Question

A particle (q = 3.0 mC, m = 20 g) has a speed of 20 m/s when it enters a region where the electric field has a constant magnitude of 80 N/C and a direction which is the same as the velocity of the particle. What is the speed of the particle 3.0 s after it enters this region?

Answer 1

Answer:

56 m/s

Explanation:

The electirc force applied on the particle by the field will be

F = q * E

F = 3*10^-3 * 80 = 0.24 N

This force will cause an acceleration:

F = m * a

a = F/m

a = 0.24 / 0.02 = 12 m/s^2

The equation for speed under constant acceleration is:

V(t) = V0 + a*t

V(3) = 20 + 12 * t = 56 m/s

The final speed will be  56 m/s.