CK-12-Physics-Concepts - Intermediate

(Marvins-Underground-K-12) #1

http://www.ck12.org Chapter 17. Electric Fields


The charge on the plates is adjustable. By measuring the terminal velocity of the oil drops with the electric field
off, Millikan could determine the mass of the drops. Millikan and his graduate assistant were able to determine the
force of the field on the drops when it was turned on by comparing the velocity of the drops with the field on to their
velocity with the field off. This is easily determined when the oil drop is stationary; namely, when the downward
gravitational force exactly equals the upward electrical force on the drop.


Example Problem: An oil drop weighs 1. 9 × 10 −^14 N. It is suspended in an electric field whose intensity is
4. 0 × 104 N/C. Since the oil drop is suspended, the gravitational force, 1. 9 × 10 −^4 N, is equal to the electrical force,
FE=Eq.


Solution:q=weightE =^1.^9 ×^10


− (^14) N
4. 0 × 104 N/C=^4.^8 ×^10
− (^19) C
The charge on this particular oil drop was 4. 8 × 10 −^19 C. In doing his experiment, however, Millikan faced a
problem. When the oil is sprayed through the atomizer, some oil drops are negatively charged, but we don’t know
how many extra electrons the drops acquire. The charge on this oil drop could be the result of having one extra
electron, or having five extra electrons.
In order to determine the charge on one electron, the oil drop experiment was carried out many times, and the charges
on many oil drops was determined. The smallest charge was found to be 1. 6 × 10 −^19 C, and all the other charges
on oil drops were found to be whole number multiples of 1. 6 × 10 −^19 C. In the example problem above, we would
conclude that the oil drop held three extra electrons.
Summary



  • Using a uniform electric field between two charged parallel plates and oil drops, Millikan determined the
    charge on a single electron.

  • The friction of the oil passing through the narrow opening puts charges on many of the oil drops.

  • When an oil drop is stopped by adjusting the charge on the plates, its weight,mg,downward equals the
    electrical force,Eq, upward.

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