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Question

A proton is released at rest 10 cm from a charged sheet (perpendicular to it) having uniform electric field around it of intensity 123 N/C towards (perpendicular to) the sheet. It will strike the sheet after the time (approximately)
(Charge on proton is 1.6×1019 C and mass of proton is 1.6×1027 kg)

A
4 μs
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B
2 μs
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C
22 μs
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D
42 μs
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Solution

The correct option is A 4 μs
Given:
Intensity of electric field E=123 N/C;
Distance of the particle perpendicular to plate is s=10 cm=10×102 m

As we know that

F=qE

F=1.6×1019×123=1.968×1017 N

From Newton's law of motion

F=ma

a=Fm

a=1.968×10171.6×1027=1.23×1010 m/s2

Further, from equation of motion

s=ut+12at2

from question, u=0

10×102=0+12×1.23×1010×t2

t=2×10×1021.23×10104×106 s

t4 μs

Hence, option (a) is the correct answer.

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