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 text2.0 T is applied at right angles to the electric field in a direction normal to both overset arrow textE and overset arrow textv . If g is text9.8 m s- 2 and the charge on the particle is 10- 6 C , then the velocity of the charged particle so that it continues to move horizontally is

Question

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 text2.0 T is applied at right angles to the electric field in a direction normal to both overset arrow textE and overset arrow textv . If g is text9.8 m s- 2 and the charge on the particle is 10- 6 C , then the velocity of the charged particle so that it continues to move horizontally is (A) 2ms- 1 (B) 20ms- 1 (C) 0.2ms- 1 (D) 100ms- 1

Tardigrade Admin · Accepted Answer

Net force on the particle should be zero.

Solution

qE = 1 0^{- 6} \times 200 = 2 \times 1 0^{- 4} N

mg = 2 \times 1 0^{- 5} \times 9.8 = 1.96 \times 1 0^{- 4} N

Since

qE > m g

, so magnetic force

q v B

should act downwards to balance the forces.

qE = mg + qvB \Rightarrow 2 \times 1 0^{- 4} = 1.96 \times 1 0^{- 4} + 1 0^{- 6} v \times 2

\Rightarrow v = 2 m / s

$2 m s^{- 1}$

$20 m s^{- 1}$

$0.2 m s^{- 1}$

$100 m s^{- 1}$

2ms−1

20ms−1

0.2ms−1

100ms−1

Net force on the particle should be zero. qE=10−6×200=2×10−4N mg=2×10−5×9.8=1.96×10−4N Since qE>mg , so magnetic force qvB should act downwards to balance the forces. qE=mg+qvB⇒2×10−4=1.96×10−4+10−6v×2 ⇒v=2m/s

$2 m s^{- 1}$

$20 m s^{- 1}$

$0.2 m s^{- 1}$

$100 m s^{- 1}$