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An electron travelling from infinity with velocity ‘v’ into an electric field due to two stationary electrons separated by a distance of 2m. If it comes to rest when it reaches the mid point of the line joining the stationary electrons. The initial velocity ‘ v ‘ of the electron is

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a

32√2m/s

b

16√2m/s

c

32m/s

d

16m/s

answer is B.

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Detailed Solution

$P E_{i} + K E_{i} = P E_{f} + K E_{f} h e r e, P E_{i} = 0, K E_{f} = 0 \Rightarrow P E_{f} = K E_{i} \frac{1}{2} {mv}^{2} = \frac{1}{4 {πε}_{0}} \frac{q_{1} q_{2}}{r} \times 2 [r = 1 m, m i d p o i n t o f e l e c t r o n s] \Rightarrow v = 32 m / s$

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