Question

The velocity of a particle (m = 10 mg, q = –4.0 μC) at t = 0 is 30 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 (in m/s) at t = 5.0 s?

Answer 1

The force acting on the charged particle moving in the uniform electric field is

F =m × a = q E ...................(1)

Given m = 10mg = 10 × 10^-6 kg = 10^-5 kg

q = -4 microcolomb = -4 × 10^-6 Colomb

E = 20 N/C

Substituting the values in equation (1)

(10^-5) a =-4 × 10^-6 × 20

Implies acceleration, a = -8

Since acceleration is the rate of change of velocity

Or dV/dt = -8

Integrating both sides,

V = -8t + Constant........(2)

At t=0, V = 30 m/s.....given

Therefore from equation (2)

30 = 0 + Constant

Or

Constant = 30

Substituting in equation (2)

V = -8t + 30................(3)

Therefore at time t = 5s , The velocity of the particle can be obtained from equation (3)

V = -8×5 + 30

Velocity, V = -10 m/s

= ANSWER