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JEE Main 2026 (24 Jan, Shift 1) · Electric Potential & Equipotentials

Electric Potential & Equipotentials question from JEE Main 2026 (24 Jan, Shift 1)

Q23JEE Main 2026 · 24 JAN · SHIFT 1Difficult

Electric Potential & Equipotentials

The electrostatic potential in a charged spherical region of radius rr varies as V=ar3+bV = ar^3 + b, where aa and bb are constants. The total charge in the sphere of unit radius is α×πaϵ0\alpha \times \pi a\epsilon_0. The value of α\alpha is ______. (permittivity of vacuum is ϵ0\epsilon_0)

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Answer: (a)

  1. E=−dVdr=−3ar2E = -\dfrac{dV}{dr} = -3ar^2
  2. Gauss's law: qin=ε0E(4πr2)=−12πaε0r4q_{in} = \varepsilon_0 E(4\pi r^2) = -12\pi a\varepsilon_0 r^4
  3. For r=1r = 1: qin=−12πaε0q_{in} = -12\pi a\varepsilon_0, so α=−12\alpha = -12.
  4. So the answer is (a) −12-12.
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