-qEx for a vertical spring in a constant Electric field

[Albertgauss]

A massless spring of spring constant k = 13 Newtons per meter hangs purely vertically. A 20 Newton, 5(10-3) Coulombs charged particle is attached to the spring and released from rest. Besides all this, there is also a constant, external Electric Field of 9000 Newtons per Coulomb pointing upwards. See figure below. After the release point, and assuming the particle only travels downwards and vertical, what is the speed of the particle as it passes through the equilibrium point? See the attached figure called "Figure_for_Problem.jpg"

Homework Equations

Conservation of energy, except that now we have a -qEx mixed in with the kinetic energy, mgy, and the spring potential energy.

The Attempt at a Solution

I am only having some trouble knowing what the sign on -qEy should be, and/or which side of the conservation of energy this term goes on.

A few thoughts: since the charge is +Q, it has lower (more negative) potential energy at the Release Point. Thus, on the equilibrium side of the equation, I would put y = 0. On the “release” point of the equation, I would write in –qEy, where q is +, E is +, and y is +. Is this correct? I can't get the latex on this website to work on my computer, so I uploaded a jpeg of where I am so I can keep all my symbols and subscripts.

 

[haruspex]

I would write in –qEy, where q is +, E is +, and y is +. Is this correct?

Looks right to me.
But are you encountering a problem when you try to solve it numerically? If so, consider what the electric force is on the particle.

 

[Albertgauss]

Ok, sounds good to me. I can take it from there. I've already worked through the numbers a few times and this was the only part I was unsure about. Thank you for your help.

 

[haruspex]

Ok, sounds good to me. I can take it from there. I've already worked through the numbers a few times and this was the only part I was unsure about. Thank you for your help.

And you didn't notice anything strange? What answer did you get? What was the value of y?

 

[Albertgauss]

The "y" was 1.9 meters. You can find that from the equilibrium position of the particle. You do Newton's Laws, ∑F = ma. That gives +QE -mg +ky = 0

So then y = ( mg - QE ) / k = (20-45)/ 13 = -25/13.

There is also a problem in that, if you keep going with the numbers in the energy conservation of the slide, the velocity turns out to be impossible because you end up, after some algebra, with something like: negative number = v^2. So, the numbers make the problem unable to be solved like this. I realize the parameters for m, Q, E, and k have to be adjusted so that the numbers make sense.

That's okay, though. I did get the help I needed, and that was to check where to place -QEy on the correct side of the conservation equation. This was the main issue I was having. I can fix the numbers later. I certainly appreciate this site's help on this.

 

[haruspex]

the parameters for m, Q, E, and k have to be adjusted so that the numbers make sense.

Yes, the electric force is greater then the gravitational force, so it will move up.