Question

A particle of mass m and charge q is placed at rest in a uniform electric field E and then released. The K.E attained by the particle after moving a distance y is

• A. $qE{y}^2$
• B. $q{E}^{2}y$
• C. $qEy$
• D. $q^{2}Ey$

Answer 1

The correct option is C $qEy$

Force on the charged particle in uniform electric field F = m a = Eq or $a=\frac{Eq}{m}$
Now equation of motion
$V^2=u^2+2as$ or $v^2=0+2\left(\frac{Eq}{m}\right)y$
Or $v^2=\left(\frac{2Eqy}{m}\right)$
Kinetic energy $E=\frac{1}{2}m{v}^2=\frac{1}{2}m\times \left(\frac{2Eqy}{m}\right)=Eqy$
Or KE = work done = Fd = (qE)y