In the famous Millikan oil-drop experiment, tiny spherical...

In the famous Millikan oil-drop experiment, tiny spherical droplets of oil are sprayed into a uniform vertical electric field. The drops get a very small charge (just a few electrons) due to friction with the atomizer as they are sprayed. The field is adjusted until the drop (which is viewed through a small telescope) is just balanced against gravity and therefore remains stationary. Using the measured value of the electric field, we can calculate the charge on the drop and from this calculate the charge $e$ of the electron. In one apparatus the drops are $1.10 \mu \mathrm{m}$ in diameter, and the oil has a density of $0.850 \mathrm{~g} / \mathrm{cm}^{3}$.(a) If the drops are negatively charged, which way should the electric field point to hold them stationary (up or down)? (b) Why? (c) If a certain drop contains four excess electrons, what magnitude electric field is needed to hold it stationary? (d) You measure a balancing field of $5183 \mathrm{~N} / \mathrm{C}$ for another drop. How many excess electrons are on this drop? Example $16-5$

Key Concepts

Equilibrium Condition

When the oil drop is stationary, it is in a state of equilibrium where the net force acting on it is zero. This means that the upward electric force exactly counterbalances the downward gravitational force, allowing one to set up the equation qE = mg to solve for the charge on the drop.

Charge Quantization

The charge quantization principle states that the charge on any object, including the oil drop, is an integer multiple of the elementary charge. This concept is fundamental in the experiment, as measuring the net charge on a drop and recognizing that it comes in discrete units helps to confirm that charge is quantized.

Gravitational Force

Gravitational force is the attractive force acting on an object due to gravity (F = mg) and always acts downward toward the center of the Earth. In the experiment, it provides the constant force that must be balanced by the electric force to keep the oil drop stationary.

Electric Force

The electric force on a charged particle is given by the product of the charge and the electric field (F = qE). The direction of this force depends on the sign of the charge relative to the field direction; for a negatively charged particle, the force is in the opposite direction to the electric field.

Uniform Electric Field

A uniform electric field is one in which the electric field strength and direction are constant in space. In the context of the experiment, it exerts a constant electric force on a charged droplet, making it possible to finely adjust the field to counteract gravity and achieve equilibrium.

Millikan Oil-Drop Experiment

This experiment is a classic demonstration of measuring the elementary charge by balancing gravitational force against the electric force acting on tiny charged oil droplets. It shows how experimental techniques can reveal fundamental quantization of electric charge by observing the behavior of these droplets in a uniform electric field.

Transcript

00:01 For this problem on the topic of electrostatics, we are considering a certain setup of the milliken oil drop experiment, and we have oil drops in this case that are 1 .1 micrometers in diameter, and the oil has a density of 0 .85 grams per cubic centimeter.

00:16 If the drops are all negatively charged, we want to know which way the electric field should point in order to hold them stationary.

00:22 We also want to know why this is so.

00:26 And then if a certain drop contains four axis electrons, we want to know why.

00:30 What magnitude electric field would be required to hold it stationary.

00:35 If you then measure a balancing field of 5 ,183 newtons per kulum for another drop, we want to know how many excess electrons are on that drop.

00:46 Now the electric field to hold the oil drop stationary should point downward.

00:56 And the reason for this is that gravity will pull the drop downward.

01:02 And so the electric force must act upward in order to keep the oil wrap stationary.

01:11 But since the drop is negatively charged, the electric field must point downward in order for the force to point upward.

01:31 Next for part c, we know that the resultant force in the vertical direction is the electric force minus the weight of the oil drop.

01:52 And by newton's second law, this must equal the mass of the oil drop times its vertical acceleration, which we know is equal to zero...

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