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} $
BETA
