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# A charged particle of charge 2 micro c and mass 10 mg moving with a velocity 1000m/s enters a uniform electric field of strength 10^3 N/C directed perpendicular to its direction of motion.Find the velocity and displacement, of the particle after 10 s.

Grade:12

## 2 Answers

Khimraj
3007 Points
3 years ago
horizontal velocity will be constant
vx = 1000m/s
vertical velocity
vy = a*t = (qE/m)*t = 0.2*10 = 2m/s
ao net velocity will be $\sqrt{1000^{2} + 2^{2}} \simeq 1000m/s$
and displacement will be 1000*10 = 10000 m
Hope it clears.
Khimraj
3007 Points
3 years ago
verticla displacement is neglected as it is negligible in compare with horizontal displacement.

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