A solid ball of radius R has a charge density | vedclass

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Question

$	ext { Charge density, } ho=ho_{0}\left(1-\frac{r}{R}ight)$

$d q=ho d v$

$q_{i n}=\int d q=ho d v$

$=ho_{0}\left(1-\frac{r}{R}\

Vedclass · Accepted Answer

$\frac{{{\rho _0}{R^3}}}{{{12\varepsilon _0}{r^2}}}$

Vedclass · Answer

$	ext { Charge density, } ho=ho_{0}\left(1-\frac{r}{R}ight)$

$d q=ho d v$

$q_{i n}=\int d q=ho d v$

$=ho_{0}\left(1-\frac{r}{R}ight) 4 \pi r^{2} d r \quad\left(\because d v=4 \pi \mathrm{r}^{2} \mathrm{dr}ight)$

$=4 \pi p_{0} \int_{0}^{R}\left(1-\frac{r}{R}ight) r^{2} d r$

$=4 \pi ho_{0} \int_{0}^{R} r^{2} d r-\frac{r^{2}}{R} d r$

$=4 \pi ho_{0}\left[\left[\frac{r^{3}}{3}ight]_{0}^{R}-\left[\frac{r^{4}}{4 R}ight]_{0}^{R}ight]$

${=4 \pi ho_{0}\left[\frac{R^{3}}{3}-\frac{R^{4}}{4 R}ight]}$

${=4 \pi ho_{0}\left[\frac{R^{3}}{3}-\frac{R^{3}}{4}ight]=4 \pi ho_{0}\left[\frac{R^{3}}{12}ight]}$

${q=\frac{\pi ho_{0} R^{3}}{3}}$

$E .4 \pi r^{2}=\left(\frac{\pi ho_{0} R^{3}}{3 \epsilon_{0}}ight)$

Electric field outside the ball, $E=\frac{ho_{0} R^{3}}{12 \epsilon_{0} r^{2}}$

A solid ball of radius $R$ has a charge density $\rho $ given by $\rho = {\rho _0}\left( {1 - \frac{r}{R}} \right)$ for $0 \leq r \leq R$. The electric field outside the ball is

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