In this part, we will be using simulation of Millikan’s famous oil drop experiment from https://www.thephysicsaviary.com/PhysicsPrograms/Labs/MillikanOilDrop.html. The idea is that we can suspend a droplet of oil in the air by balancing the gravitational force with the electrostatic (Coulomb) force. Write an algebraic expression for the gravitational force on a spherical drop of radius “r” and density “rho” in the Earth’s gravitational field “g”. (4/3)pir^3grho The electrostatic force applied by two parallel plates (similar to a capacitor) with potential difference between them. Write an algebraic expression for the electric field between two such plates separated by distance “d”. E = V/d Write an algebraic expression for the electrostatic force that acts on a charge due to this electric field in terms of potential difference “V”. (qV) Combine these two expressions to write an algebraic expression for the charge in terms of “r”, “d”, “g”, “rho”, and “V”. (4/3)pir^3grho(d/V) Now you are going to perform the experiment. Read through the instructions below and click “Begin” to start the experiment. You will need to collect data from 10 drops and complete the table below: Millikan Oil Drop Lab
The Correct Answer and Explanation is:
Final Algebraic Expression for Charge:
q=43πr3ρg⋅dVq = \frac{4}{3} \pi r^3 \rho g \cdot \frac{d}{V}q=34πr3ρg⋅Vd
Explanation
Millikan’s oil drop experiment was a pivotal moment in physics that led to the quantization of electric charge. In this simulation, the goal is to determine the charge qqq on a tiny oil droplet suspended in an electric field. To do this, we balance the gravitational force pulling the droplet downward with the electrostatic force pushing it upward. This equilibrium allows us to derive a formula for qqq using measurable quantities like the droplet’s radius, the electric field, and the physical constants.
First, we calculate the gravitational force acting on a spherical oil droplet. The mass mmm of the droplet is the product of its volume and density:m=43πr3ρm = \frac{4}{3} \pi r^3 \rhom=34πr3ρ
So, the gravitational force becomes:Fgravity=mg=43πr3ρgF_{\text{gravity}} = m g = \frac{4}{3} \pi r^3 \rho gFgravity=mg=34πr3ρg
Next, the electric field EEE between two parallel plates separated by distance ddd and with a potential difference VVV is:E=VdE = \frac{V}{d}E=dV
The electrostatic force acting on a charged droplet in this field is:Felectric=qE=q⋅VdF_{\text{electric}} = qE = q \cdot \frac{V}{d}Felectric=qE=q⋅dV
At equilibrium, these two forces are equal in magnitude but opposite in direction:43πr3ρg=q⋅Vd\frac{4}{3} \pi r^3 \rho g = q \cdot \frac{V}{d}34πr3ρg=q⋅dV
Solving for qqq:q=43πr3ρg⋅dVq = \frac{4}{3} \pi r^3 \rho g \cdot \frac{d}{V}q=34πr3ρg⋅Vd
This final equation allows you to compute the charge on each oil droplet using the measured radius rrr, the known density ρ\rhoρ of oil, gravitational acceleration ggg, plate separation ddd, and applied voltage VVV. Repeating this for 10 droplets will give you several values of qqq, from which you can estimate the elementary charge by finding a common factor or analyzing the smallest quantum of charge observed.
