Finding Displacement: Solve for Particle's Position

[AmandaJoy]

I am trying to figure out the displacement from this question:

A particle of charge 12x10^-6 C and mass 3.8x10^-5 kg is released from rest in a region where there is a constant electric field of 480 N/C. What is the displacement of the particle after a time of 1.6x10^-2 s.

I have tried several different ways to approach this but for whatever reason I just can't get it. Please help!

 

[kuruman]

Hi AmandaJoy, welcome to PF. Please use the template for posting homework questions. You need to show us what you tried first before we can help you.

 

[kerimek]

I understand your frustration with this problem. Finding displacement in this situation requires using the equations of motion and the principles of electromagnetism. Here is the step-by-step solution to help you solve for the particle's position:

1. First, we need to identify the initial conditions of the particle. In this case, the particle is released from rest, which means its initial velocity is 0 m/s.

2. Next, we can use the equation of motion, x = x0 + v0t + 1/2at^2, to find the final position of the particle after 1.6x10^-2 s. Since the initial position (x0) is not given, we can assume it to be 0. Therefore, the equation becomes x = 0 + 0 + 1/2at^2.

3. Now, we need to find the acceleration (a) of the particle. We can use the equation F = ma, where F is the force and m is the mass of the particle. In this case, the force acting on the particle is the electric force, given by F = qE, where q is the charge of the particle and E is the electric field. So, we have a = F/m = qE/m = (12x10^-6 C)(480 N/C)/(3.8x10^-5 kg) = 1.5 m/s^2.

4. Substituting the value of acceleration in the equation from step 2, we get x = 1/2(1.5 m/s^2)(1.6x10^-2 s)^2 = 2.4x10^-4 m.

Therefore, the displacement of the particle after a time of 1.6x10^-2 s is 2.4x10^-4 m. I hope this helps you understand how to approach and solve this type of problem. Keep practicing and you will become more comfortable with these concepts.