Question

In a Millikan oil drop experiment, an oil drop is suspended between the plates a parallel plate capacitor. (This means that the oil drop has zero velocity and the downward force of gravity is balanced by the upward electrostatic force F-qE.) The plates are separated by a distance d-2.25 cm and have an area A= 16.0 cm². The oil drop has a mass of 4.29 x 10 kg and the voltage across the plates is 98.5 Volts. How many excess electrons are contained within the oil drop? (Find the total charge, divide by the charge of the electron to get the number of excess electrons in the drop.)

Answer 1

In the Millikan oil drop experiment, the number of excess electrons on an oil drop is calculated by equating the gravitational force to the electrostatic force, then dividing the total charge by the charge of an electron.

Step-by-step explanation:

In the Millikan oil drop experiment, the number of excess electrons on an oil drop can be determined by balancing the gravitational force and the electrostatic force on the drop. Given a mass for the oil drop of 4.29 x 10-15 kg and voltage across the plates of 98.5 V, with plate separation of 2.25 cm, the total charge on the oil drop can be calculated as follows:

• First, calculate the gravitational force (Fg) using Fg = mg where m is the mass and g is the acceleration due to gravity (approximated as 9.81 m/s2).
• Next, use the electrostatic force formula Fq = qE, where q is the charge and E is the electric field. The electric field can be found from E = V/d where V is the voltage and d is the distance between the plates.
• The charge on the oil drop is found when Fg = Fq, thus q = mg/E.

To determine the number of excess electrons, divide the total charge (q) by the fundamental charge of an electron (-1.602 x 10-19 C).

For the values given, and assuming gravity is the only downward force:

• Fg = (4.29 x 10-15 kg)(9.81 m/s2)
• E = 98.5 V / 0.0225 m
• q = Fg / E
• Number of excess electrons = q / (1.602 x 10-19 C)