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A particle of mass 2 × 10 - 5 kg moves horizontally between the plates of a parallel plate capacitor which produce an electric field of 200 N C - 1 in the vertically upward direction. A magnetic induction of 2.0 T is applied at right angles to the electric field in a direction normal to both E → and v → . If g is 9.8 m s - 2 and the charge on the particle is 10 - 6 C , then the velocity of the charged particle so tha

Options

  1. A2   m   s - 1
  2. B20   m   s - 1
  3. C0.2   m   s - 1
  4. D100   m   s - 1

Correct answer

A. 2   m   s - 1

Step-by-step solution

Net force on the particle should be zero. qE = 1 0 - 6 × 2 0 0 = 2 × 1 0 - 4 N mg = 2 × 1 0 - 5 × 9.8 = 1.96 × 1 0 - 4 N Since qE > mg , so magnetic force qvB should act downwards to balance the forces. qE = mg + qvB ⇒ 2 × 1 0 - 4 = 1.96 × 1 0 - 4 + 1 0 - 6 v × 2 ⇒    v = 2 m/s

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