A particle of mass $$2$$×$$10$$−$$5$$ kg moving horizontally between two horizontal plates of a charged parallel plate capacitor between which there is an electric field of $$200$$ NC acting upward. A magnetic induction of $$2.0$$ T is applied at right angles to the electric field in a direction normal to both B¯ and V¯. If g is $$9.8$$ $$m/sec2$$ and the charge on the particle is $$10$$−$$6C$$ , then find the velocity of charge particle, so that it continues to move horizontally.

Question

A particle of mass $$2$$×$$10$$−$$5$$ kg moving horizontally between two horizontal plates of a charged parallel plate capacitor between which there is an electric field of $$200$$ NC acting upward. A magnetic induction of $$2.0$$ T is applied at right angles to the electric field in a direction normal to both B¯ and  V¯. If g is $$9.8$$ $$m/sec2$$ and the charge on the particle is $$10$$−$$6C$$ , then find the velocity of charge particle, so that it continues to move horizontally.

Infinity Learn · Accepted Answer

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) ⇒$$v=0.04$$×$$10-410-6$$×$$2.0=2$$ $$m/s$$ .

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