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Question: Answered & Verified by Expert
A solid spherical conductor of radius $\mathrm{R}$ has a spherical cavity of radius $a(a < R)$ at its centre. A charge $+Q$ is kept at the centre. The charge at the inner surface, outer surface and at a position $r(a < r < R)$ are respectively
PhysicsNuclear PhysicsVITEEEVITEEE 2008
Options:
  • A $+Q,-Q, 0$
  • B $-Q,+Q, 0$
  • C $0,-\mathrm{Q}, 0$
  • D $+Q, 0,0$
Solution:
1743 Upvotes Verified Answer
The correct answer is: $-Q,+Q, 0$
A charge $Q$ will be induced on the inner surface of the solid spherical conductor. An equal but opposite charge will be induced on the outer surface of the conductor. There will be no charge at a position between the inner and outer surface.

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