {\rho }_{0}r}{3{\epsilon }_0}(\frac{5}{3}-\frac{r}{R})$
• C. $\frac{{\rho }_{0}r}{4{\epsilon }_0}(\frac{5}{3}-\frac{r}{R})$
• D. $\frac{4{\rho }_{0}r}{3{\epsilon }_0}(\frac{5}{4}-\frac{r}{R})$

Answer 1

The correct option is C $\frac{{\rho }_{0}r}{4{\epsilon }_0}(\frac{5}{3}-\frac{r}{R})$

Charge enclosed by a Gaussian surface sphere of radius r(< R) is
$q=\int \rho dV={\int }_0^r{\rho }_0(\frac{5}{4}-\frac{r}{R})4\pi r^{2}dr=4\pi {\rho }_0{\int }_0^r(\frac{5}{4}{r}^2-\frac{r^3}{R})dr=4\pi {\rho }_0[\frac{5}{4}{\left|\frac{r^3}{3}\right|}_0^r-{\left|\frac{r^4}{4R}\right|}_0^r]=4\pi {\rho }_0[\frac{5}{4}.\frac{r^3}{3}-\frac{r^4}{4R}]UsingGaus{s}^{\prime }slaw,E\times \pi r^2=\frac{q}{{\epsilon }_0}=\frac{4\pi {\rho }_0}{{\epsilon }_0}[\frac{5}{4}.\frac{r^3}{3}-\frac{r^4}{4R}]\Rightarrow E=\frac{{\rho }_{0}r}{4{\epsilon }_0}[\frac{5}{3}-\frac{r}{R}]$