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(A) To find the electric field at a distance $r_1$ from the center,we use Gauss's Law: $\oint E \cdot dA = \frac{q_{enclosed}}{\epsilon_0}$.The charge

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$\frac{Qr_1^2}{4\pi \epsilon_0 R^4}$

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(A) To find the electric field at a distance $r_1$ from the center,we use Gauss's Law: $\oint E \cdot dA = \frac{q_{enclosed}}{\epsilon_0}$.The charge enclosed $q_{enclosed}$ in a sphere of radius $r_1$ is given by the integral of the charge density $\rho(r)$ over the volume:$q_{enclosed} = \int_0^{r_1} \rho(r) \cdot 4\pi r^2 dr$Substitute $\rho(r) = \frac{Q}{\pi R^4} r$:$q_{enclosed} = \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$$q_{enclosed} = \frac{4Q}{R^4} \left[ \frac{r^4}{4} \right]_0^{r_1} = \frac{Q r_1^4}{R^4}$.Now,applying Gauss's Law for a spherical surface of radius $r_1$:$E(4\pi r_1^2) = \frac{q_{enclosed}}{\epsilon_0} = \frac{Q r_1^4}{\epsilon_0 R^4}$.Solving for $E$:$E = \frac{Q r_1^4}{4\pi \epsilon_0 r_1^2 R^4} = \frac{Q r_1^2}{4\pi \epsilon_0 R^4}$.

The charge density of a solid sphere of radius $R$ and total charge $Q$ is given by $\rho(r) = \frac{Q}{\pi R^4} r$. The value of the electric field at a point $P$ inside the sphere at a distance $r_1$ from the center is .......

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