Two parallel plates separated by a distance of 5 mm are kept at a potential difference of 50 V. A particle of mass 10-15 kg and charge 10-11 C enters in it with a velocity 107 m / s. The acceleration of the particle will be

Question

Two parallel plates separated by a distance of 5 mm are kept at a potential difference of 50 V. A particle of mass 10-15 kg and charge 10-11 C enters in it with a velocity 107 m / s. The acceleration of the particle will be (A) 108 m / s 2 (B) 5 × 105 m / s 2 (C) 105 m / s 2 (D) 2 × 103 m / s 2

Tardigrade Admin · Accepted Answer

a=(q E/m)=(q/m)((V/d)) =(10-11/10-15) × (50/5 × 10-3) =108 m / sec 2

Q. Two parallel plates separated by a distance of $5 mm$ are kept at a potential difference of $50 V$ . A particle of mass $1 0^{- 15} k g$ and charge $1 0^{- 11} C$ enters in it with a velocity $1 0^{7} m / s$ . The acceleration of the particle will be

$1 0^{8} m / s^{2}$

$5 \times 1 0^{5} m / s^{2}$

$1 0^{5} m / s^{2}$

$2 \times 1 0^{3} m / s^{2}$

$a = \frac{qE}{m} = \frac{q}{m} (\frac{V}{d})$
$= \frac{1 0 ^{- 11}}{1 0 ^{- 15}} \times \frac{50}{5 \times 1 0 ^{- 3}}$
$= 1 0^{8} m / sec^{2}$

Q. Two parallel plates separated by a distance of 5mm are kept at a potential difference of 50V. A particle of mass 10−15kg and charge 10−11C enters in it with a velocity 107m/s. The acceleration of the particle will be

108m/s2

5×105m/s2

105m/s2

2×103m/s2