Electrostatics

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



1) $\frac{{E{q^2}m}}{{2{t^2}}}$
2) $\frac{{2{E^2}{t^2}}}{{mq}}$
3) $\frac{{{E^2}{q^2}{t^2}}}{{2m}}$
4) $\frac{{Eqm}}{t}$
Solution: Explanation: Explanation
electric-field-potential

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