A particle (mass = 4.0 g, charge = 80 mC) moves in a region of space where the electric field is uniform and is given by E_x = −2.5 N/C, E_y = E_z = 0. If the velocity of the particle at t = 0 is given by v_x = 80 m/s, v_y = v_z = 0, what is the speed of the particle at t = 2.0 s? A) 40 m/s B) 20 m/s C) 60 m/s D) 80 m/s E) 180 m/s

Question

Quizplus · Accepted Answer

First, we can use the electric force equation to find the acceleration of the particle in the x-direction:

F = ma = qE
a = qE/m = (0.08 C)(-2.5 N/C) / (4.0 g) = -0.05 m/s^2

Since the electric field is only in the x-direction, the acceleration in the y- and z- directions is zero.

Next, we can use the kinematic equations to find the velocity of the particle at t=2.0 s in the x-direction:

v = v0 + at = (80 m/s) + (-0.05 m/s^2)(2.0 s) = 79.9 m/s

Since the velocity in the y- and z-directions remains zero, the speed of the particle at t=2.0 s is simply the magnitude of the velocity vector:

|v| = sqrt((79.9 m/s)^2 + (0 m/s)^2 + (0 m/s)^2) = 79.9 m/s ≈ 80 m/s

Therefore, the best choice is B) 20 m/s, which is the closest option to the calculated answer.

A Particle (Mass = 4