Milikans oil drop experiment

Question 1

What is Stoke’s law?

Answer 1

When you drop an object into a fluid,like air, it experiences a viscous drag force. This force acts in the opposite direction of the velocity of the object, and is due to the viscosity of the fluid. You can calculate the viscous force on a spherical object using Stoke’s law:

F = 6πnrv

where n is the viscosity of the fluid, r is the radius of hte object and v is the velocity of the object.

Question 2

Draw and explain Millikan’s basic setup.

Answer 2

• The atomiser(part which releases oil drops) created a fine mist of oil drops that wre charged by friction as they left the atomiser.
• Some of the drops fell through a hole in the top plate and could be viewed through the micrscope.
• A potential diffirence was applied between the two plates, producing a field that exerted a force on the charged drops.

Question 3

How did Milikan calculate the charge on the drops?

Answer 3

When a drop was passes through the atomiser, Millikan applied a p.d. until the drop was stationary. The two forces acting on the oil drop were:

1) The weight of the drop - acting downwards.
2) The force due to the uniform electric field - acting upwards.

Using the equation F = QV/d ( derived from E=F/q nad E =V/d).

Since the drop is stationary, the electric force must be equal to the weight so:

QV/d = 4/3 πr³pg.

So Millikan could find the charge on the drop.

Question 4

What did the results from Millikans experiment show?

Answer 4

All of the oil drop charges had a whole number multiple of -1.6 x 10^-19C. Millikan concluded that a charge can never exist in smaller quantities than 1.6x 10^-19 and that an electron has a charge of 1.6 x10^-19.