In the Millikan oil drop experiment, an atomizer (a sprayer with a fine nozzle) is used to introduce many tiny droplets of oil between two oppositely charged parallel metal plates. Some of the droplets pick up one or more excess electrons. The charge on the plates is adjusted so that the electric force on the excess electrons exactly balances the weight of the droplet. The idea is to look for a droplet that has the smallest electric force and assume that it has only one excess electron. This strategy lets the observer measure the charge on the electron. Suppose we are using an electric field of 3.0 104 N/C. The charge on one electron is about 1.6 10-19 C. Estimate the radius of an oil drop of density 774 kg/m3 for which its weight could be balanced by the electric force of this field on one electron. (This problem courtesy of E.F. Reddish.)

To solve this problem, we need to analyze the forces acting on the oil drop and set them equal to each other.

The weight of the oil drop can be calculated using the formula:

Weight = Mass x Gravity

The mass of the oil drop can be calculated using the formula:

Mass = Density x Volume

The volume of a sphere can be calculated using the formula:

Volume = (4/3) x π x Radius^3

The electric force on the droplet can be calculated using the formula:

Electric force = Charge x Electric field

Given:
Electric field = 3.0 x 10^4 N/C
Charge on one electron = 1.6 x 10^-19 C
Density of oil drop = 774 kg/m^3

We need to find the radius of the oil drop.

Now, let's set up the equations:

Weight = Electric force

Mass x Gravity = Charge x Electric field

Density x Volume x Gravity = Charge x Electric field

Substituting the formulas above:

Density x ((4/3) x π x Radius^3) x Gravity = Charge x Electric field

Now, let's plug in the given values:

774 kg/m^3 x ((4/3) x π x Radius^3) x 9.8 m/s^2 = 1.6 x 10^-19 C x 3.0 x 10^4 N/C

Simplifying the equation:

Radius^3 = (1.6 x 10^-19 C x 3.0 x 10^4 N/C) / (774 kg/m^3 x ((4/3) x π x 9.8 m/s^2))

Radius^3 = 1.6 x 3.0 x (10^-19 C / kg) x (10^4 N / C) x (m^3 / kg) / (774 x (4/3) x π x 9.8)

Radius^3 = 2.4684547061 x 10^-24

Now, let's solve for the radius:

Radius ≈ ∛(2.4684547061 x 10^-24)

Using a calculator or a math software, we find that the radius is approximately 1.076 x 10^-8 meters.

Therefore, the estimated radius of the oil drop is approximately 1.076 x 10^-8 meters.