Contents

Preface
1What is Physics?
2Motion in One Dimension
3Acceleration
4Forces and Newton’s Laws of Motion
5Motion in Two Dimensions
6Circular and Rotational Motion
7Newton's Law of Gravitation
8Momentum
9Work, Energy, and Simple Machines
10Special Relativity
11Thermal Energy, Heat, and Work
12Thermodynamics
13Waves and Their Properties
14Sound
15Light
16Mirrors and Lenses
- Introduction
- 16.1Reflection
- 16.2Refraction
- 16.3Lenses
- Key Terms
- Section Summary
- Key Equations
- Chapter Review
17Diffraction and Interference
18Static Electricity
19Electrical Circuits
20Magnetism
21The Quantum Nature of Light
22The Atom
23Particle Physics
- Introduction
- 23.1The Four Fundamental Forces
- 23.2Quarks
- 23.3The Unification of Forces
- Key Terms
- Section Summary
- Chapter Review
AReference Tables
Index

Coulomb's law

Section Learning Objectives

By the end of this section, you will be able to do the following:

Describe Coulomb’s law verbally and mathematically
Solve problems involving Coulomb’s law

Section Key Terms

Coulomb’s law

inverse-square law

More than 100 years before Thomson and Rutherford discovered the fundamental particles that carry positive and negative electric charges, the French scientist Charles-Augustin de Coulomb mathematically described the force between charged objects. Doing so required careful measurements of forces between charged spheres, for which he built an ingenious device called a torsion balance.

This device, shown in Figure 18.15, contains an insulating rod that is hanging by a thread inside a glass-walled enclosure. At one end of the rod is the metallic sphere A. When no charge is on this sphere, it touches sphere B. Coulomb would touch the spheres with a third metallic ball (shown at the bottom of the diagram) that was charged. An unknown amount of charge would distribute evenly between spheres A and B, which would then repel each other, because like charges repel. This force would cause sphere A to rotate away from sphere B, thus twisting the wire until the torsion in the wire balanced the electrical force. Coulomb then turned the knob at the top, which allowed him to rotate the thread, thus bringing sphere A closer to sphere B. He found that bringing sphere A twice as close to sphere B required increasing the torsion by a factor of four. Bringing the sphere three times closer required a ninefold increase in the torsion. From this type of measurement, he deduced that the electrical force between the spheres was inversely proportional to the distance squared between the spheres. In other words,

F \propto \frac{1}{r^{2,}} F \propto \frac{1}{r^{2,}}

18.5

where r is the distance between the spheres.

An electrical charge distributes itself equally between two conducting spheres of the same size. Knowing this allowed Coulomb to divide an unknown charge in half. Repeating this process would produce a sphere with one quarter of the initial charge, and so on. Using this technique, he measured the force between spheres A and B when they were charged with different amounts of charge. These measurements led him to deduce that the force was proportional to the charge on each sphere, or

F \propto q_{A} q_{B,} F \propto q_{A} q_{B,}

18.6

where $q_{A} q_{A}$ is the charge on sphere A, and $q_{B} q_{B}$ is the charge on sphere B.

This figure consists of seven drawings, the largest of which is labeled “Fig. 1”. The largest structure appears as follows: there is a cap at the top, and below that are a disc, a long cylinder, a wider disc, and a shorter cylinder. The six smaller drawings show the parts from which the large structure is assembled. One part, labeled “Fig. 5” and placed nearest the upper left corner, consists of a vertical rod with a hook at the top, a sphere at the bottom, and a disc over a hemisphere near the middle. Another part, labeled “Fig. 4”, lies parallel to the bottom edge of the figure. It consists of a rod with a short line and a small sphere attached to one end. Three parts, labeled “Fig. 2”, are arranged in a column on the right side of the figure. Of these, the upper part consists of a cap-shaped structure with a horizontal projection and a vertical projection; the middle part consists of a cylinder with a wider discover it; and the lower part consists of another cylinder with a lip. The final part, labeled “Fig. 3” and placed near the bottom right corner of the figure, is a T-shaped structure with a small sphere at each end of the horizontal line.

Figure 18.15 A drawing of Coulomb’s torsion balance, which he used to measure the electrical force between charged spheres. (credit: Charles-Augustin de Coulomb)

Combining these two proportionalities, he proposed the following expression to describe the force between the charged spheres.

F = \frac{k q_{1} q_{2}}{r^{2}} F = \frac{k q_{1} q_{2}}{r^{2}}

18.7

This equation is known as Coulomb’s law, and it describes the electrostatic force between charged objects. The constant of proportionality k is called Coulomb’s constant. In SI units, the constant k has the value $k = 8.99 \times 10^{9} N \cdot m^{2} {/C}^{2.} k = 8.99$

The direction of the force is along the line joining the centers of the two objects. If the two charges are of opposite signs, Coulomb’s law gives a negative result. This means that the force between the particles is attractive. If the two charges have the same signs, Coulomb’s law gives a positive result. This means that the force between the particles is repulsive. For example, if both $q_{1} q_{1}$ and $q_{2} q_{2}$ are negative or if both are positive, the force between them is repulsive. This is shown in Figure 18.16(a). If $q_{1} q_{1}$ is a negative charge and $q_{2} q_{2}$ is a positive charge (or vice versa), then the charges are different, so the force between them is attractive. This is shown in Figure 18.16(b).

