Uniform elecric fields- force and kinetic energy

AI Thread Summary
A particle with a mass of 3x10^-12 kg and a charge of +4.0x10^-14 C is released between parallel conducting plates in a vacuum. The net force on the particle is calculated using the electric force and gravitational weight. As the separation of the plates increases, the electric field strength and force on the particle decrease, assuming constant potential difference. However, the distance the particle travels also increases, which affects the kinetic energy gained. Ultimately, while the force decreases, the increase in distance means the kinetic energy may not necessarily decrease.
chanderjeet
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Homework Statement


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.


Homework 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)

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?
 
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chanderjeet said:
I'm thinking that since F= VQ/d if the distance increases then the force would decrease. I could be wrong.
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)
chanderjeet said:
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?
You forgot that the distance traveled by the particle, which is the distance between the two plates, also increases.
 
ohh, i did miss that. Thanks very much
 
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