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?

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

d is the correct answer

the moon affects the earth gravitation ally, causing certain motions of that body and also producing tides in its ocean and atmosphere. the tidal wave in our atmosphere is, however, far too minute to be an effective cause in any known meteorological phenomenon.

explanation: