Question

A ball of radius R carries a positive charge whose volume density depends only on a separation r from the ball’s centre as$\rho ={{\rho }_{0}}\left( 1-\dfrac{r}{R} \right)$, where ${{\rho }_{0}}$ is a constant. Assuming the permittivities of the ball and the environment to be equal to unity, find:
(A) The magnitude of the electric field strength as a function of the distance r both inside and outside the ball;
(B) The maximum intensity ${{E}_{\max }}$ and the corresponding distance${{r}_{m}}$.

Answer 1

You could find the charge enclosed within volume under consideration and then you could apply Gauss’s law and then perform the integration. Follow the same steps for both specified regions. Then you could find the maximum electric field intensity and also the corresponding distance ${{r}_{m}}$.
Formula used:
Gauss’s Law,
$E.dA=\dfrac{{{q}_{en}}}{{{\varepsilon }_{0}}}$

Complete answer:
In the question we are given a ball of radius R with volume density, $\rho ={{\rho }_{0}}\left( 1-\dfrac{r}{R} \right)$
We are asked to assume that the permittivity of ball and environment to be unity and asked to find the following:
(A) The magnitude of the electric field strength as the function of r both inside and outside the ball:
For $r\langle R$ , enclosed charge would be,
${{q}_{en}}=\int\limits_{0}^{r}{4\pi {{r}^{2}}}dr\rho =\int\limits_{0}^{r}{4\pi {{r}^{2}}{{\rho }_{0}}\left( 1-\dfrac{r}{R} \right)}dr$ ………………………………………… (1)
Now we have Gauss’s theorem given by,
${{E}_{r}}4\pi {{r}^{2}}=\dfrac{{{q}_{en}}}{{{\varepsilon }_{0}}}=\dfrac{{{\rho }_{0}}}{{{\varepsilon }_{0}}}\int\limits_{0}^{r}{4\pi {{r}^{2}}\left( 1-\dfrac{r}{R} \right)dr}$
$\therefore {{E}_{r}}=\dfrac{{{\rho }_{0}}}{3{{\varepsilon }_{0}}}\left[ {{r}^{2}}-\dfrac{3{{r}^{2}}}{4R} \right]$
Now for $r\rangle R$ ,
${{q}_{en}}=\int\limits_{0}^{R}{4\pi {{r}^{2}}}dr{{\rho }_{0}}\left( 1-\dfrac{r}{R} \right)$
$\Rightarrow {{E}_{r}}=\dfrac{{{\rho }_{0}}}{{{r}^{2}}{{\varepsilon }_{0}}}\left[ \dfrac{{{R}^{3}}}{3}-\dfrac{{{R}^{4}}}{4R} \right]$
$\therefore {{E}_{r}}=\dfrac{{{\rho }_{0}}{{R}^{3}}}{12{{r}^{2}}{{\varepsilon }_{0}}}$
(B) We see that the electric field magnitude decreases for increasing value of r for $r\rangle R$ and it would maximum for$r\langle R$ . We could now say that,
$\dfrac{d}{dr}\left( r-\dfrac{3{{r}^{2}}}{4R} \right)=0$
$\Rightarrow 1-\dfrac{3r}{2R}=0$
$\Rightarrow r={{r}_{m}}=\dfrac{2R}{3}$
$\therefore {{E}_{\max }}=\dfrac{{{\rho }_{0}}R}{9{{\varepsilon }_{0}}}$
Therefore, we found the maximum electric field intensity to be ${{E}_{\max }}=\dfrac{{{\rho }_{0}}R}{9{{\varepsilon }_{0}}}$and the corresponding distance to be${{r}_{m}}=\dfrac{2R}{3}$.

Note:
Gauss’s law actually gives us the net electric flux through the area under consideration. The statement of the Gauss’s law goes like this: the net electric flux through a certain area would be $\dfrac{1}{{{\varepsilon }_{0}}}$ times the charge enclosed. We also have the expression that gives electric flux as the dot product of electric field intensity and area. We have used these two formulas to find the solution.