Question

In the early 1900s, Robert Millikan used small charged droplets of oil, suspended in an electric field, to make the first quantitative measurements of the electron’s charge. A 0.68-μm-diameter droplet of oil, having a charge of +e, is suspended in midair between two horizontal plates of a parallel-plate capacitor. The upward electric force on the droplet is exactly balanced by the downward force of gravity. The oil has a density of 860 kg/m3, and the capacitor plates are 4.0 mm apart. Part A What must the potential difference between the plates be to hold the droplet in equilibrium?

Answer 1

The gravitational force on the oil droplet is given by:

F = mg

F = gravitational force, m = mass, g = gravitational acceleration

The mass of the droplet is given by:

m = pV

m = mass, p = density, V = volume

The volume of the droplet is given by:

V = 4π(d/2)³/3

V = volume, d = diameter

Make some substitutions:

F = 4pgπ(d/2)³/3

Given values:

p = 860kg/m³, g = 9.81m/s², d = 0.68×10⁻⁶m

Plug in and solve for F:

F = 4(860)(9.81)π(0.68×10⁻⁶/2)³/3

F = 1.389×10⁻¹⁵N

The magnitude of the electric force is equal to the magnitude of the gravitational force. The electric force is given by:

F = Eq

F = electric force, E = electric field, q = droplet charge

The electric field between the capacitor plates is given by:

E = ΔV/d

E = electric field, ΔV = potential difference, d = plate separation

Make a substitution:

F = ΔVq/d

Given values:

F = 1.389×10⁻¹⁵N, q = +e = 1.6×10⁻¹⁹C, d = 4.0×10⁻³m

Plug in and solve for ΔV:

1.389×10⁻¹⁵ = ΔV(1.6×10⁻¹⁹)/(4.0×10⁻³)

ΔV = 34.73V

Round ΔV to 2 significant figures:

ΔV = 35V