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 :

Question

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

Tardigrade Admin · Accepted Answer

Distance travelled by particle before stopping ( V 2/2 a )= S ⇒ ( v 2 m /2 qE ) ⇒ ((200)2 × 100 × 10-6/2 × 40 × 10-6 × 105)=0.5 m

Q. A positive charge particle of 100mg is thrown in opposite direction to a uniform electric field of strength 1×105NC−1. If the charge on the particle is 40μC and the initial velocity is 200ms−1, how much distance it will travel before coming to the rest momentarily :

1 m

5 m

10 m

0.5 m

Distance travelled by particle before stopping 2aV2​=S⇒2qEv2m​ ⇒2×40×10−6×105(200)2×100×10−6​=0.5m

Distance travelled by particle before stopping
$\frac{V ^{2}}{2 a} = S \Rightarrow \frac{v ^{2} m}{2 qE}$
$\Rightarrow \frac{( 200 ) ^{2} \times 100 \times 1 0 ^{- 6}}{2 \times 40 \times 1 0 ^{- 6} \times 1 0 ^{5}} = 0.5 m$