Gauss LawMedium
Question
The electrostatic potential in a charged spherical region of radius $r$ varies as $V = ar^{3} + b$, where $a$ and $b$ are constants. The total charge in the sphere of unit radius is $\alpha \times \pi a \in_{0}$. The value of $\alpha$ is $\_\_\_\_$ .
(permittivity of vacuum is $\in_{0}$ )
Options
A.-12
B.-6
C.-9
D.
-8
Solution
(4) -8
Sol. $v = {ar}^{3} + b$
$$\begin{matrix} & E = - \frac{dv}{dr} = - 3{ar}^{2} \\ & \phi_{\text{closed~}} = \frac{q_{enc}}{\varepsilon_{0}} \\ & q_{enc} = \varepsilon_{0} \cdot E \cdot \text{ }A \\ & \ = \varepsilon_{0}\left( - 3a \cdot (1)^{2} \right)4\pi(1)^{2} \\ & \ = - 12\pi a\varepsilon_{0} \\ & \ \therefore x = - 12 \end{matrix}$$
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