Let P(r) = frac{Q}{pi R^4} r be the charge de | vedclass

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(C) According to Gauss's Law,the electric field $E$ at a distance $r_1$ from the center is given by $\oint E \cdot dA = \frac{q_{enclosed}}{\varepsilo

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$\frac{Q r_1^2}{4\pi \varepsilon_0 R^4}$

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(C) According to Gauss's Law,the electric field $E$ at a distance $r_1$ from the center is given by $\oint E \cdot dA = \frac{q_{enclosed}}{\varepsilon_0}$.For a spherical surface of radius $r_1$,this becomes $E(4\pi r_1^2) = \frac{1}{\varepsilon_0} \int_0^{r_1} \rho(r) (4\pi r^2) dr$.Substituting $\rho(r) = \frac{Q}{\pi R^4} r$,we get $E(4\pi r_1^2) = \frac{1}{\varepsilon_0} \int_0^{r_1} \left( \frac{Q}{\pi R^4} r \right) (4\pi r^2) dr$.$E(4\pi r_1^2) = \frac{4\pi Q}{\pi R^4 \varepsilon_0} \int_0^{r_1} r^3 dr$.$E(4\pi r_1^2) = \frac{4 Q}{R^4 \varepsilon_0} \left[ \frac{r^4}{4} \right]_0^{r_1} = \frac{Q r_1^4}{R^4 \varepsilon_0}$.Solving for $E$,we get $E = \frac{Q r_1^4}{4\pi \varepsilon_0 R^4 r_1^2} = \frac{Q r_1^2}{4\pi \varepsilon_0 R^4}$.

Let $P(r) = \frac{Q}{\pi R^4} r$ 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 the electric field is

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