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 need to find the acceleration of the particle using the electric field:

a = F/m = (qE)/m

Plugging in the values, we get:

a = (0.08*C * 102.5 N/C) / 0.004 kg = 2.05 x 10^3 m/s^2

Next, we can use the acceleration to find the velocity at t = 2.0 s:

v = v0 + at

v = 80 m/s + (2.05 x 10^3 m/s^2)(2.0 s) = 4,160 m/s

However, this is the velocity, not the speed. Since the initial velocity was only in the x-direction, the final velocity will also only have an x-component. Therefore, the speed is simply the magnitude of the x-component, which is:

|v| = |vx| = |v0x + axt| = |80 m/s + (2.05 x 10^3 m/s^2)(2.0 s)| = 4,160 m/s

Thus, the answer is (B) 20 m/s.

A Particle (Mass = 4