JEE Main 2022 · 29 Jun · Shift 1 · Physics Q48

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PhysicsSingle correct+4 / -1Question 48 of 90

A positive charge particle of 100 mg is thrown in opposite direction to a uniform electric field of strength $1\times10^5$ NC $^{-1}$ . If the charge on the particle is 40 $\mu$ C and the initial velocity is 200 ms $^{-1}$ , how much distance it will travel before coming to the rest momentarily :

1 m

5 m

10 m

0.5 m

Answer and worked solution

Official answer: D

Deceleration. $a=\frac{qE}m=\frac{40\times10^{-6}\times10^5}{10^{-4}}=4\times10^4$ m s $^{-2}$ .

Distance. $\frac{v^2}{2a}=\frac{4\times10^4}{8\times10^4}=0.5$ m.

(D)

Topics:Electric field Motion in a straight line

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JEE Main 2022 · 29 Jun · Shift 1, Physics question 48