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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. What will be the speed of the particle after travelling this distance?
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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
Using the third equation of motion, we get
\[v^2 = u^2 + 2as\]
\[ \Rightarrow v^2 = 0 + 2 \times 3 \times {10}^3 \times 4 \times {10}^{- 1} \]
\[ \Rightarrow v = 4 . 9 \times 10 = 49 \text{ m/s }\]
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