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 0.67-um-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/m^3, and the capacitor plates are 4.0 mm apart.
What must the potential difference between the plates be to hold the droplet in equilibrium?
To find the potential difference (V) between the plates required to hold the droplet in equilibrium, we need to consider the forces acting on the droplet.
1. Gravity Force (Fg):
The force of gravity (Fg) acting on the droplet is given by the equation Fg = mg, where m is the mass of the droplet and g is the acceleration due to gravity.
Given:
Density of oil (ρ) = 860 kg/m^3
Diameter of droplet (d) = 0.67 μm = 0.67 × 10^-6 m
To find the mass (m) of the droplet, we can use the equation m = ρ × V × d, where V is the volume of the droplet.
The volume of the droplet (V) can be calculated using the equation V = (π/6) × d^3.
Substituting the given values, we can calculate the mass (m) of the droplet.
2. Electric Force (Fe):
The electric force (Fe) acting on the droplet is given by the equation Fe = qE, where q is the charge on the droplet and E is the electric field between the plates.
Given:
Charge on the droplet (q) = +e (where e is the elementary charge)
Distance between the plates (d) = 4.0 mm = 4.0 × 10^-3 m
To find the electric field (E), we can use the equation E = V/d, where V is the potential difference between the plates.
Since the droplet is in equilibrium, the electric force (Fe) must be equal to the gravity force (Fg), i.e., Fe = Fg.
Substituting the expressions for Fe and Fg, we get:
qE = mg
Now, substituting the expressions for q, E, and m in terms of the given values, we can solve for V.
Note: The value of the elementary charge (e) is 1.6 × 10^-19 coulombs.
Let's put all the values together and calculate the potential difference (V).
To find the potential difference between the plates necessary to hold the droplet in equilibrium, we can use the principle of equilibrium, where the electric force is balanced by the gravitational force acting on the droplet.
The electric force on the charged droplet is given by the formula:
F_electric = qE
where q is the charge of the droplet and E is the electric field strength between the plates.
The electric field strength can be calculated using the formula:
E = V/d
where V is the potential difference between the plates and d is the distance between the plates.
The gravitational force acting on the droplet is given by the formula:
F_gravity = mg
where m is the mass of the droplet and g is the acceleration due to gravity.
Now, let's calculate each of these quantities:
1. Charge of the droplet (q):
The charge of the droplet is given as +e, where e is the elementary charge, which is approximately 1.60 x 10^-19 Coulombs.
q = + e = + 1.60 x 10^-19 C
2. Electric field strength (E):
We need to find the electric field strength using the given potential difference (V) and distance between the plates (d).
E = V/d = V/0.004 m
3. Mass of the droplet (m):
The mass of the droplet can be calculated using the given diameter (0.67 μm) and the density (860 kg/m^3).
The volume of the droplet can be calculated using the formula:
V_droplet = (4/3) π r^3
where r is the radius of the droplet. Since the diameter is given, we can find the radius using the formula:
r = diameter/2 = 0.67 μm/2
Now, we can calculate the mass of the droplet:
mass = density x volume
4. Gravitational force (F_gravity):
The gravitational force acting on the droplet can be calculated using the mass (m) and the acceleration due to gravity (g).
F_gravity = m x g
5. Equilibrium condition:
For the droplet to remain in equilibrium, the electric force (F_electric) and gravitational force (F_gravity) must be equal. Therefore:
F_electric = F_gravity
Now, we can equate the two forces and solve for the potential difference (V):
qE = mg
V/d * q = m * g
Rearranging the equation:
V = (m * g * d)/q
Substituting the values we calculated earlier, we can now solve for V.
Actually, in the Millikan experiment, there can be one or several charges per drop. What was observed was that the charges were a small multiple of e.
In your case, go along with the assumption that the charge is +e and equate the electric force
e V/d to the particle's weight. d is the plate separation. Solve for V.
Ignore the small effect of buoyancy, the error will be less than 1%. Millikan had to take it into account