question_answer

Let there be a spherically symmetric charge distribution with charge density varying as \[\rho (r)={{\rho }_{0}}\left( \frac{5}{4}-\frac{r}{R} \right)\]for \[r\le R\], and \[\rho (r)=0\]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) \[\frac{{{\rho }_{0}}r}{3{{\varepsilon }_{0}}}\left( \frac{5}{4}-\frac{r}{R} \right)\]          

B) \[\frac{4\pi {{\rho }_{0}}r}{3{{\varepsilon }_{0}}}\left( \frac{5}{3}-\frac{r}{R} \right)\]

C) \[\frac{{{\rho }_{0}}r}{4{{\varepsilon }_{0}}}\left( \frac{5}{3}-\frac{r}{R} \right)\]          

D)   \[\frac{4{{\rho }_{0}}r}{3{{\varepsilon }_{0}}}\left( \frac{5}{4}-\frac{r}{R} \right)\]

Correct Answer: C

Solution :

[c] : Consider a thin spherical shell of radius x and thickness dx as shown in the figure. Volume of the shell,\[dV=4\pi {{x}^{2}}dx\] Let us draw a Gaussian surface of radius r(r < -R) as shown in the figure above. Total charge enclosed inside the Gaussian surface is \[{{Q}_{in}}=\int\limits_{0}^{r}{\rho dV}=\int\limits_{0}^{r}{{{\rho }_{0}}}\left( \frac{5}{4}-\frac{x}{R} \right)4\pi {{x}^{2}}dx\] \[=4\pi {{\rho }_{0}}\int\limits_{0}^{r}{\left( \frac{5}{4}{{x}^{2}}-\frac{{{x}^{3}}}{R} \right)}dx=\pi {{\rho }_{0}}\left[ \frac{5}{3}{{r}^{3}}-\frac{{{r}^{4}}}{R} \right]\] According to Gauss s law, \[E4\pi {{r}^{2}}=\frac{{{Q}_{in}}}{{{\varepsilon }_{0}}}=\frac{\pi {{\rho }_{0}}}{{{\varepsilon }_{0}}}\left[ \frac{5}{3}{{r}^{3}}-\frac{{{r}^{4}}}{R} \right];\] \[\therefore \]\[E=\frac{{{\rho }_{0}}r}{4{{\varepsilon }_{0}}}\left[ \frac{5}{3}-\frac{r}{R} \right]\]