# The magnitude of electric filed as a function of the distance r inside the sphere is given by

$(a)\;\large\frac{\rho_{0}}{\in_{0}}\;[\large\frac{r}{3}-\large\frac{r^2}{4R}]\qquad(b)\;\large\frac{\rho_{0}}{\in_{0}}\;[\large\frac{r}{4}-\large\frac{r^2}{3R}]\qquad(c)\;\large\frac{\rho_{0}}{\in_{0}}\;[\large\frac{r}{3}+\large\frac{r^2}{4R}]\qquad(d)\;\large\frac{\rho_{0}}{\in_{0}}\;[\large\frac{r}{4}+\large\frac{r^2}{3R}]$

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• 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 $\;\rho=\rho_{0}\;(1-\large\frac{r}{R})\;.$Where $\rho_{0}\;$ is a constant .
Answer : (a) $\;\large\frac{\rho_{0}}{\in_{0}}\;[\large\frac{r}{3}-\large\frac{r^2}{4R}]$
Explanation :
Due to symmetry , if electric field exists then it will be radial
Let the electric field inside the sphere at a distance r from the centre is $\;E_{r}$
Then flux through a shell of radius r is :
By Gauss 's law
$E_{r}\;.4 \pi r^2=\phi$
$\phi=E_{r} 4 \pi r^2$
$\phi=\large\frac{q_{inc}}{\in_{0}}$
$E_{r} 4 \pi r^2=\large\frac{q_{inc}}{\in_{0}}---(1)$
$q_{inc}\;$= charge enclosed in the shell of radius r
$q_{inc}= \int \; \rho dV$
Consider a shell of radius r and thichness dr
$q_{inc}= \int _{0}^{r}\;\rho_{0}\;(1-\large\frac{r}{\rho})\;4 \pi r^2 dr$
$q_{inc}=4 \pi \rho_{0}\; \int_{0}^{r}\;(r^2-\large\frac{r^3}{R}) dr$
$=4 \pi \rho_{0} \;(\large\frac{r^3}{3}-\large\frac{r^4}{4R})$
Putting the value of $\;q_{inc}\;$ in equation (1) we get
$E_{r} 4 \pi r^2=\large\frac{4 \pi \rho_{0}}{\in_{0}}\;(\large\frac{r^3}{3}-\large\frac{r^4}{4R})$
$E_{r}=\large\frac{\rho_{0}}{\in_{0}}\;(\large\frac{r}{3}-\large\frac{r^2}{4R})$
answered Feb 18, 2014 by