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How many electron charges are there on the drop? (Millikan)

  1. Oct 25, 2016 #1
    1. The problem statement, all variables and given/known data
    In a measurement of the electron charge by Millikan's method, a potential difference of 1.5 kV can be applied between horizontal parallel metal plates 12 mm apart. With the field switched off, a drop of oil of mass 10-14 kg is observed to fall with constant velocity 400 μm s-1. When the field is switched on, the drop rises with constant velocity 80 μm s-1. How many electron charges are there on the drop? (You may assume that the air resistance is proportional to the velocity of the drop, and that air buoyancy may be neglected.)

    (The electronic charge = 1.6 * 10-19 C, the acceleration due to gravity = 10 m s-2.)

    Answer: 6.

    2. The attempt at a solution
    As I understand I need to find Q?

    In my book I have this formula: Q = 6 π r η v / E, where η is the coefficient of viscosity of air. The formula is applicable "when an electric field has been applied such that the drop is stationary." There are also formulas when there is no electric field, but I think in this case we have electric field.

    But we don't have the radius and have the distance between plates instead.

    I am also not sure whether I need to find Q, or some other value. Also how to find η?
     
  2. jcsd
  3. Oct 25, 2016 #2

    gneill

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    Yes you need to find Q. In particular you need to find out how many elementary charges it takes to total up to Q.

    You are not given enough information to calculate the drag force directly. However, you are told to assume that it is proportional to velocity. Can you determine one value for the drag force at a particular velocity (other than zero velocity, of course)? Can you then scale it for other velocities?
     
  4. Oct 25, 2016 #3
    Q = (6 π η / E) * (9 η v / 2 ρo g)1 / 2 v

    E = V / d = 1500 / 0.012 = 125 000 V m-1.

    Now η (coefficient of viscosity of air) and ρo (density of oil) are required.

    Update
    As I understand when there is no electric field velocity is 400 μm s-1. Weight = Upthrust due to air + Viscous drag. Upthrust is zero. So: (4/3) π r3 po g = 6 π r η v. Simplify them to find radius: r = √ 4.5 η v / ρo g.

    Then we find radius and turn the field on. We already found E = 125 000 V m-1. And then we use Q = (4/3) π r3 ρo g / E. Q = 4 π (√ 4.5 η v / ρo g)3 ρo g / 3 E.

    Only need to find η and ρo somehow...
     
    Last edited: Oct 25, 2016
  5. Oct 25, 2016 #4

    gneill

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    You can forget the formula for drag and all of its constants that you don't know. Just find the force (in Newtons) that it creates when the oil drop is falling at a constant speed without the electric field in place.
     
  6. Oct 25, 2016 #5
    Hm, is it F = m g? We have the mass present.
     
  7. Oct 25, 2016 #6

    gneill

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    Yes. The drop is falling with constant velocity so the net force acting is zero.
     
  8. Oct 25, 2016 #7
    F = m g
    E = V / d

    m g = Q (V / d)
    Q = m g d / V = 10-14 * 10 * 0.012 / 1500 = 8 * 10-19 C.
     
  9. Oct 25, 2016 #8

    gneill

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    You want to specify that that force F above is the force due to drag for that one velocity only.
    Can you explain in words what the first equation above represents? I can see an electric force, I can see the gravitational force, but I don't see the drag accounted for.
     
  10. Oct 25, 2016 #9
    Weigth = Upthrust + Electric force
    Weight, as you say, shouldn't be 4/3 π r3 ρo g but F = m g.

    Upthrust is zero.

    Electric force = Q E, Q = charge on the drop, E = the electric field strength.
     
  11. Oct 25, 2016 #10

    gneill

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    There will be a different drag force acting when the drop is moving upwards because its moving at a different speed than when it was falling. You need to account for that drag force. Use the given assumption about the drag force and velocity.
     
  12. Oct 25, 2016 #11
    What is drag force?
     
  13. Oct 25, 2016 #12

    gneill

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    Air resistance, or viscous drag. You calculated a value for the drag force for the falling oil drop in posts #5 and #7.
     
  14. Oct 25, 2016 #13
    I'm sorry, I don't get it.

    I need to find Q. Assuming everything I did is not correct. What do I need to do then?

    Update
    I think I understand what you mean. If we have no electric field then the drop is falling and therefore weight = m g. And therefore we have weight = upthrust due to air + viscous drag (in contrast to when there is an electric field and weight = upthrust + electric force). I used to formula for the situation when an electric field is present.

    So now we have weight = viscous drag. m g = 6 π r η v. Though I don't see how it helps -- we need the radius and the coefficient of viscosity of air η.
     
    Last edited: Oct 25, 2016
  15. Oct 25, 2016 #14

    gneill

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    Yes, you need to find Q. You do that by finding out what force that charge experiences due to the known electric field. But there are several forces acting at once and you need to factor out their contributions. Gravity is easy because you have the mass of the oil drop. Air resistance also acts on the drop whenever it is moving through air. It's a type of friction.

    The idea is to establish what this friction force is for the upward moving drop so that you can "remove it" from the sum of forces, just as you will "remove" the gravitational force.

    That's why you went to the trouble of finding what the air resistance force is for the drop while it was falling at constant speed. The only two forces acting then were gravity and this friction force.

    Now you need to find the friction force acting on the upward moving drop so that you can account for it in the net force sum.
     
  16. Oct 25, 2016 #15
    Not sure whether you saw this:
    Is this correct logic?
     
  17. Oct 25, 2016 #16

    gneill

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    Yes the logic is fine as far as it goes. But you don't need to know any of the constants for the air resistance formula; it's not needed here since you can use the value for the force at a particular velocity to find the force for other velocities thanks to the given assumption about how the air resistance varies with velocity.
     
  18. Oct 25, 2016 #17
    OK, so no electric field: Weight = Upthrust (zero) + Air resistance → m g = Air resistance.

    With electric field: Weight = Upthrust (zero) + Electric force → 4 / 3 π r3 ρo g = Q E → 4 / 3 π r3 ρo g = Q V / d.

    I think this should be correct.
     
  19. Oct 25, 2016 #18

    gneill

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    You're still plugging in the useless air resistance formula. You need to use the first air resistance force to find a value for the second. Then write your force balance using that.
     
  20. Oct 25, 2016 #19
    In that case I'm missing some force.

    This is what I have my book:
    dbbde07626c3.jpg

    From book:
    No EF: Weight = Upthrust due to air + Viscous drag.
    EF: Weight = Upthrust + Electric force.

    So:
    No EF: Viscous drag = 10-14 * 10 = 10-13 N.
    EF: Q = m g / (V / d) = 10-13 / (1500 / 0.012) = 8 * 10-19 C.

    So we have the answer Q = 8 * 10-19 C. I don't understand why we need all the other numbers given in the problem and why we need to calculate the no EF force.
     
  21. Oct 25, 2016 #20

    gneill

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    In the book's figure the drop on the right hand side is stationary, so no air resistance force, just the electric force balancing the gravitational force.

    In the present problem the oil drop is moving in both cases.
     
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