Millikans Oil Drop Experiment To Determine The Charge Of The Electron Success Criteria Falling through a fluid Measuring the viscosity of water
McGraw, Jay, Executive Producer has reference to this Academic Journal, PHwiki organized this Journal Millikans Oil Drop Experiment To Determine The Charge Of The Electron Success Criteria Be able to state in addition to explain Stokes Law as long as bodies moving through fluids Be able to identify the as long as ces acting on a stationary charged object in an electric field Recall in addition to explain how Millikan was experimentally able to determine the charge on the electron Give a detailed description of the as long as ces acting on an object moving in a fluid How could the drag as long as ce on the object be calculated Falling through a fluid
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Measuring the viscosity of water Measurements: Mass of ball bearing Radius of ball bearing Terminal velocity of ball bearing Stokes Law When an object is dropped through a fluid, it experiences a as long as ce called viscous drag. This as long as ce acts in the opposite direction to the velocity of the object, in addition to is due to the viscosity of the fluid. The as long as ce on a spherical object can be calculated using Stokes Law: F=6rv =coeff of viscosity of fluid, r=radius of object v=velocity of object, =Pi Assumptions Small Reynolds number (small particles satisfy this) Laminar flow Smooth, spherical particles Homogeneous fluid Particles do not interfere with each other
Millikans Apparatus Atomiser created a fine mist of oil droplets, charged by friction (Later experiments used x-rays). Some droplets fell through hole in addition to could be viewed through the microscope with a scale to measure distances in addition to hence velocities. Millikan applied an electric field between the plates which would exert a as long as ce on the droplets. He could adjust the p.d. in addition to hence field strength to get the oil droplets to hover. Forces without a field Be as long as e the electric field is turned on, the as long as ces on the droplets are: a) the weight of the drop; b) the viscous as long as ce from the air. Viscous Force Weight The drop will reach terminal velocity when the two as long as ces are equal. mg=6rv The mass of the drop is the volume multiplied by the density of the oil, , so the equation can be written: 4/3r3g = 6rv => r2 = 9v/2g Millikan in earlier experiments measured in addition to , in addition to now can calculate r. With An Electric Field Millikan adjusted p.d. till the drop was stationary. Since as long as viscous as long as ce to act the drop needed to be moving, this as long as is now not there. Two as long as ces now are: Weight of droplet; Force due to uni as long as m electric field.
For a charged oil droplet The electric as long as ce is given by: F = QV/d Where Q=charge on oil drop V=p.d. between plates d=distance between plates If drop is stationary then electric as long as ce must be equal to the weight. QV/d = 4/3r3g Earlier, r was calculated so the only unknown is Q. Millikan could calculate the charge on the droplet. He repeated his experiment many times in addition to found that the smallest value calculated was -1.6X10-19 C in addition to all other values were a multiple of it. He concluded that charge could never exist in smaller quantities than this, in addition to this was the charge carried by a single electron. Practical Task Find the coefficient of viscosity of glycerine
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