Problem 91 - University Physics with Modern Physics

Question

The Millikan Oil-Drop Experiment. The charge of an electron was first measured by the American physicist Robert Millikan during \(1909-1913 .\) In his experiment, oil is sprayed in very fine drops (around \(10^{-4} \mathrm{mm}\) in diameter) into the space between two parallel horizontal plates separated by a distance \(d . \mathrm{A}\) potential difference \(V_{A B}\) is maintained between the parallel plates, causing a downward electric field between them. Some of the oil drops acquire a negative charge because of frictional effects or because of ionization of the surrounding air by \(x\) rays or radioactivity. The drops are observed through a microscope. (a) Show that an oil drop of radius \(r\) at rest between the plates will remain at rest if the magnitude of its charge is $$q=\frac{4 \pi}{3} \frac{\rho r^{3} g d}{V_{A B}}$$ where \(\rho\) is the density of the oil. (Ignore the buoyant force of the air.) By adjusting \(V_{A B}\) to keep a given drop at rest, the charge on that drop can be determined, provided its radius is known. (b) Millikan's ooil drops were much too small to measure their radii directly. Instead, Millikan determined \(r\) by cutting off the electric field and measuring the terminal speed \(v_{t}\) of the drop as it fell. (We discussed the concept of terminal speed in Section \(5.3 . )\) The viscous force \(F\) on a sphere of radius \(r\) moving with speed \(v\) through a fluid with viscosity \(\eta\) is given by Stokes's law: \(F=6 \pi \eta r v .\) When the drop is falling at \(v_{1}\) , the viscous force just balances the weight \(w=m g\) of the drop. Show that the magnitude of the charge on the drop is $$q=18 \pi \frac{d}{V_{A B}} \sqrt{\frac{\eta^{3} v_{t}^{3}}{2 \rho g}}$$ Within the limits of their experimental error, every one of the thousands of drops that Millikan and his coworkers measured had a charge equal to some small integer multiple of a basic charge \(e\) . That is, they found drops with charges of \(\pm 2 e, \quad \pm 5 e,\) and so on, but none with values such as 0.76\(e\) or 2.49\(e .\) A drop with charge \(-e\) has acquired one extra electron; if its charge is \(-2 e,\) it has acquired two extra electrons, and so on. (c) A charged oil drop in a Millikan oil-drop apparatus is observed to fall 1.00 mm at constant speed in 39.3 s if \(V_{A B}=0 .\) The same drop can be held at rest between two plates separated by 1.00 \(\mathrm{mm}\) if \(V_{A B}=9.16 \mathrm{V} .\) How many excess electrons has the drop acquired, and what is the radius of the drop? The viscosity of air is \(1.81 \times 10^{-5} \mathrm{N} \cdot \mathrm{s} / \mathrm{m}^{2},\) and the density of the oil is 824 \(\mathrm{kg} / \mathrm{m}^{3} .\)

Step-by-Step Solution

Answer

The drop has a radius of about \(4.45 \times 10^{-7} \text{m}\) and has 3 excess electrons.

Step 1: Equation for Charge Balance

To find the charge of an oil drop at rest, we balance the forces acting on it: electric force and gravitational force. The electric force on a charged particle is given by\[ F_{electric} = q \cdot E = q \cdot \frac{V_{AB}}{d} \]where \( q \) is the magnitude of the charge, \( V_{AB} \) is the potential difference, and \( d \) is the separation between the plates.The gravitational force on the drop is\[ F_{gravity} = m \cdot g = \rho \cdot \frac{4 \pi}{3} r^3 \cdot g \]where \( \rho \) is the density of the oil, and \( r \) is the radius of the drop.For the drop to be at rest, \( F_{electric} = F_{gravity} \). Therefore,\[ q \cdot \frac{V_{AB}}{d} = \rho \cdot \frac{4 \pi}{3} \cdot r^3 \cdot g \]

Step 2: Expression for Charge q

Rearranging the equation from the previous step for \( q \):\[ q = \frac{4 \pi}{3} \cdot \frac{\rho \cdot r^3 \cdot g \cdot d}{V_{AB}} \]

Step 3: Calculating Radius using Terminal Speed

Since we cannot measure the radius directly, we use the concept of terminal velocity.According to Stokes's law, the viscous force is:\[ F_{viscous} = 6 \pi \eta r v_t \]For terminal velocity, viscous force equals gravitational force:\[ 6 \pi \eta r v_t = \rho \cdot \frac{4 \pi}{3} r^3 g \]Rearranging, we find:\[ r = \sqrt{\frac{9 \eta v_t}{2 \rho g}} \]

Step 4: Expression for Charge q with Terminal Speed

Substitute the expression for the radius \( r \) obtained in the previous step into the charge expression.We know:\[ q = \frac{4 \pi}{3} \frac{\rho r^3 g d}{V_{AB}} \]Substituting the expression for \( r \):\[ r = \sqrt{\frac{9 \eta v_t}{2 \rho g}} \]\[ q = \frac{4 \pi}{3} \frac{\rho \left( \sqrt{\frac{9 \eta v_t}{2 \rho g}} \right)^3 g d}{V_{AB}} \]This simplifies to:\[ q = 18 \pi \frac{d}{V_{AB}} \sqrt{\frac{\eta^3 v_t^3}{2 \rho g}} \]

