Question

A particle of mass m carrying charge q is released from rest in uniform electric field of intensity E. The kinetic energy acquired by the particle after moving a distance of x is (neglect gravitational force)

• A. $q\, E\, x$
• B. $ q \, E \, x^2 $
• C. $ q \, E^2 \, x $
• D. $ q^2 \, E \, x $
• E. $ q^2 \, E^2 \, x $

Answer 1

Answer: (A)

Velocity acquired by particle in electric field
$ v^2 = u^2 + 2as $
or $ v^2 = 0 + 2a x $
$ v^2 = 2ax $
Hence, kinetic energy of particle
KE = $ \frac{ 1}{2} mv^2 $
= $ \frac{ 1}{2} m ( 2ax) $
= amx
but F = q E
or ma = qE
or a = $ \frac{ qE }{ m} $
Hence, KE = $ \frac{ qE }{ m} \times mx = q \, E \, x $