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JEE Advanced2011PhysicsElectrostaticsActual

A spherical metal shell A of radius R_A and a solid metal sphere B of radius R_B < (R_A ) are kept far apart and each is given charge +Q . Now, they are connected by a thin metal wire. Then

Options

  1. AE_A^ inside =0
  2. BQ_A>Q_B
  3. C_ A _B = R_B R_A
  4. DE_A^ on surface < E_B^ on surface

Correct answer

A. E_A^ inside =0

Step-by-step solution

Inside a conducting shell, electric field is always zero. Therefore, option (a) is correct. When the two are connected, their potentials become the same. array ll & V_A=V_B or & Q_A R_A = Q_B R_B ( V= 1 4 ₀ Q R ) array Since, R_A>R_B Q_A>Q_B Option (b) is correct. Potential is also equal to, array rlrl V & = R ₀ V_A & =V_B _A R_A & = _B R_B or _B & _A _B & = R_B R_A array Option (c) is correct. Electric field on surface, E= E₀ or E or _A < _B Since, _A < _B E_A < E_B Option (d) is also correct. Correct options are

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