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JEE Main 2011 (AIEEE) · Electric Potential & Equipotentials

Electric Potential & Equipotentials question from JEE Main 2011 (AIEEE)

Q118JEE Main 2011 · AIEEEMedium

Electric Potential & Equipotentials

The electrostatic potential inside a charged spherical ball is given by ϕ=ar2+b\phi = ar^2 + b where rr is the distance from the center and a,ba, b are constants. Then the charge density inside the ball is:

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Answer: (c)

  1. E=−dϕdr=−2arE = -\frac{d\phi}{dr} = -2ar (radial).
  2. Gauss's law: E⋅4πr2=qinε0⇒qin=−8πε0ar3E\cdot 4\pi r^2 = \frac{q_{in}}{\varepsilon_0} \Rightarrow q_{in} = -8\pi\varepsilon_0 a r^3.
  3. ρ=dqin4πr2 dr=−24πε0ar24πr2=−6aε0\rho = \frac{dq_{in}}{4\pi r^2\,dr} = \frac{-24\pi\varepsilon_0 a r^2}{4\pi r^2} = -6a\varepsilon_0
  4. So the answer is (c) −6aε0-6a\varepsilon_0.
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