A) \[\frac{E{{q}^{2}}M}{2{{t}^{2}}}\]
B) \[\frac{2{{E}^{2}}{{t}^{2}}}{mq}\]
C) \[\frac{{{E}^{2}}{{q}^{2}}{{t}^{2}}}{2m}\]
D) \[\frac{Eqt}{t}\]
Correct Answer: C
Solution :
When charge q is released in uniform electric field E then its acceleration \[a=\frac{qE}{m}\] (constant) So, its-motion will be uniformly accelerated motion and its velocity after t second is given by \[v=at=\frac{qE}{m}t\] The KE of charged particle \[KE=\frac{1}{2}m{{v}^{2}}=\frac{1}{2}m{{v}^{2}}=\frac{1}{2}m{{\left( \frac{qE}{m}t \right)}^{2}}\] \[=\frac{{{q}^{2}}{{E}^{2}}{{t}^{2}}}{2m}\]
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