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(B) The electrostatic potential $V$ inside a charged sphere is given by $V = A r^2 + B$.The electric field $E$ is related to the potential $V$ by the

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$-6 A \varepsilon_0$

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(B) The electrostatic potential $V$ inside a charged sphere is given by $V = A r^2 + B$.The electric field $E$ is related to the potential $V$ by the relation $E = -\frac{dV}{dr}$.Substituting the given expression for $V$:$E = -\frac{d}{dr}(A r^2 + B) = -2 A r$.According to Gauss's Law in differential form,the volume charge density $ho$ is related to the electric field $E$ by the equation $
abla \cdot E = \frac{ho}{\varepsilon_0}$.For a spherically symmetric distribution,this becomes $\frac{1}{r^2} \frac{d}{dr}(r^2 E) = \frac{ho}{\varepsilon_0}$.Substituting $E = -2 A r$:$\frac{ho}{\varepsilon_0} = \frac{1}{r^2} \frac{d}{dr}(r^2 \cdot (-2 A r)) = \frac{1}{r^2} \frac{d}{dr}(-2 A r^3) = \frac{1}{r^2} (-6 A r^2) = -6 A$.Therefore,the charge density is $ho = -6 A \varepsilon_0$.

The electrostatic potential inside a charged sphere is given as $V = A r^2 + B$,where $r$ is the distance from the centre of the sphere,$A$ and $B$ are constants. Then,the charge density in the sphere is

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