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Uniform elecric fields- force and kinetic energy

  1. Nov 25, 2009 #1
    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.
  2. jcsd
  3. Nov 25, 2009 #2
    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 Fx= (dW/dx)= (1/2)Q2/e0A
    So if Q is an invariant, the force is independent of separation.
    Also Fx = 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
  4. Nov 25, 2009 #3
    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)
  5. Nov 25, 2009 #4
    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
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