A particle (mass = 5.0 g, charge = 40 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_y = 50 m/s, v_x = v_z = 0, what is the speed of the particle at t = 2.0 s? A) 81 m/s B) 72 m/s C) 64 m/s D) 89 m/s E) 25 m/s

Question

Quizplus · Accepted Answer

First, we need to calculate the force acting on the particle using the electric field: 
$F = qE$
$F = (40 	imes 10^{-3}	ext{ C})(2.5	ext{ N/C})$
$F = 0.1	ext{ N}$

Next, we can use the formula for acceleration in the x-direction:

$a_x = \frac{F}{m} = \frac{0.1	ext{ N}}{5.0	ext{ g}} = 20	ext{ m/s}^2$

Since the acceleration is constant, we can use kinematic equations to find the velocity of the particle at t = 2.0 s:

$v_x = v_{0,x} + a_xt$
$v_x = (0	ext{ m/s}) + (20	ext{ m/s}^2)(2.0	ext{ s}) = 40	ext{ m/s}$

The velocity in the y and z directions remains constant, so the speed of the particle at t = 2.0 s is:

$V = \sqrt{v_x^2 + v_y^2 + v_z^2} = \sqrt{(40	ext{ m/s})^2 + (50	ext{ m/s})^2 + (0	ext{ m/s})^2} = 64	ext{ m/s}$

Therefore, the correct answer is choice C) 64 m/s.

A Particle (Mass = 5