The velocity of a particle (m = 10 mg, q = - 4.0 μC) at t = 0 is 20 m/s in the positive x direction. If the particle moves in a uniform electric field of 20 N/C in the positive x direction, what is the particle's speed at t = 5.0 s?

Vf = - 20 m/s (-ve sign shows that the particle is moving opposite to positive x - direction).

Explanation:

Given:

Vi = 20 m/s, m = 10 mg = 1 * 10⁻⁵ kg, q = - 4.0 * 10⁻⁶ C, E = 20 N/C. t = 5.0 s

first to find Electric Force

F = Eq = 20 * - 4.0 10⁻⁶ C = - 8 * 10⁻⁵ N (-ve sign shows that the field will push the particle opposite to positive x - direction)

We also have F=ma

⇒ a = F/m = - 8 * 10⁻⁵ N / 10 * 10⁻⁵ kg = - 8 m/s² (-ve sign shows that the particle is accelerated opposite to positive x - direction)

Now according the first equation of Motion.

Vf = Vi + at

Vf = 20 m/s + - 8 m/s² * 5 s

Vf = - 20 m/s (-ve sign shows that the particle is moving opposite to positive x - direction)