MillikanOilDropExperiment

- MILLIKAN OIL DROP EXPERIMENT INTRODUCTION In this experiment you will experimentally determine the quantum nature of charge Robert Millikan was

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1 MILLIKAN OIL DROP EXPERIMENT INTRODUCTION In this experiment you will experimentally determine the quantum nature of charge. Robert Millikan was awarded the Nobel Prize in physics in 1923 for this brilliant experiment. A simplistic schematic of his apparatus is shown below. 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. EQUIPMENT 1. Millikan Oil Drop Apparatus (see diagrams below) 2. High voltage power supply (0-500VDC) 3. Digital multimeter 4. 4 banana leads 5. Stopwatch 6. Micrometer 7. Barometer
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2 Millikan Oil Drop Apparatus
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3 THEORY The charge carried by an oil droplet can be obtained by analyzing the forces acting on the drop under different conditions. Figure 1 shows the forces acting on the drop when it is falling in air and has reached its terminal velocity. kv f velocity d o w n w a r d mg Figure 1 In this experiment the terminal velocity of the drops is reached in a few milliseconds. Applying Newton’s 2 nd Law to the falling oil drop yields: k v f - mg = 0 (E-field off) (1) Figure 2 shows the forces acting on the drop when it is rising under the influence of an electric field. + v e l o c i t y q n E upward mg kv r __ Figure 2
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4 Applying Newton’s 2 nd Law again we have: q n E - mg - kv r = 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 q n on the oil drop. The result is: E v v v mg q f r f n ) ( + = (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: V v v v mgd q f r f n ) ( + = (4) The mass ‘ m ’of the oil drop is given by: m = (4/3 ) π a 3 σ (4) where ‘ a’ is the radius of the drop and ‘ is the density of the oil drop. To calculate the radius of the oil drop you will need to use Stokes’ Law, which relates the radius of a spherical falling body in a viscous medium to its terminal velocity. According to Stokes’ Law the viscous force on a spherical falling body in a viscous medium is given by: F = 6 πη av f (5) where η is the coefficient of viscosity, which in our case will be the viscosity of air. Using this expression and applying Newton’s 2
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This note was uploaded on 09/02/2011 for the course PHYS 4D taught by Professor Luna during the Spring '11 term at DeAnza College.

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- MILLIKAN OIL DROP EXPERIMENT INTRODUCTION In this experiment you will experimentally determine the quantum nature of charge Robert Millikan was

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