# A sphere of charge of radius R carries a positive charge whose volume charge density depends only on the distance…..

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 = ρ0 (1 − r/R). Where ρ0 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)

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

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

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

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

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

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

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