In an experiment replicating Millikan’s oil drop experiment, a pair of parallel plates are placed 0.0200 m apart and the top plate is positive. When the potential difference across the plates is 240.0 V, an oil drop of mass 2.0 × 10-11 kg gets suspended between the plates. (e = 1.6 × 10-19 C) (___/ 5)
(a) Describe in words what a free-body diagram for the charge Would look like.
(b) What is the charge on the oil drop?
(c) Is there an excess or deficit of electrons on the oil drop? How many electrons are in excess or deficit?
(a) The free-body diagram for the charge would show two forces acting on the oil drop: the gravitational force pulling it downwards and the electric force pushing it upwards.
(b) We can use Millikan's formula to determine the charge on the oil drop: q = (mgd)/(Vρg), where m is the mass of the drop, g is the acceleration due to gravity, d is the distance between the plates, V is the potential difference across the plates, ρ is the density of the oil, and q is the charge on the drop. Plugging in the given values, we get:
q = [(2.0 × 10^-11 kg)(9.81 m/s^2)(0.0200 m)]/(240.0 V)(860 kg/m^3)(9.81 m/s^2)
q ≈ 1.6 × 10^-19 C
Thus, the charge on the oil drop is approximately 1.6 × 10^-19 C.
(c) We know that the oil drop is suspended between the plates, which means that the upward electric force is equal in magnitude to the downward gravitational force. Since we know the charge on the oil drop (from part (b)), we can calculate the electric force using F = Eq, where E is the electric field between the plates. Rearranging this equation, we get:
E = F/q
Plugging in the known values, we get:
E = (mg)/q
E = [(2.0 × 10^-11 kg)(9.81 m/s^2)]/(1.6 × 10^-19 C)
E ≈ 1.2 × 10^5 N/C
We know that the electric field between the plates is due to the electric potential difference across them (V), so we can use the equation E = V/d to find V:
V = Ed
V = (1.2 × 10^5 N/C)(0.0200 m)
V = 2400 V
This is the same as the given potential difference across the plates, which confirms our calculation.
From the fact that the oil drop is suspended, we know that it has either lost or gained electrons to become charged. Since the charge on the drop is positive (from part (b)), it has lost electrons and has a deficit of electrons. We can calculate the number of electrons in the deficit using the equation:
q = Ne
where N is the number of electrons and e is the charge of an electron. Plugging in the known values, we get:
(1.6 × 10^-19 C) = N(1.6 × 10^-19 C)
N = 1
Thus, the oil drop has a deficit of one electron.