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

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(C) To find the electric field at a distance $r_1$ from the center,we use Gauss's Law: $\oint E \cdot dA = \frac{q_{enc}}{\varepsilon_0}$.For a spheri

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

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(C) To find the electric field at a distance $r_1$ from the center,we use Gauss's Law: $\oint E \cdot dA = \frac{q_{enc}}{\varepsilon_0}$.For a spherical Gaussian surface of radius $r_1$,$E(4\pi r_1^2) = \frac{q_{enc}}{\varepsilon_0}$.The charge enclosed $q_{enc}$ is the integral of the charge density $\rho(r)$ over the volume of the sphere of radius $r_1$:$q_{enc} = \int_0^{r_1} \rho(r) (4\pi r^2) dr = \int_0^{r_1} \left( \frac{Q}{\pi R^4} r \right) (4\pi r^2) dr = \frac{4Q}{R^4} \int_0^{r_1} r^3 dr = \frac{4Q}{R^4} \left[ \frac{r^4}{4} \right]_0^{r_1} = \frac{Q r_1^4}{R^4}$.Substituting this into Gauss's Law:$E(4\pi r_1^2) = \frac{Q r_1^4}{\varepsilon_0 R^4} \implies E = \frac{Q r_1^4}{4\pi \varepsilon_0 R^4 r_1^2} = \frac{Q r_1^2}{4\pi \varepsilon_0 R^4}$.

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

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