Question

A space is filled up with volume charge density $\rho ={\rho }_0{e}^{-\alpha r^3}$ where ${\rho }_0$ and $\alpha$ are positive constants, r is the distance from the centre of this system. The magnitude of electric field strength as a function of r is given as $\frac{{\rho }_0}{x{\epsilon }_0\alpha r^2}(1-e^{-\alpha r^3})$. Find $x$ :

Answer 1

Consider a spherical gaussian surface of radius $r$ as shown in the figure, with centre located at origin.

By Gauss' Law, the net flux through the surface is $\oint \overrightarrow{E}.\overrightarrow{dS}=\frac{q_{\text{encl}}}{{ϵ}_0}$

The total amount of charge enclosed in the gaussian surface is $q_{\text{encl}}=\underset{0}{\overset{r}{\int }}\rho \left(4\pi r^2\right)dr=\underset{0}{\overset{r}{\int }}4\pi r^2{\rho }_0{e}^{-\alpha r^3}dr=4\pi {\rho }_0[\frac{-e^{-\alpha r^3}}{3\alpha }{]}_0^r=4\pi {\rho }_0\frac{1-e^{-\alpha r^3}}{3\alpha }$

The total flux is $\oint \overrightarrow{E}.\overrightarrow{ds}=E\left(4\pi r^2\right)$

Thus, by Gauss' Law, $E=\frac{{\rho }_0}{3\alpha {ϵ}_0{r}^2}(1-e^{-\alpha r^3})$