Step 5: Observed Drop Information

The drop falls at terminal speed with \( V_{AB} = 0 \), which means it travels 1.00 mm (0.001 m) in 39.3 s. This gives us the terminal velocity \( v_t \):\[ v_t = \frac{0.001 \, \text{m}}{39.3 \, \text{s}} = 2.54 \times 10^{-5} \, \text{m/s} \]

Step 6: Calculate Radius and Charge

Use the expressions derived:1. Calculate the radius using the terminal velocity \[ r = \sqrt{\frac{9 \eta v_t}{2 \rho g}} \] \[ r = \sqrt{\frac{9 \times 1.81 \times 10^{-5} \, \text{Pa} \cdot \text{s} \times 2.54 \times 10^{-5} \, \text{m/s}}{2 \times 824 \, \text{kg/m}^3 \times 9.8 \, \text{m/s}^2}} = \text{approximately} \, 4.45 \times 10^{-7} \, \text{m} \]2. Calculate charge using the known \( V_{AB} = 9.16 \, \text{V} \) and radius. \[ q = 18 \pi \frac{0.001}{9.16} \sqrt{\frac{(1.81 \times 10^{-5})^3 (2.54 \times 10^{-5})^3}{2 \times 824 \times 9.8}} \approx -4.8 \times 10^{-19} \text{C} \]

Step 7: Determine Excess Electrons

The number of excess electrons is given by the charge \( q \) divided by the elementary charge \( e \approx 1.6 \times 10^{-19} \, \text{C} \).\[ n = \frac{|q|}{e} = \frac{4.8 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 3 \]

Key Concepts

Electric Charge Measurement

The concept of electric charge measurement is fundamental to understanding how forces interact in electric fields. In the Millikan Oil-Drop Experiment, the quantity of electric charge on a small oil drop can be measured by scrutinizing the balance of forces acting on it. The two main forces are the electric force, exerted by the field between charged plates, and the gravitational force pulling the drop downward.

For a drop to stay at rest, these forces must be equal. The electric force is calculated using the formula:
- Electric Force: \( F_{electric} = q \cdot E = q \cdot \frac{V_{AB}}{d} \)
Here, \( q \) is the charge, \( V_{AB} \) is the potential difference, and \( d \) is the space between the plates. Meanwhile, the gravitational force is determined by:
- Gravitational Force: \( F_{gravity} = m \cdot g = \rho \cdot \frac{4 \pi}{3} r^3 \cdot g \)
where \( \rho \) is the oil density, and \( r \) is the radius of the oil drop.

By setting these two expressions equal, we can solve for \( q \), the charge of the drop, illustrating the precise methodology of measuring electric charge in an experimental setup like Millikan's.

This careful balance and measurement process was pivotal, as it beautifully exposed the quantization of electric charge, showing it as a basic, indivisible unit, or a small integer multiple of electrons.

Terminal Velocity

Terminal velocity is a concept deeply integrated into the Millikan Oil-Drop Experiment illustrating how objects in fluid environments reach a constant speed. This terminal speed occurs when the forces acting on a falling object—gravitational and viscous forces—are in perfect equilibrium.

The oil drop achieves terminal velocity when the downward gravitational pull is precisely counterbalanced by the upward drag or viscous force exerted on it by the air it moves through. This is crucial because you can calculate the drop's radius and charge once it's known.

In the context of the experiment, when the oil drop is left to fall freely without the impact of an electric field, it attains terminal velocity given by:
- Terminal Speed Calculation: \( v_t = \frac{0.001 \, \text{m}}{39.3 \, \text{s}} = 2.54 \times 10^{-5} \, \text{m/s} \)
This speed was gathered by observing the drop over a known distance and time. Through this, Millikan could effectively measure very small particles, leading to a bigger picture of measuring charges. Terminal velocity allowed precise determination of the oil drop’s radius, thus the electric charge when tied with Stokes's Law.

Stokes's Law

Stokes's Law is a pivotal idea, especially in the context of the Millikan Oil-Drop Experiment. It provides a means to understand the viscous forces experienced by a sphere moving through a fluid. Stokes's Law expresses the drag force on a sphere through:
- Viscous Force: \( F = 6 \pi \eta r v \)
Here, \( \eta \) represents the fluid's viscosity, \( r \) the radius of the sphere, and \( v \) the sphere's velocity. In this experiment, the viscous force effectively opposes the gravitational force when the drop moves at terminal velocity, allowing the measurement of the drop's size and charge.

In Millikan's work, once the drop reaches terminal velocity, the viscous force counteracts gravity perfectly, enabling precise computation of the drop's properties. By utilizing Stokes's Law and established formulas, the exact measurement of the electric charge was possible through an indirect method.

Stokes's Law, therefore, not only facilitated understanding interactions at micro levels but also bolstered the profound implication that all charges were multiples of a fundamental unit, the electron's charge. This mathematical relationship was fundamental in reinforcing the quantized nature of electric charge ubiquitously in physics.