# Milikan Oil Drop Experiment

Pages: 4 (1188 words) Published: June 24, 2013
Milikan Oil Drop Experiment
Objective
To experimentally determine the quantum nature of charge by using the equipment especially designed for this experiment. Theory
A spherical drop of oil, falling through a viscous medium like air, will quickly reach a constant velocity. When it reaches this equilibrium state, the viscous force is balanced by other forces acting on the drop, such as gravity, buoyant forces from the air, electrical forces, etc. In this experiment an electrical force of varying magnitude is introduced to change the motion of the falling drop by an ionization source. By measuring the velocity of the oil drop under different conditions the amount of charge on the drop may be determined. If the charge on the drop is an integer multiple of the fundamental unit of charge (the electron), then one will be able to confirm the quantization of charge. The charge carried by an oil droplet can be obtained by analyzing the forces acting on the drop under different conditions. In this experiment the terminal velocity of the drops is reached in a few milliseconds. Applying Newton’s 2 Law to the falling oil drop yields: Kvf - mg = 0 (E-field off) (1)

qnE - mg – kvr = 0 (E-field on) (2)
In Equations (1) and (2) we have neglected the buoyant force exerted by the air on the droplet. This is reasonable since the density of air is only about one-thousandth that of oil. The equations can now be used to obtain an expression for the charge qn on the oil drop. The result is: qn = (mg(vf+vr))/vfE¬ (3)

The electric field in the region of the oil drop is produced by to two parallel plates maintained at a potential difference V and separated by a distance d. The relation is given by E = V/d. Thus, equation (3) becomes: qn = (mgd(vf+vr))/vfV¬ (4)

The mass ‘m’ of the oil drop is given by:
m = (4/3)πa3σ (4)
where ‘a’ is the radius of the drop and ‘σ’ is the density of the oil drop. To calculate the radius of the oil drop you will need to use...