Unraveling Millikan's Experiment: Calculating Electrons & Water Molecules

[Cosmo Cookie]

Homework Statement

Millikan: A tiny, charged, sphere of water, with a radius of 1.25 microns is suspended at rest, in a vacuum, between the parallel plates of a capacitor. If the plates are 1 cm. apart, and have a potential difference of 500 volts between them, find the number of individual electrons required on the charged drop (to keep it suspended). What would happen if a radioactive source of beta particles (electrons) were introduced? Why did Millikan not use water? BEFORE you do this problem, first calculate the number of water molecules in the 1.25 micron drop, and write down your GUESS aas to the number of electrons that you will find is needed.
Why all the "ones"?

Homework Equations

1 cc=1 ml=1 gr for water, but the (mks) density for water is NOT one.

Im not quite sure how to even get started on this one. Can someone help me out? toss a few bones at me such as equations i'll need? and perhaps some hints on the non mathematical aspect?

 

[Jhero]

Hi there,

I can help you with this problem. First, let's start by calculating the number of water molecules in the 1.25 micron drop. The radius of the drop is given as 1.25 microns, which is equal to 1.25 x 10^-6 meters. We can use the formula for the volume of a sphere to find the volume of the drop: V = (4/3)πr^3. Plugging in the radius, we get V = (4/3)π(1.25 x 10^-6)^3 = 8.18 x 10^-19 m^3. Now, we can use the density of water to find the mass of the drop: ρ = m/V, where ρ is the density of water (1 g/cm^3 or 1000 kg/m^3 for mks units). Solving for m, we get m = ρV = (1000 kg/m^3)(8.18 x 10^-19 m^3) = 8.18 x 10^-16 kg. We can then use the mass of the water drop to find the number of water molecules using Avogadro's number (6.022 x 10^23 molecules/mol): n = m/M, where M is the molar mass of water (18.015 g/mol or 0.018015 kg/mol). Plugging in the values, we get n = (8.18 x 10^-16 kg)/(0.018015 kg/mol) = 4.54 x 10^7 water molecules.

Now, let's move on to the main problem. We have a charged drop of water suspended between parallel plates of a capacitor with a potential difference of 500 volts between them. The force acting on the drop is given by F = qE, where q is the charge on the drop and E is the electric field between the plates. Since the drop is at rest, the force must be balanced by the gravitational force acting on the drop, which is given by F = mg, where m is the mass of the drop and g is the acceleration due to gravity. Equating these two forces, we get qE = mg. We can rearrange this equation to find the charge on the drop: q = mg/E. Plugging in the values, we get q = (8.