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Question
A particle of mass 1 g and charge 2.5 × 10−4 C is released from rest in an electric field of 1.2 × 10 4 N C−1. How long will it take for the particle to travel a distance of 40 cm?
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Solution
Given:
Charge of the particle, q = 2.5 × 10−4 C
Initial velocity, u = 0
Electric field intensity, E = 1.2 × 104 N/C
Mass of the particle, m = 1 g = 10−3 kg
Distance travelled, s = 40 cm = 4 × 10−1 m
Acceleration of the particle,
\[a = \frac{F_e}{m} = \frac{3}{{10}^{- 3}} = 3 \times {10}^3 \text{ m/ s}^2\]
Let t be the time taken by the particle to cover the distance s = 40 cm. Then,
\[s = \frac{1}{2}a t^2 \]
\[ \Rightarrow t = \sqrt{\frac{2s}{a}} = 1 . 63 \times {10}^{- 2} s\]
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