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 $$\alpha \times \pi a\in_{\circ}$$. the value of $$\alpha$$ is_____. (permittivity of vacuum is $$\in_{\circ}$$)

Potential in a spherical region is given by $$V = ar^3 + b$$. We seek the total charge enclosed in a sphere of unit radius.

Since the electric field is related to the potential by $$E = -\frac{dV}{dr}$$, it follows that $$E = -3ar^2$$. The negative sign indicates the field points inward when $$a>0$$.

Using Gauss’s law for a sphere of radius $$r$$, we have $$\oint \vec{E}\cdot d\vec{A} = \frac{Q_{\text{enc}}}{\varepsilon_0}$$, which simplifies to $$E\cdot 4\pi r^2 = \frac{Q}{\varepsilon_0}$$. At $$r = 1$$, the field becomes $$E = -3a$$ (with magnitude $$|E| = 3|a|$$, directed radially). Substituting these values gives $$(-3a)\,(4\pi)\,(1)^2 = \frac{Q}{\varepsilon_0}$$, and hence $$Q = -12\pi a\varepsilon_0$$.

Comparing this result with the form $$Q = \alpha \pi a \varepsilon_0$$ shows that $$\alpha = -12$$.

The correct answer is Option (3): $$-12$$.