Milikans Oil Drop

Question 1

Stoke’s Law: Viscous Force Equation

Answer 1

𝐹 = 6𝜋ηrv

η = viscoscity of fluid, r = radius of object, v = velocity

Question 2

Milikans experiment

Explain the Experiment

Answer 2

• Two plates approx 1.5cm apart (one over other) were connected to a system that could apply a pd across them and create an electric field between them.
• An atomiser created a fine mist of oil drops that were charged by friction as they left it (positively charged if they lost electrons and negativly charged if they gained electrons)
• Some of the oil drops fell through the a hole in the top plate and could be viewed by the microscope (eyepiece had scale to measure more accurately)
• Before field turned on, only forces acting on drops was Weight of drop (downwards) and Viscous Force from air (upwards).
• Drop reaches** terminal velocity **(no acceleration) when the two forces are equal : mg = 6𝜋ηrv
• To find mass of drop do volume multiplied by density then subsitute back and rearange to find r : 4/3𝜋r³ρg = 6𝜋ηrv
• He then turns on the electric field to introduce the electric force on the drop. He adjusts the electric field until the drop is stationary, this would mean the viscous force is gone as it was proportional to velocity and now the drop has 0 velocity.
• Forces now acting on drop: Weight (Downwards) and Force due to uniform electric field (Upwards)
• Electric Force given by :** F = QV / d** (Q = Charge on drop, V = pd between plates, d = distance between plates)
• Since drop is stationary Electric Force must be equal to the Weight : QV / d = mg = 4/3𝜋r³ρg
• Only uknown now is Q. Millikan could now find the charge of the drop.
• Millikan repeated the experiment on 100s of drops. The charge on the drop was always a multiple of 1.6x10^-19C
• He concluded that charge can never exist in smaller quantities than 1.6x10^-19C and that this was the charge carried by an electron.