The electric field intensity on the surface of a solid charged sphere of radius ' $r$ ' and volume charge…
The electric field intensity on the surface of a solid charged sphere of radius ' $r$ ' and volume charge density ' $\rho$ ' is ( $\varepsilon_0=$ permittivity of free space $)$
$\frac{\rho r}{3 \varepsilon_0}$
$\frac{\rho}{4 \pi \varepsilon_0 r}$
zero
$\frac{5 \rho \mathrm{r}}{6 \varepsilon_0}$
Solution
According to Gauss' theorem,
$\phi=\mathrm{E} \cdot \mathrm{A}=\frac{\mathrm{q}_{\mathrm{enc}}}{\varepsilon_0}$
But, $\mathrm{q}_{\text {enc }}=\rho \times$ volume
$\therefore \quad \mathrm{q}_{\mathrm{enc}}=\rho\left(\frac{4}{3} \pi \mathrm{r}^3\right)$
And area is $A=4 \pi r^2$
$\begin{aligned}
& \therefore \quad \mathrm{E}\left(4 \pi \mathrm{r}^2\right)=\frac{\rho\left(\frac{4}{3} \pi \mathrm{r}^3\right)}{\varepsilon_0} \\
& \therefore \quad \mathrm{E}=\frac{\rho \mathrm{r}}{3 \varepsilon_0}
\end{aligned}$
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