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)

the answer would be c