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 $a \times \pi \varepsilon_{0}$. The value of $a$ is ___.
(Permittivity of vacuum is $\varepsilon_{0}$)

• A. $-8$
• B. $-12$
• C. $-9$
• D. $-6$

Hint:

Electric field can always be obtained by taking the negative gradient of electrostatic potential.

Answer 1

The Correct Option is B

Step 1: Finding electric field from potential.
Electric field is related to potential by:
\[ E = -\dfrac{dV}{dr} \] Given,
\[ V = ar^{3} + b \] \[ E = -\dfrac{d}{dr}(ar^{3} + b) = -3ar^{2} \] Step 2: Using Gauss’s law.
For a spherical surface of radius $r$:
\[ \oint \vec{E}\cdot d\vec{A} = \dfrac{Q_{\text{enc}}}{\varepsilon_0} \] \[ E \cdot 4\pi r^{2} = \dfrac{Q(r)}{\varepsilon_0} \] Step 3: Substituting value of $E$.
\[ (-3ar^{2})(4\pi r^{2}) = \dfrac{Q(r)}{\varepsilon_0} \] \[ Q(r) = -12\pi a \varepsilon_0 r^{4} \] Step 4: Charge enclosed in unit sphere.
For $r = 1$:
\[ Q = -12\pi a \varepsilon_0 \] Given, total charge $= a \times \pi \varepsilon_0$.
Step 5: Comparing both expressions.
\[ -12a = a \Rightarrow a = -12 \] Step 6: Final conclusion.
The value of $a$ is $-12$, corresponding to option (2).