# Uniform elecric fields- force and kinetic energy

1. Nov 27, 2009

### chanderjeet

1. The problem statement, all variables and given/known data
A particle of mass 3x 10^-12kg carrying a charge of +4.0x 10^-14C is released at the upper plate of a pair of horizontal parallel conducting plates placed in a vacuum.

2. Relevant equations
(1) Calculate the net force on the particle: I added the electric force VQ/d and the weight W=mg
(2) Calculate the KE of the particle when it reaches the lower plate: work done =gain in KE= Fd
(3) Explain the effect on force and the KE if the separation of the plates is increased. (this i'm not sure about)

3. The attempt at a solution

I'm thinking that since F= VQ/d if the distance increases then the force would decrease. I could be wrong.

Since, the gain in KE is equivalent to the work done =Fd and since F decreases (I assume) then the KE would also decrease. Am I wrong?

2. Nov 27, 2009

### Fightfish

Yup, you are right. Assuming that the potential difference between the plates remains the same, the larger the plate separation, the lower the electric field strength (and electric force consequently)
You forgot that the distance travelled by the particle, which is the distance between the two plates, also increases.

3. Nov 27, 2009

### chanderjeet

ohh, i did miss that. Thanks very much