A positive charge particle of 100 mg is thrown in opposite direction to a uniform electric field of strength $1 \times 10^{5} \mathrm{NC}^{-1}$. If the charge on the particle is $40 \mu \mathrm{C}$ and the initial velocity is $200 \mathrm{~ms}^{-1}$, how much distance it will travel before coming to the rest momentarily :

JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Work, Power & Energy

A positive charge particle of 100 mg is thrown in opposite direction to a uniform electric field of strength

1 \times 1 0^{5} 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✓ Correct

Answer: (D)

Solution

The particle experiences a retarding force

F = q E

opposite to its motion, giving acceleration

a = \frac{q E}{m}

.

Using

v^{2} = u^{2} - 2 a s

with final velocity

v = 0

s = \frac{u ^{2}}{2 a} = \frac{u ^{2} m}{2 q E}

s = \frac{( 200 ) ^{2} \times 100 \times 1 0 ^{- 6}}{2 \times 40 \times 1 0 ^{- 6} \times 1 0 ^{5}}

s = \frac{40000 \times 1 0 ^{- 4}}{80 \times 1 0 ^{- 1}} = \frac{4}{8} = 0.5

JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Single correct option

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