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Question

Let us consider a spherical shell of radius $x$ and thickness $dx.$

Charge on this shell

$\mathrm{dq}=\rho 4 \pi \mathrm{x}^{2} \mathrm{dx}=\rho

Vedclass · Accepted Answer

$\frac{{{\rho _0}}}{{{\varepsilon _0}}}\left( {\frac{r}{3} - \frac{{{r^2}}}{{4R}}} \right)$

Vedclass · Answer

Let us consider a spherical shell of radius $x$ and thickness $dx.$

Charge on this shell

$\mathrm{dq}=ho 4 \pi \mathrm{x}^{2} \mathrm{dx}=ho_{0}\left(1-\frac{\mathrm{x}}{\mathrm{R}}ight) .4 \pi \mathrm{x}^{2} \mathrm{dx}$

$	herefore$ Total charge in the spherical region from centre to $r(r < R)$

$q=\int d q=4 \pi ho_{0} \int_{0}^{T}\left(1-\frac{x}{R}ight) x^{2} d x$

$=4 \pi ho_{0}\left[\frac{x^{3}}{3}-\frac{x^{4}}{4 R}ight]_{0}^{t}$

$=4 \pi ho_{0}\left[\frac{r^{3}}{3}-\frac{r^{4}}{4 R}ight]$

$=4 \pi ho_{0} r^{3}\left[\frac{1}{3}-\frac{r}{4 R}ight]$

$	herefore$ Electric field atr, $\mathrm{E}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{\mathrm{q}}{\mathrm{r}^{2}}$

$ = \frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{{4\pi {ho _0}{r^3}}}{{{r^2}}} \cdot \left[ {\frac{1}{3} - \frac{r}{{4R}}} ight]$

$=\frac{ho_{0}}{\varepsilon_{0}}\left[\frac{	au}{3}-\frac{	au^{2}}{4 R}ight]$

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