Q. A sphere of charge of radius R carries a positive charge whose volume charge density depends only on the distance *r* from the ball’s centre as ρ = ρ_{0} (1 − r/R). Where ρ_{0} is a constant. Assume ε as the permittivity of space. The magnitude of electric field as a function of the distance *r* inside the sphere is given by

(a) $ \displaystyle \frac{\rho_0}{\epsilon}[\frac{r}{3}-\frac{r^2}{4R}]$

(b) $\displaystyle \frac{\rho_0}{\epsilon}[\frac{r}{4}-\frac{r^2}{3R}]$

(c) $ \displaystyle \frac{\rho_0}{\epsilon}[\frac{r}{3}+ \frac{r^2}{4R}]$

(d) $\displaystyle \frac{\rho_0}{\epsilon}[\frac{r}{4}-\frac{r^2}{3R}]$

Ans: (a)

Sol: charge inside a sphere of radius r (r < R)

$\displaystyle q = \int_{0}^{r} \rho dv $

$ \displaystyle = \int_{0}^{r} \rho_0 (1-\frac{r}{R})4\pi r^2 dr $

$ \displaystyle q = 4\pi \rho_0 \int_{0}^{r}(r^2 -\frac{r^3}{R}) $

$ \displaystyle q = 4\pi \rho_0 (\frac{r^3}{3}-\frac{r^4}{4R}) $

$\displaystyle E = \frac{1}{4\pi \epsilon}\frac{q}{r^2} $

$ \displaystyle E = \frac{1}{4\pi \epsilon}.4\pi \rho_0 (\frac{r}{3}-\frac{r^2}{4R}) $

$ \displaystyle \frac{\rho_0}{\epsilon}(\frac{r}{3}-\frac{r^2}{4R}) $