$(a)\;\large\frac{\rho_{0} R}{9 \in_{0}}\qquad(b)\;\large\frac{\rho_{0} R}{12 \in_{0}}\qquad(c)\;\large\frac{\rho_{0} R}{3 \in_{0}}\qquad(d)\;\large\frac{\rho_{0} R}{6 \in_{0}}$

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Answer : (a) $\;\large\frac{\rho_{0} R}{9 \in_{0}}$

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 $\;r=R\;.$

Electric field is max . at $\;r=\large\frac{2R}{3}$

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

$=\large\frac{\rho_{0}}{\in_{0}}\;[\large\frac{2R}{9}-\large\frac{4}{9}\;\large\frac{R^2}{4R}]$

$=\large\frac{\rho_{0} R}{\in_{0}}\;[\large\frac{2}{9}-\large\frac{1}{9}]$

$E_{r}|_{max} = \large\frac{\rho_{0} R}{9 \in_{0}}\;.$

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