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

  • #26
gneill
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v1 = 4 * 10-4 m s-1.
F1 = m g = 10-14 * 10 = 10-13 N.

v2 = 8 * 10-5 m s-1.
F2 = ?

F1 = k v1, where k = F1 / v1 = 10-13 / 4 * 10-4 = 2.5 * 10-10.

So F2 = k v2 = 2.5 * 10-10 * 8 * 10-5 = 2 * 10-14 N.
Okay, good so far!
Weight = Upthrust + EF + ARRising or F2
10-13 N = zero + Q E + 2 * 10-14 N
Q E = 8 * 10-14 N
Q = 8 * 10-14 / 125 000 = 6.4 * 10-19 Q.

Then 6.4 * 10-19 / 1.6 * 10-19 = 4, not 6 as in the answer. What's wrong? I recalculated everything and checked the book answer, it's 6. So, don't think that it's the calculation.
Friction always works against the direction of motion. You've added the friction force to the "Upthrust" and electric force, which is not correct. You might find it helpful to write out the force balance so that it sums to zero. Since this mysterious "Upthrust" is always zero, just drop it from your equation:

##F_E - Weight - F_2 = 0##

Where ##F_E## is the electric force.
 
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  • #27
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Okay, good so far!

Friction always works against the direction of motion. You've added the friction force to the "Upthrust" and electric force, which is not correct. You might find it helpful to write out the force balance so that it sums to zero. Since this mysterious "Upthrust" is always zero, just drop it from your equation:

##F_E - Weight - F_2 = 0##

Where ##F_E## is the electric force.
Yes, I think I lost that moment. Air resistance is against motion. Since the drop is moving upwards, therefore the air resistance is directed downwards in pair with the weight.

So it's Q E = Weight + Air resistance.
Q E = 2 * 10-14 N + 10-13 N.
Q E = 1.2 * 10-13 N
Q = 1.2 * 10-13 / 125 000 = 9.6 * 10-19 Q.

Then 9.6 * 10-19 / 1.6 * 10-19 = 6.

Thank you! : )
 

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