Electrostatics Question 457

Question: A charged particle of mass $ m $ and charge $ q $ is released from rest in a uniform electric field $ E. $ Neglecting the effect of gravity, the kinetic energy of the charged particle after t second is [KCET 2003]

Options:

A) $ \frac{Eq^{2}m}{2t^{2}} $

B) $ \frac{2E^{2}t^{2}}{mq} $

C) $ \frac{E^{2}q^{2}t^{2}}{2m} $

D) $ \frac{Eqm}{t} $

Show Answer

Answer:

Correct Answer: C

Solution:

When charge q is released in uniform electric field $ E $ then its acceleration $ a=\frac{qE}{m} $ (is constant) So its motion will be uniformly accelerated motion and its velocity after time t is given by $ v=at $

$ =\frac{qE}{m}t $

therefore KE $ =\frac{1}{2}mv^{2}=\frac{1}{2}{{( \frac{qE}{m}t )}^{2}}=\frac{q^{2}E^{2}t^{2}}{2m} $



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