A positive charge particle of 100 mg is thrown in opposite direction to a uniform electric field of strength 1 × 10 5 N C - 1 . If the charge on the particle is 40 μ C and the initial velocity is 200

Question

A positive charge particle of 100 mg is thrown in opposite direction to a uniform electric field of strength 1 × 10 5 N C - 1 . If the charge on the particle is 40 μ C and the initial velocity is 200 m s - 1 , how much distance it will travel before coming to the rest momentarily

TestHub · Accepted Answer

Force experienced by a particle with charge $q$ in an electric field $F = q E$ .

Now, the acceleration produced is given as $a = \frac{F}{m} = \frac{q E}{m}$ $= \frac{40 \times 10^{- 6} \times 10^{5}}{1 \times 10^{- 4}}$

$= 4 \times 10^{4} m s^{- 2}$ (As the particle is projected against the electric field, hence it is decelerated)

Using, $v^{2} = u^{2} - 2 a s$

We have, $0^{2} = u^{2} - 2 a s$

$s = \frac{v^{2}}{2 a} = \frac{{(200)}^{2}}{2 \times 4 \times 10^{4}} = 0.5 m$

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