Final answer:
The Millikan oil-drop experiment requires equating the electrical force to the gravitational force to find the charge on the oil drop, which can be expressed in terms of electrons by dividing by the elementary charge.
Step-by-step explanation:
In Millikan's oil-drop experiment, the charge on an oil drop held motionless between two plates can be found by equating the electrical force to the gravitational force. Given the voltage between the plates is 2090 V and the plate separation is 2.06 cm, we can calculate the electric field (E) as E = Voltage / Separation. The electrical force (Felectric) on the drop is equal to this electric field (E) times the charge (q) on the drop, that is, Felectric = E * q.
To find the gravitational force (Fgravity), we need to calculate the mass of the oil drop. The volume (V) of the sphere is given by V = (4/3) * π * (diameter/2)^3, and the mass (mdrop) is the volume times the density of the oil. The gravitational force is Fgravity = mass * g, where g is the acceleration due to gravity (9.8 m/s^2).
Setting Felectric equal to Fgravity and solving for the charge (q) gives us the charge on the oil drop in coulombs. Then, to find the number of excess electrons, we divide the charge by the elementary charge (approximately 1.602 x 10^-19 C).