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
The potential difference between the plates be to hold the droplet in equilibrium ΔV = 35V
Newton’s Law of Universal Gravitation is used to explain gravitational force. This law states that every massive particle in the universe attracts every other massive particle with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them
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:
[tex]V = \dfrac{4\pi\dfrac{(d}{2})^3}{3}[/tex]
V = volume, d = diameter
Make some substitutions:
[tex]F = \dfrac{4pg\pi \dfrac{d}{2}^3}{3}[/tex]
Given values:
p = 860kg/m³, g = 9.81m/s², d = 0.68×10⁻⁶m
Plug in and solve for F:
[tex]F = \dfrac{4(860)(9.81)\pi\dfrac{(0.68\times10^{-6}}{2})^3}{3}}[/tex]
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
Thus the potential difference between the plates be to hold the droplet in equilibrium ΔV = 35V
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