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(D) The electric field $E$ is related to the potential $V$ by the relation $E = -\frac{dV}{dr}$.Given $V = ar^2 + b$,we have $E = -\frac{d}{dr}(ar^2 +

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$-6 a \varepsilon_0$

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(D) The electric field $E$ is related to the potential $V$ by the relation $E = -\frac{dV}{dr}$.Given $V = ar^2 + b$,we have $E = -\frac{d}{dr}(ar^2 + b) = -2ar$.According to Gauss's law in differential form,the volume charge density $ho$ is related to the electric field by $
abla \cdot \vec{E} = \frac{ho}{\varepsilon_0}$.In spherical coordinates,for a radial field $E(r)$,the divergence is given by $\frac{1}{r^2} \frac{d}{dr}(r^2 E) = \frac{ho}{\varepsilon_0}$.Substituting $E = -2ar$ into the equation:$\frac{ho}{\varepsilon_0} = \frac{1}{r^2} \frac{d}{dr}(r^2 \cdot (-2ar)) = \frac{1}{r^2} \frac{d}{dr}(-2ar^3) = \frac{1}{r^2} (-6ar^2) = -6a$.Therefore,$ho = -6a\varepsilon_0$.

The electrostatic potential inside a charged spherical ball is given by $V = ar^2 + b$,where $r$ is the distance from its centre and $a$ and $b$ are constants. The volume charge density of the ball is [$\varepsilon_0$ = permittivity of free space].

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