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JEE Advanced 2016 Paper 2 · Physics · Electrostatics · medium

A thick conducting spherical shell has inner radius 10 cm and outer radius 20 cm and carries a net charge of +3 nC. A point charge of +2 nC is held…

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A thick conducting spherical shell has inner radius 10 cm10\ \mathrm{cm} and outer radius 20 cm20\ \mathrm{cm} and carries a net charge of +3 nC+3\ \mathrm{nC}. A point charge of +2 nC+2\ \mathrm{nC} is held at its centre. Use 1/(4πε0)=9×109 N m2 C−21/(4\pi\varepsilon_0)=9\times10^{9}\ \mathrm{N\,m^2\,C^{-2}}. Choose the correct statement(s).
  1. A. The charge on the inner surface of the shell is −2 nC-2\ \mathrm{nC}.
  2. B. The charge on the outer surface of the shell is +3 nC+3\ \mathrm{nC}.
  3. C. The potential of the shell is 225 V225\ \mathrm{V}.
  4. D. If the shell is now earthed, the charge on its outer surface vanishes and the potential at a point 5 cm5\ \mathrm{cm} from the centre is 90 V90\ \mathrm{V}.
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Answer: A, C

Gauss's law inside the conductor requires zero field, so the inner surface carries −q=−2 nC-q=-2\ \mathrm{nC}; A is true. Conservation of charge then puts 3−(−2)=+5 nC3-(-2)=+5\ \mathrm{nC} on the outer surface, not +3 nC+3\ \mathrm{nC}; B is false. The outside field is that of a total charge +5 nC+5\ \mathrm{nC} at the centre, so the shell potential is 9×109×5×10−9/0.20=225 V9\times10^9\times5\times10^{-9}/0.20=225\ \mathrm{V}; C is true. On earthing, the potential of the shell becomes zero, which removes the outer charge, leaving only −2 nC-2\ \mathrm{nC} on the inner surface. At r=5 cmr=5\ \mathrm{cm} the potential is kq(1r−1a)=9×2×(20−10)=180 Vkq\left(\dfrac1r-\dfrac1a\right)=9\times2\times(20-10)=180\ \mathrm{V}, not 90 V90\ \mathrm{V}; D is false.
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