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