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$R$ ત્રિજ્યા ધરાવતા ઘન ગોળની વિજભાર ઘનતા $0 \leq r \leq R$ માટે $\rho = {\rho _0}\left( {1 - \frac{r}{R}} \right)$ મુજબ આપવામાં આવે છે. તો બોલની બહાર વિદ્યુતક્ષેત્ર કેટલું હશે?
$\frac{{{\rho _0}{R^3}}}{{{\varepsilon _0}{r^2}}}$
$\frac{{{4\rho _0}{R^3}}}{{{3\varepsilon _0}{r^2}}}$
$\frac{{{3\rho _0}{R^3}}}{{{4\varepsilon _0}{r^2}}}$
$\frac{{{\rho _0}{R^3}}}{{{12\varepsilon _0}{r^2}}}$
Solution
$\text { Charge density, } \rho=\rho_{0}\left(1-\frac{r}{R}\right)$
$d q=\rho d v$
$q_{i n}=\int d q=\rho d v$
$=\rho_{0}\left(1-\frac{r}{R}\right) 4 \pi r^{2} d r \quad\left(\because d v=4 \pi \mathrm{r}^{2} \mathrm{dr}\right)$
$=4 \pi p_{0} \int_{0}^{R}\left(1-\frac{r}{R}\right) r^{2} d r$
$=4 \pi \rho_{0} \int_{0}^{R} r^{2} d r-\frac{r^{2}}{R} d r$
$=4 \pi \rho_{0}\left[\left[\frac{r^{3}}{3}\right]_{0}^{R}-\left[\frac{r^{4}}{4 R}\right]_{0}^{R}\right]$
${=4 \pi \rho_{0}\left[\frac{R^{3}}{3}-\frac{R^{4}}{4 R}\right]}$
${=4 \pi \rho_{0}\left[\frac{R^{3}}{3}-\frac{R^{3}}{4}\right]=4 \pi \rho_{0}\left[\frac{R^{3}}{12}\right]}$
${q=\frac{\pi \rho_{0} R^{3}}{3}}$
$E .4 \pi r^{2}=\left(\frac{\pi \rho_{0} R^{3}}{3 \epsilon_{0}}\right)$
Electric field outside the ball, $E=\frac{\rho_{0} R^{3}}{12 \epsilon_{0} r^{2}}$
