Q. A particle of mass m and charge q is placed at a rest in a uniform electric field $E$ and then released. The kinetic energy attained by the particle after moving a distance y is
A) $q\ Ey^{2}$
B) $q\ E^{2}y$
C) $q\ Ey$
D) $q^{2}Ey$
Explanation
(c) Force on the charged particle in electric field, $F = qE;$ acceleration of particle, $a = F/m$= $q\ E/m$; using the relation $v^{2} = u^{2} + 2a,$ we have $v^{2} = 0 + 2(qE/m$)y
Or $\frac{1}{2}mv^{2} = q\ E\ y;\ \ $so KE is $q\ Ey.$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.