JEE Main 2026 — Jan 24, Shift 1 — Physics solution

JEE Main2026Jan 24, Shift 1Physics

The electrostatic potential in a charged spherical region of radius r varies as $V = a r^{3} + b$ , where $a$ and $b$ are constants. The total charge in the sphere of unit radius is $α \times π a ϵ_{0}$ . The value of $α$ is _____.

(permittivity of vacuum is $ϵ_{0}$ )

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

Solution

Radial electric field from the potential: $E = - \frac{d V}{d r} = - 3 a r^{2}$ .

At $r = 1$ : $E = - 3 a$ , directed radially.

By Gauss's law, the flux through a sphere of radius 1:

$Φ = E \cdot (4 π r^{2}) = - 3 a \cdot 4 π = - 12 π a$ .

And $Φ = \frac{q _{enc}}{ϵ _{0}}$ :

$q_{enc} = - 12 π a ϵ_{0} = α π a ϵ_{0} \Rightarrow α = - 12$ .

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