question_answer

Let there be a spherically symmetric charge distribution with charge density varying as \[\rho (X)={{\rho }_{0}}\left( \frac{5}{4}-\frac{r}{R} \right)\] upto r = R where, r is the distance from the origin. Find the electric field at a distance r(r < R) from the origin.

Answer:

            Due to spherically symmetric charge distribution, the charge on the spherical surface of radius r in figure is \[q=\int{\rho dV=\int_{0}^{r}{{{\rho }_{0}}\left( \frac{5}{4}-\frac{x}{R} \right)4\pi {{x}^{2}}dx}}\] \[={{\rho }_{0}}\left[ \frac{5}{4}4\pi \frac{{{x}^{3}}}{3}-\frac{4\pi {{x}^{4}}}{4R} \right]_{0}^{r}\] \[={{\rho }_{0}}\left( \frac{5}{3}\pi {{r}^{3}}-\frac{\pi {{r}^{4}}}{R} \right)\] Now, electric field intensity at a point on this spherical surface is             \[E=\frac{1}{4\pi {{\varepsilon }_{0}}}\cdot \frac{q}{{{r}^{2}}}\]             \[=\frac{1}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}\times {{\rho }_{0}}\left( \frac{5}{3}\pi {{r}^{3}}-\frac{\pi {{r}^{4}}}{R} \right)\]             \[=\frac{{{\rho }_{0}}r}{4{{\varepsilon }_{0}}}\left( \frac{5}{3}-\frac{r}{R} \right)\]