Questions
A particle of mass moving horizontally between two horizontal plates of a charged parallel plate capacitor between which there is an electric field of acting upward. A magnetic induction of 2.0 T is applied at right angles to the electric field in a direction normal to both and . If g is and the charge on the particle is , then find the velocity of charge particle, so that it continues to move horizontally.
detailed solution
Correct option is A
Net force on the particle should be zero. ∴ Eq=200×10−6N=2×10−4N mg=2×10−5×9.8 =1.96×10−4NSince Eq>mg, force qvB should act downward to balance the forces. ∴ Eq=mg+qvB ⇒2×10−4 = 1.96×10−4 + (10-6×v×2.0) ⇒v=0.04×10-410-6×2.0=2 m/s .Talk to our academic expert!
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A potential difference of 600 V is applied across the plates of a parallel plate capacitor. The separation between the plates is 3 mm. An electron projected vertically, parallel to the plates, with a velocity of moves undeflected between the plates. What is the magnitude of the magnetic field between the capacitor plates?
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