A charge particle of charge 2μC and mass 10 miligram moving with a velocity of 1000 ms^–1enters a uniform electric field of strength 10² NC^–1 directed perpendicular to its direction of motion. Find the velocity and acceleration of the particle after 10 s.

Solution

Given,
Charge on the particle, q= 2 μC
Mass of the particle, m= 10 mg
Velocity with which the particle is moving, v= 1000 ms^–1Strength of uniform electric field, E= 102 NC^–1

Therfore,
Acceleration experienced by the particle after 10 sec,

$a_{y} = \frac{qE}{m}$
$= \frac{2 \times 10^{- 6} \times 10^{2}}{10 \times 10^{- 6}} = 20 {ms}^{- 2}$

And, velocity attained by the particle after 10 sec, perpendicular to the direction of motion is,
v_y = u_y + a_yt
= 0 + 20 x 10
= 200 ms^–1Hence, there is no force acting along the direction of motion.
∴ velocity along the direction of motion remains same i.e., v_n = 1000 ms^–1∴ Net velocity after 10 s
$v = \sqrt{v_{x}^{2} + v_{y}^{2}} = \sqrt{(1000)^{2} + (200)^{2}} = 100 \sqrt{104} {ms}^{- 1} .$

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A charge particle of charge 2μC and mass 10 miligram moving with a velocity of 1000 ms^–1enters a uniform electric field of strength 10² NC^–1 directed perpendicular to its direction of motion. Find the velocity and acceleration of the particle after 10 s.

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A charge particle of charge 2μC and mass 10 miligram moving with a velocity of 1000 ms–1 enters a uniform electric field of strength 102 NC–1 directed perpendicular to its direction of motion. Find the velocity and acceleration of the particle after 10 s.

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A charge particle of charge 2μC and mass 10 miligram moving with a velocity of 1000 ms^–1enters a uniform electric field of strength 10² NC^–1 directed perpendicular to its direction of motion. Find the velocity and acceleration of the particle after 10 s.