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Shown in the figure are two point charges + Q and - Q inside the cavity of a spherical shell. The charges are kept near the surface of the cavity on opposite sides of the centre of the shell. If σ 1 is the surface charge on the inner surface and Q 1 net charge on it and σ 2 the surface charge on the outer surface and Q 2 net charge on it then:

Options

  1. Aσ 1 = 0 , Q 1 = 0 , σ 2 = 0 , Q 2 = 0
  2. Bσ 1 ≠ 0 , Q 1 = 0 , σ 2 ≠ 0 , Q 2 = 0
  3. Cσ 1 ≠ 0 , Q 1 ≠ 0 , σ 2 ≠ 0 , Q 2 ≠ 0
  4. Dσ 1 ≠ 0 , Q 1 = 0 , σ 2 = 0 , Q 2 = 0

Correct answer

D. σ 1 ≠ 0 , Q 1 = 0 , σ 2 = 0 , Q 2 = 0

Step-by-step solution

By the property of electrostatic shielding in the conductors E = 0 in the conductor. So, electric flux, = 0 through a dotted Gaussian surface as shown The net enclosed charge through the Gaussian surface = 0 , ⇒ The net charge Q 1 on the inner surface = 0 , but the equal and opposite induced charge on the surface will be distributed non uniformly on the inner surface So, σ 1 = 0 ∵ Q 1 = 0 on the inner surface So, net charge Q 2 = 0 on the outer surface as conductor is neutral but ∵ outer surf

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