Question

A particle of mass m and charge q is placed at rest in a uniform electric field E and then released. The kinetic energy attained by the particle after moving a distance y is(a) qEy²(b) qE²y(c) qEy(d) q²Ey

Answer 1

c option should be the correct answer

Answer 2

Answer:

C) qEy

Explanation:

Mass of particle = m ( Given)

Charge of particle = q ( Given)

Uniform electric field = E ( Given)

Kinetic Energy = 1/2mv²

To determine the speed - v²=u²+2as

where u = 0, and s = y

= v² = 2ay  (a=f/m)  (F=Eq)

Therefore, a=Eq/m

Substituting in the equation

v²= 2(Eq/m)y

KE = 1/2 x m x 2(Eqy/m)  ( KE = electric field times the charge times the distance it travels)

KE = m x (Eqy/m)

KE = Eqy

Thus, The kinetic energy attained by the particle after moving a distance y is qEy