Let rho (r) =frac{Q}{{pi {R^4}}}r be the char | vedclass

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Question

Let us consider a spherical shell of thickness $d x$ andradius $x$.

The volume of this spherical shell $=4 \pi r^{2} d r$.

The charge enclosed w

Vedclass · Accepted Answer

$\;\frac{{Q{r_1}^2}}{{3\pi {\varepsilon _0}{R^4}}}$

Vedclass · Answer

Let us consider a spherical shell of thickness $d x$ andradius $x$.

The volume of this spherical shell $=4 \pi r^{2} d r$.

The charge enclosed within shell

$=\frac{Q r}{\pi R^{4}}\left[4 \pi r^{2} d right]$

The charge enclosed in a sphere of radius $r_{1}$ is

$=\frac{4 Q}{R^{4}} \int_{0}^{r_{1}} r^{3} d r=\frac{4 Q}{R^{4}}\left[\frac{r^{4}}{4}ight]_{0}^{r}=\frac{Q}{R^{4}} r_{1}^{4}$

$	herefore $ The electric field at point $p$ inside the sphere at a distance $r_{1}$ from the centre of the sphere is

$E=\frac{1}{4 \pi \epsilon_{0}}-\frac{\left[\frac{Q}{R^{4}} r_{1}^{4}ight]}{r_{1}^{2}}=\frac{1}{4 \pi \epsilon_{0}} \frac{Q}{R^{4}} r^{2}$

Let $\rho (r) =\frac{Q}{{\pi {R^4}}}r$ be the charge density distribution for a solid sphere of radius $R$ and total charge $Q$. For a point '$p$' inside the sphere at distance $r_1$ from the centre of the sphere, the magnitude of electric field is

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