Question

Consider an oil droplet of mass andcharge . We want to determine the charge on thedroplet in a Millikan-type experiment. We will do this in several steps. Assume, for simplicity, that the charge is positive and that the electric field betweenthe plates points upward.

The field $E_0$ is easily determined by knowing thespacing between the plates and measuring the potential difference applied to them. The larger problem is to determine the mass of a microscopic droplet. Consider amass falling through a viscous medium in which there is aretarding or drag force. For very small particles, the retarding force is given by , where is a constant and is the droplet's speed. The negative sign tells us that the drag force vectorpoints upward when the droplet is falling. A falling droplet quickly reaches a constant speed, called theterminal speed. Write anexpression for the terminal speed $v_term$ .Express your answer in terms of , , and .

Answer 1

from newtonslaw F_drag = mg6πηrv_term = mg6πηrv_term = ρ*V*gwhere V =(4/3)*π*r³ is the volume of thesphere6πηrv_term =ρ*(4/3)*π*r³*gv_term = [ ρ*(4/3)*π*r³*g ]/( 6πη r )

Answer 2

Vterm = mg/b

Answer 3

Initially the oil drops are allowed to fall between the plates with the electric field turned off. They very quickly reach aterminal velocitybecause of friction with the air in the chamber.The field is then turned on and, if it is large enough, some of the drops (the charged ones) will start to rise. (This is because the upwards electricforceF_Eis greater for them than the downwards gravitational forceg, in the same way bits of paper can bepicked by a charged rubber rod). A likely looking drop is selected and kept in the middle of the field of view by alternately switching off the voltageuntil all the other drops have fallen. The experiment is then continued with this one drop.

The drop is allowed to fall and its terminal velocity v₁in the absence of an electric field is calculated. Thedragforce acting on the drop can then be worked out usingStokes' law:

wherev₁is the terminal velocity (i.e. velocity in the absence of an electric field) of the fallingdrop,ηis theviscosityof the air,andris theradiusof the drop.

The weightWis the volumeVmultiplied by the densityρand the acceleration due togravityg. However, what is needed is the apparent weight. The apparent weight in air is the true weight minus theupthrust(which equals the weight of air displaced by the oildrop). For a perfectly spherical droplet the apparent weight can be written as:

At terminal velocity the oil drop is notaccelerating. Therefore thetotal force acting on it must be zero and the two forcesFandWmust cancel one another out (thatis,F=W). This implies

Onceris calculated,Wcan easily be worked out.

Now the field is turned back on, and the electric force on the drop is

whereqis the charge on the oil drop andEis the electric field between the plates. For parallel plates

whereVis the potential difference anddis the distance between the plates.

One conceivable way to work outqwould be to adjustVuntil the oil drop remained steady. Then we couldequateF_EwithW. But in practice this is extremely difficult to do precisely. Also,determiningF_Eproves difficult because the mass of the oil drop is difficult to determine without reverting back tothe use of Stokes' Law. A more practical approach is to turnVup slightly so that the oil drop rises with a new terminalvelocityv₂. Then