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The magnitude of the electric field as a function of the distance r outside the ball is given by

$(a)\;\large\frac{\rho_{0} R^3}{8 \in_{0} r^2}\qquad(b)\;\large\frac{\rho_{0} R^3}{12 \in_{0} r^2}\qquad(c)\;\large\frac{\rho_{0} R^3}{8 \in_{0} r}\qquad(d)\;\large\frac{\rho_{0} R^3}{12 \in_{0} r}$

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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 : (b) $\;\large\frac{\rho_{0} R^3}{12 \in_{0} r^2}$
Explanation :
Electric field outside the ball at a distance x from centre will be equal to the sum of electric fields due to shell having radii vary from 0 to R
$E_{net}=\int_{R}\; \large\frac{k\;dq}{r^2}$
$=\int_{0}^{R}\;\large\frac{k}{x^2} \; \rho_{0} \;(1-\large\frac{r}{R})\;4 \pi r^2\;dr$
$=\large\frac{4 \pi k \rho_{0}}{x^2} \;\int_{0}^{R} r^2\;(1-\large\frac{r}{R})\;dr$
$=\large\frac{\rho_{0}}{\in_{0} x^2} \;[\large\frac{r^3}{3}-\large\frac{r^4}{4R}]_{0}^{R}$
$=\large\frac{\rho_{0} R^3}{\in_{0} x^2} \;[\large\frac{1}{3}-\large\frac{1}{4} ]$
$E_{net}=\large\frac{\rho_{0} R^3}{12 \in_{0} x^2}$
Electric field at a distance x from centre and outside the ball = $\;\large\frac{\rho_{0} R^3}{12 \in_{0} x^2}$
answered Feb 18, 2014 by yamini.v

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