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Question

Let us consider a spherical shell of radius $x$ and thickness $dx$

Charge on this shell

$d q=\rho \cdot 4 \pi x^{2} d x=\rho_{0}\left(\frac{5}{4

Vedclass · Accepted Answer

$\frac{{{\rho _o}r}}{{4{\varepsilon _0}}}\;\left( {\frac{5}{3} - \frac{r}{R}} \right)$

Vedclass · Answer

Let us consider a spherical shell of radius $x$ and thickness $dx$

Charge on this shell

$d q=ho \cdot 4 \pi x^{2} d x=ho_{0}\left(\frac{5}{4}-\frac{x}{R}ight) .4 \pi x^{2} d x$

$	herefore$ Total charge in the spherical region from centre to $r$ $(r < R)$ is

$q=\int d q=4 \pi ho_{0} \int_{0}^{r}\left(\frac{5}{4}-\frac{x}{R}ight) x^{2} d x$

$=4 \pi ho_{0}\left[\frac{5}{4} \cdot \frac{r^{3}}{3}-\frac{1}{R} \cdot \frac{r^{4}}{4}ight]=\pi ho_{0} r^{3}\left(\frac{5}{3}-\frac{r}{R}ight)$

Electric field at $r, E=\frac{1}{4 \pi \epsilon_{0}} \cdot \frac{q}{r^{2}}$

$=\frac{1}{4 \pi \epsilon_{0}} \cdot \frac{\pi ho_{0} r^{3}}{r^{2}}\left(\frac{5}{3}-\frac{r}{R}ight)=\frac{ho_{0} r}{4 \epsilon_{0}}\left(\frac{5}{3}-\frac{r}{R}ight)$

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