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Motion in Fields (two parallel plates when voltage is applied)

  1. Jan 14, 2010 #1
    1. The problem statement, all variables and given/known data

    Two parallel metallic plates are placed horizontally with a separation of 2.2mm. When the voltage is reduced to zero (so that the plates are neutral), the particle (or sphere) is observed to fall with constant velocity. It is observed to travel 0.85 mm in 8.9s.
    A new p.d. V volts is now applied to the plates, which causes the particle to move upwards with a constant velocity of 4 x 10^-5 m/s. You may assume for small velocities like this, the frictional force is proportional to the speed.
    Calculate the value of V.

    Mass of Particle: 3.5 x 10^-15 kg
    Charge of Particle: 6.4 x 10^-19 C
    g : 9.81ms^-2

    2. Relevant equations

    3. The attempt at a solution

    KE1= Vq = .5mv^2 0= .5 x 3.5 x 10^-15 x (.00085 / 8.9)^2 = 1.596 x 10^-23J
    KE2= Vq = .5mv^2 0= .5 x 3.5 x 10^-15 x (4 x 10^-5)^2 = 2.8 x 10^-24

    KEtotal= KE1 + KE2 = 1.876 x 10^-23J = Vq
    V = 2.9x10^-5 volts

    Isn't this value faaar too small?
    Last edited: Jan 14, 2010
  2. jcsd
  3. Jan 14, 2010 #2
    Any answers or suggestions are highly appreciated!
  4. Jan 14, 2010 #3


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    Re: Interesting yet challenging question of motion of particle in parallel plates fie

    You forgot to consider air resistance, which robs the particle of energy.

    Try calculating this using forces instead of energies. You know that in both cases, the particle is moving at constant velocity, meaning net force = 0.

    This problem is extremely easy. You'll see this when you get the answer.
  5. Jan 14, 2010 #4


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    These equations don't apply here. This would be true IF the particle started from rest at one plate and had a constant acceleration until it reached the other plate.

    Instead, use the fact that
    F = -bv
    where b is a constant.
  6. Jan 15, 2010 #5
    well for the first one (no voltage), v = 9.55 x 10^-5

    F = -bv so 9.81 x 3.5 x 10^-15 = -b x 9.55 x 10^-5

    is it correct to use 9.81 x 3.5 x 10^-15 as the force? if so then you could find
    -b = 9-81x3.5 x 10^-15 /v

    then u find F(2) = -b x (4 x 10^-5)

    and since F = Vq / D, V can be found. Is this the correct method? But have I taken into account that it is travelling at constant velocity (net force= 0)?

    Thanks !!!
    Last edited: Jan 15, 2010
  7. Jan 15, 2010 #6
    but then i get V to be 49.14 voltswhich is the wrong answer, as it says that when V = 118V, the particle is stationary (not moving up or downwards!)...what have i done wrong?
    i.e. what do i use as the force?
    Last edited: Jan 15, 2010
  8. Jan 15, 2010 #7


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    You're on the right track. Don't forget that gravity always acts on the particle, whether it is moving upward or downward.
  9. Jan 15, 2010 #8
    oh so the total force would be

    F = (-b x v) + (mg)= (3.59 x 10^-10 x 4 x 10^-5 ) + (9.81 x 3.5 x 10^-15)

    = 4.87 x 10^-14

    then, by using F = Vq / D, I get V to be 167V.

    This must be the correct answer (at least I hope it is!)

    thanks so much!
    Last edited: Jan 15, 2010
  10. Jan 15, 2010 #9


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    Looks good (at least that's what I got too) :smile:
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