How Does Changing Voltage Affect Particle Descent Time?

[stod]

Homework Statement

Two parallel plates are positioned horizontally and when the potential te?difference between the plates is 400 V, a charged particle is suspended freely 10 cm above the lower plate. If the potential difference is suddenly changed to 200 V, how long will it take the particle to reach the lower pla

Homework Equations

-dV/dr=E
U=qV
V=kq/r

The Attempt at a Solution

If the difference between Va and Vb=400 V,

1. 400=kq/r
q=4.44 x 10^-9

2. U=qV, V=200
U=8.89 x 10^-7

3. U=KE



Since there's no mass, this is where I got stuck. Honestly, I've looked at this problem several different ways, and I just really don't understand electric potential well enough to know if what I'm doing even makes sense. Any feedback would be appreciated.

 

[kuruman]

First off, V = kq/r is the potential for a point charge. This is not what you have here. Here the electric field between the parallel plates is constant, so the potential difference over distance d is given by V = E d, where E is the (constant) electric field between the plates.

Can you find an expression for the mass in terms of the other parameters if you know that the charged particle is suspended freely when V = 400 Volts?

 

[stod]

Okay, so if I understand what you're saying, mg=Fcoulomb, and we should be able to get Fcoulomb from E(q)? How do we get q if V=kq/r is for a point charge? Basically, I don't remember finding V for parallel plates. Am I on the right track yet, though?

 

[kuruman]

You don't really need q. All you need to understand is that, for a given potential difference between the plates, the electric field is uniform and proportional to that potential difference. Now, when the potential difference is 400 V, the electric force is equal to the gravitational force. What do you think the electric force becomes when the potential difference is reduced to 200 V?

 

[stod]

Ohh okay, I think I got it. Thanks for spelling it out for me.