Electrostatics

JEE Main level, single correct, medium

91 of 311

Question 91 of 311

A positive charge particle of 100mg100mg is thrown in opposite direction to a uniform electric field of strength 1×105NC−11\times {10}^{5}N{C}^{-1}. If the charge on the particle is 40μC40\mu C and the initial velocity is 200ms−1200{ms}^{-1}, how much distance it will travel before coming to the rest momentarily

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Answer: (A) 0.5m0.5m

Force experienced by a particle with charge qq in an electric field F=qEF=qE.
Now, the acceleration produced is given as a=Fm=qEma=\dfrac{F}{m}=\dfrac{qE}{m}=40×10−6×1051×10−4=\dfrac{40\times {10}^{-6}\times {10}^{5}}{1\times {10}^{-4}}
=4×104ms−2=4\times {10}^{4}m{s}^{-2} (As the particle is projected against the electric field, hence it is decelerated)
Using, v2=u2−2as{v}^{2}={u}^{2}-2as
We have, 02=u2−2as{0}^{2}={u}^{2}-2as
s=v22a=(200)22×4×104=0.5ms=\dfrac{{v}^{2}}{2a}=\dfrac{{\left(200\right)}^{2}}{2\times 4\times {10}^{4}}=0.5m

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