$(a)\;r_{m}=\large\frac{R}{3}\qquad(b)\;r_{m}=\large\frac{3R}{2}\qquad(c)\;r_{m}=\large\frac{2R}{3}\qquad(d)\;r_{m}=\large\frac{4R}{3}$

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Answer : $\;r_{m}=\large\frac{2R}{3}$

Explanation :

Outside the ball electric field = $\;\large\frac{\rho_{0} R^3}{12 R^2 \in_{0}}$

It will be maximum when $r=R$

Then ,

$E_{r}=\large\frac{\rho R^3}{12 R^2 \in_{0}}=\large\frac{\rho_{0} R}{12 \in_{0}}$

Inside the ball $\;E_{r}=\large\frac{\rho_{0}}{\in_{0}}\;[\large\frac{r}{3}-\large\frac{r^2}{4R}]$

For $\;E_{r}\;$ to be max

$\large\frac{d E_{r}}{dr}=0$

$\large\frac{\rho_{0}}{\in_{0}}\;[\large\frac{1}{3}-\large\frac{2r}{4R}]=0$

$r=\large\frac{2R}{3}$

$\large\frac{d^2 E_{r}}{dr^2}=-\large\frac{\rho_{0}}{2 \in_{0} R} < 0$

Hence at $\;r=\large\frac{2R}{3}\;E_{r}\;$ is maximum and is greater than at $\;r=R\;.$

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