A particle of mass m and charge q is released from rest in uniform electric field of intensity E. Calculate the kinetic energy it attains afect moving a distance x between the plates.

Answer

qEx

Explanation

As `F = qE`
`a = (F)/(m) = (qE)/(m)`
As `v^(2) - u^(2) = 2` as, and `u = 0 :. V^(2) = 2as`
`implies v^(2) = 2 ((qE)/(m))x xx = (2qEx)/(m)`
`K.E. = (1)/(2) mv^(2) = (1)/(2) m ((2q Ex)/(m)) = qEx`