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The electrostatic potential in a charged spherical region of radius r varies as V=ar³+b, where a and b are constants. The total charge inside a sphere of unit radius is α×π aε₀. The value of α is (the permittivity of vacuum is ε₀)

Asked in JEE Main 24th Jan 1st Shift 2026 · Relation between field and potential

Answer: (2) -12

Step-by-step solution

The field follows from the potential: Eᵣ=-(dV)/(dr)=-3ar².

Apply Gauss's law over the sphere of unit radius: Φ=Eᵣ(4π r²) evaluated at r=1, so Φ=(-3a)(4π)=-12π a.

q_enc=ε₀Φ=-12π aε₀.

Comparing with α×π aε₀ gives α=-12.

Why the other options are wrong

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