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 Ex = 2.5 N/C, Ey = Ez = 0.If the velocity of the particle at t = 0 is given by vy = 50 m/s, vx = vz = 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 find the acceleration of the particle due to the electric field. We can use the equation F = qE, where F is the force on the particle, q is its charge, and E is the electric field.

F = qE = (40 × 10^−3 C)(2.5 N/C) = 0.1 N

The force on the particle is in the y-direction (perpendicular to the electric field), so we can use the equation F = ma in the y-direction to find the acceleration.

Fy = may

0.1 N = (5.0 × 10^−3 kg)ay

ay = 20 m/s^2

Since there is no acceleration in the x or z directions (the electric field is zero in those directions), the velocity components in those directions will remain zero.

vy = ayt = (20 m/s^2)(2.0 s) = 40 m/s

Using the Pythagorean theorem, we can find the speed of the particle at t = 2.0 s:

v = sqrt(vx^2 + vy^2 + vz^2) = sqrt(0 + (40 m/s)^2 + 0) = 64 m/s

A Particle (Mass = 5