Millikan\'s experiment measures the elementary charge e by the observation of th
ID: 2050097 • Letter: M
Question
Millikan's experiment measures the elementary charge e by the observation of the motion of small oil droplets in an electric field. The oil droplets are charged with one or several elementary charges, and if the (vertical) electric field has the right magnitude, the electric force on the droplet will balance its weight, holding the drop suspended in midair. Suppose that an oil droplet of radius 1.0 X 10^-4 cm carries a single elementary charge. What electric field is required to balance the weight? The density of oil is 0.80 g/cm^3.Explanation / Answer
To solve this, the Force due to the electric field must be balanced by the force of the weight on the electron
Force by the electric field = qE
Force by weight = mg
To find the mass, we are given the density and radius of the oild drop.
Density is known and the formula for density is = m/V (Density is mass / Volume)
So mass can be found as m = V
The volume of a sphere, V, is 4/3(r3)
Doing the substitution
qE = mg
qE = Vg
Lets get all the units into SI units
Density is given as 0.80 g/cm3. With 1000 g per kg, and 100 cm per m, the density can be expressed as...
800kg/m3
The radius is 1 X 10-4 cm, which is 1 X 10-6 m
The volume is 4/3()(1 X 10-6)3 = 4.19 X 10-18 m3
A single elementary charge is 1.6 X 10-19 C
Now solve for E with all of the numerical substitutions
E = Vg/q
E = (800)(4.19 X 10-18)(9.8)/(1.6 X 10-19)
E = 2.05 X 105 N/C