How Far Does a Charged Particle Travel in an Electric Field?

[floridianfisher]

Homework Statement

A 1-gram particle with a charge of 1 milliC starts from rest in a uniform electric field of magnitude E = 10 V/m. How
far (in m) does the particle travel in 2 seconds?
I know the answere is 10 but I don't know how to find it

Homework Equations

PE/q=-Ex

The Attempt at a Solution

I converted all units but I have PE and change in x left and I am stumped can anyone help

I have PE/.oo1= -10* x

I know the answere is 10m already but I can't figure out how

 

[Dick]

Don't use PE. Use force F=E*q. Then use F=ma to find the constant acceleration. Now use kinematics to find the distance travelled. I don't think the answer is 10m.

 

[Moetasim]

to get there.

Thank you for your question. To find the distance traveled by the particle, we need to use the equation for electric potential energy:

PE = q*ΔV

Where PE is the electric potential energy, q is the charge of the particle, and ΔV is the change in potential energy.

In this case, we have a 1-gram particle with a charge of 1 milliC, so we can substitute these values into the equation:

PE = (0.001 C)*(10 V/m) = 0.01 J

Now, we can use the fact that the particle starts from rest and assume that all of its initial potential energy is converted into kinetic energy:

KE = 0.5*m*v^2

Where KE is the kinetic energy, m is the mass of the particle, and v is its velocity.

We know that the particle travels for 2 seconds, so we can use the equation for average velocity:

v = Δx/Δt

Substituting the values we know, we get:

0.01 J = 0.5*(0.001 kg)*[(Δx/2 s)^2]

Solving for Δx, we get:

Δx = 2*sqrt(0.01/0.001) = 2*sqrt(10) = 2*3.162 = 6.324 m

Therefore, the particle travels 6.324 meters in 2 seconds in a uniform electric field with a magnitude of 10 V/m. I hope this helps!