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(B) According to Gauss's Law, the electric flux through a spherical surface of radius $R$ is given by $\oint \vec{E} \cdot d\vec{A} = \frac{q_{enclose

Vedclass · Accepted Answer

$\frac{k R^{2}}{4 \varepsilon_{0}}$

Vedclass · Answer

(B) According to Gauss's Law, the electric flux through a spherical surface of radius $R$ is given by $\oint \vec{E} \cdot d\vec{A} = \frac{q_{enclosed}}{\varepsilon_{0}}$.For a sphere of radius $R$, the electric field $E$ at the surface is uniform, so $E(4 \pi R^{2}) = \frac{q_{enclosed}}{\varepsilon_{0}}$.The total charge $q_{enclosed}$ is calculated by integrating the volume charge density $\rho = k r$ over the volume of the sphere:$q_{enclosed} = \int_{0}^{R} \rho (4 \pi r^{2}) dr = \int_{0}^{R} (k r) (4 \pi r^{2}) dr = 4 \pi k \int_{0}^{R} r^{3} dr = 4 \pi k \left[ \frac{r^{4}}{4} \right]_{0}^{R} = \pi k R^{4}$.Substituting this into Gauss's Law:$E(4 \pi R^{2}) = \frac{\pi k R^{4}}{\varepsilon_{0}}$.Solving for $E$:$E = \frac{\pi k R^{4}}{4 \pi R^{2} \varepsilon_{0}} = \frac{k R^{2}}{4 \varepsilon_{0}}$.

$A$ sphere of radius $R$ has a volume charge density $\rho = k r$, where $r$ is the distance from the center of the sphere and $k$ is a constant. The magnitude of the electric field at the surface of the sphere is given by ($\varepsilon_{0} =$ permittivity of free space):

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