Uniform elecric fields- force and kinetic energy

[chanderjeet]

what is the effect on the force as well as the gain in kinetic energy if the separation of the conducting plates is increased?

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.

 

[Bob S]

The stored energy in a capacitor is

W = (1/2)CV²= (1/2)Q²/C = (1/2)xQ²/e₀A
where x = separation and A = area.
So F_x= (dW/dx)= (1/2)Q²/e₀A
So if Q is an invariant, the force is independent of separation.
Also F_x = dW/dx = d/dx(1/2)CV²=(1/2)e₀AV² d/dx(1/x)= -(1/2)e₀AV²/x²
where V is held constant.
Bob S

 

[chanderjeet]

sorry, i didn't specify. I was referring to, say, a particle being released at the upper plate of horizontal parallel conducting plates. (in a vacuum)

 

[Bob S]

In this case, Q=CV
so V = Q/C = Qd/e₀A
where A = area and d the separation
so if Q = constant, and d is increased, the voltage between the plates increases linearly.

One way to generate high voltage between the plates is to charge them up when d is small, then separate the two plates.
Bob S