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

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Question

$\mathrm{dq}=\frac{\mathrm{Qr}}{\pi \mathrm{R}^{4}} 	imes 4 \pi \mathrm{r}^{2} \cdot \mathrm{dr}$

${m{q}} = \frac{{4{m{Q}}}}{{{{m{R}}^4}}}\i

Vedclass · Accepted Answer

$\frac{{Qr_{_1}^2}}{{4\pi {\varepsilon _0}{R^4}}}\,$

Vedclass · Answer

$\mathrm{dq}=\frac{\mathrm{Qr}}{\pi \mathrm{R}^{4}} 	imes 4 \pi \mathrm{r}^{2} \cdot \mathrm{dr}$

${m{q}} = \frac{{4{m{Q}}}}{{{{m{R}}^4}}}\int_0^{{{m{r}}_1}} {{{m{r}}^3}} \quad {m{dr}} = \frac{{{m{Qr}}_1^4}}{{{{m{R}}^4}}}$

Now the electric field at a distance $\mathrm{r}_{1}$ from the centre (inside)

$\mathrm{E}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}}{\mathrm{r}_{1}^{2}}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{Qr}_{1}^{2}}{\mathrm{R}^{4}}$

Let $\rho (r)\, = \frac{Q}{{\pi {R^4}}}\,r$ be the volume 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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