A positive charge particle of 100 mg is thrown in opposite direction to a uniform electric field of strength 1 × 105 NC ⁻¹. If the charge on the particle is 40 μ C and the initial velocity is 200 ms ⁻¹, how much distance it will travel before coming to the rest momentarily: (a) 1 m (b) 5 m (c) 10 m (d) 0.5 m

Answer: (d). Retardation a = qE/m = 40 × 10⁻⁶ × 10⁵100 × 10⁻⁶ = 4 × 10⁴ m/s² Distance to stop: s = u²/2a = (200)²2 × 4 × 10⁴ s = 0.5 m So the answer is (d) 0.5 m.

A positive charge particle of 100 mg is thrown in opposite direction… | JEE Main 2022 (29 Jun, Shift 1)

Q20JEE Main 2022 · 29 JUN · SHIFT 1Medium

Motion of Charges in an Electric Field

A positive charge particle of 100 mg is thrown in opposite direction to a uniform electric field of strength 1 $\times$ 105 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:

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Answer: (d)

Retardation $a = \frac{qE}{m} = \frac{40 \times 10^{-6} \times 10^{5}}{100 \times 10^{-6}} = 4 \times 10^{4}\text{ m/s}^2$
Distance to stop: $s = \frac{u^2}{2a} = \frac{(200)^2}{2 \times 4 \times 10^{4}}$
$s = 0.5\text{ m}$
So the answer is (d) 0.5 m.

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