JEE Main · 2026 · 24 Jan 2026 · Shift 1 · Electrostatics

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}

)

Answer: (A)

Solution

The radial field is

E = - \frac{d V}{d r} = - 3 a r^{2}

. Applying Gauss's law to a sphere of radius

r = 1

, the enclosed charge is

q = ε_{0} E (4 π r^{2}) = ε_{0} (- 3 a \cdot 1^{2}) (4 π \cdot 1^{2}) = - 12 π a ε_{0}

. Comparing with

α π a ε_{0}

gives

α = - 12

JEE Main · 2026 · 24 Jan 2026 · Shift 1 · Single correct option