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JEE Main 2024 · Physics · Electrostatics · medium

A conducting spherical shell A of radius 10 cm is placed concentrically inside another thin conducting shell B of radius 20 cm, which carries a total…

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A conducting spherical shell AA of radius 10 cm10\ \mathrm{cm} is placed concentrically inside another thin conducting shell BB of radius 20 cm20\ \mathrm{cm}, which carries a total charge of +6 μC+6\ \mu\mathrm{C}. Shell AA is then connected to the ground. The potential of shell BB is (14πε0=9×109 N m2C−2\dfrac{1}{4\pi\varepsilon_0}=9\times10^9\ \mathrm{N\,m^2C^{-2}})
  1. A. 0.9×105 V0.9\times10^5\ \mathrm{V}
  2. B. 1.35×105 V1.35\times10^5\ \mathrm{V}
  3. C. 5.4×105 V5.4\times10^5\ \mathrm{V}
  4. D. 2.7×105 V2.7\times10^5\ \mathrm{V}
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Answer: (B) 1.35×105 V1.35\times10^5\ \mathrm{V}

Let the grounded inner shell acquire charge qq. Its potential is zero: k(q0.1+6 μC0.2)=0k\left(\dfrac{q}{0.1}+\dfrac{6\ \mu\mathrm{C}}{0.2}\right)=0, so q=−3 μCq=-3\ \mu\mathrm{C}. The potential of the outer shell is due to total enclosed charge: VB=k(q+6 μC)0.2=9×109×3×10−60.2=1.35×105 VV_B=\dfrac{k(q+6\ \mu\mathrm{C})}{0.2}=\dfrac{9\times10^9\times3\times10^{-6}}{0.2}=1.35\times10^5\ \mathrm{V}. Forgetting the induced charge gives 2.7×105 V2.7\times10^5\ \mathrm{V}.
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