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(B) The electrostatic potential is given by $\Phi = a r^2 + b$.Using the relation between electric field $E$ and potential $\Phi$,we have $E = -\frac{

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$-6 a \varepsilon_0$

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(B) The electrostatic potential is given by $\Phi = a r^2 + b$.Using the relation between electric field $E$ and potential $\Phi$,we have $E = -\frac{d\Phi}{dr}$.$E = -\frac{d}{dr}(a r^2 + b) = -2ar$.According to Gauss's Law in differential form,the charge density $ho$ is related to the electric field by $
abla \cdot E = \frac{ho}{\varepsilon_0}$.In spherical coordinates,for a radial field $E(r)$,the divergence is $\frac{1}{r^2} \frac{d}{dr}(r^2 E) = \frac{ho}{\varepsilon_0}$.Substituting $E = -2ar$:$\frac{1}{r^2} \frac{d}{dr}(r^2 (-2ar)) = \frac{ho}{\varepsilon_0}$.$\frac{1}{r^2} \frac{d}{dr}(-2a r^3) = \frac{ho}{\varepsilon_0}$.$\frac{1}{r^2} (-6a r^2) = \frac{ho}{\varepsilon_0}$.$-6a = \frac{ho}{\varepsilon_0}$.Therefore,$ho = -6a \varepsilon_0$.

The electrostatic potential inside a charged spherical ball is given by $\Phi = a r^2 + b$,where $r$ is the distance from the centre and $a, b$ are constants. Then,the charge density inside the ball is ($\varepsilon_0 =$ permittivity in free space).

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