This figure has two parts. Part a shows two small spheres of equal size, labeled “q subscript 1” and “q subscript 2”, separated by a distance marked as “r”. Each sphere contains a minus sign. Two arrows, labeled “F subscript 1,2” and “F subscript 2,1”, point from the spheres away from each other. Part b also shows two small spheres of equal size, labeled “q subscript 1” and “q subscript 2”, separated by a larger distance marked as “r”. One of these spheres contains a minus sign and the other contains a plus sign. Two arrows, labeled 7F subscript 1,2” and “F subscript 2,1”, point from the spheres toward each other.

Figure 18.16 The magnitude of the electrostatic force F between point charges q₁ and q₂ separated by a distance r is given by Coulomb’s law. Note that Newton’s third law (every force exerted creates an equal and opposite force) applies as usual—the force (F₁,₂) on q₁ is equal in magnitude and opposite in direction to the force (F₂,₁) it exerts on q₂. (a) Like charges. (b) Unlike charges.

Note that Coulomb’s law applies only to charged objects that are not moving with respect to each other. The law says that the force is proportional to the amount of charge on each object and inversely proportional to the square of the distance between the objects. If we double the charge $q_{1} q_{1}$ , for instance, then the force is doubled. If we double the distance between the objects, then the force between them decreases by a factor of $2^{2} = 4 2^{2} = 4$ . Although Coulomb’s law is true in general, it is easiest to apply to spherical objects or to objects that are much smaller than the distance between the objects (in which case, the objects can be approximated as spheres).

Coulomb’s law is an example of an inverse-square law, which means the force depends on the square of the denominator. Another inverse-square law is Newton’s law of universal gravitation, which is $F = G m_{1} m_{2} / r^{2} F = G m_{1} m_{2} / r^{2}$ . Although these laws are similar, they differ in two important respects: (i) The gravitational constant G is much, much smaller than k
( $G = 6.67 \times 10^{- 11} m^{3} /kg \cdot s^{2} G = 6.67$ ); and (ii) only one type of mass exists, whereas two types of electric charge exist. These two differences explain why gravity is so much weaker than the electrostatic force and why gravity is only attractive, whereas the electrostatic force can be attractive or repulsive.

Finally, note that Coulomb measured the distance between the spheres from the centers of each sphere. He did not explain this assumption in his original papers, but it turns out to be valid. From outside a uniform spherical distribution of charge, it can be treated as if all the charge were located at the center of the sphere.

Watch Physics

Electrostatics (part 1): Introduction to charge and Coulomb's law

This video explains the basics of Coulomb’s law. Note that the lecturer uses d for the distance between the center of the particles instead of r.

Electrostatics (part 1): Introduction to Charge and Coulomb's Law

Grasp Check

True or false—If one particle carries a positive charge and another carries a negative charge, then the force between them is attractive.

true
false

Snap Lab

Hovering plastic

In this lab, you will use electrostatics to hover a thin piece of plastic in the air.

Balloon
Light plastic bag (e.g., produce bag from grocery store)

Instructions

Procedure

Cut the plastic bag to make a plastic loop about 2 inches wide.
Inflate the balloon.
Charge the balloon by rubbing it on your clothes.
Charge the plastic loop by placing it on a nonmetallic surface and rubbing it with a cloth.
Hold the balloon in one hand, and in the other hand hold the plastic loop above the balloon. If the loop clings too much to your hand, recruit a friend to hold the strip above the balloon with both hands. Now let go of the plastic loop, and maneuver the balloon under the plastic loop to keep it hovering in the air above the balloon.

Grasp Check

How does the balloon keep the plastic loop hovering?

The balloon and the loop are both negatively charged. This will help the balloon keep the plastic loop hovering.
The balloon is charged, while the plastic loop is neutral.This will help the balloon keep the plastic loop hovering.
The balloon and the loop are both positively charged. This will help the balloon keep the plastic loop hovering.
The balloon is positively charged, while the plastic loop is negatively charged. This will help the balloon keep the plastic loop hovering.

Worked Example

Using Coulomb’s law to find the force between charged objects

Suppose Coulomb measures a force of $20 \times 10^{−6} N 20$ between the two charged spheres when they are separated by 5.0 cm. By turning the dial at the top of the torsion balance, he approaches the spheres so that they are separated by 3.0 cm. Which force does he measure now?

Strategy

Apply Coulomb’s law to the situation before and after the spheres are brought closer together. Although we do not know the charges on the spheres, we do know that they remain the same. We call these unknown but constant charges $q_{1} q_{1}$ and $q_{2} q_{2}$ . Because these charges appear as a product in Coulomb’s law, they form a single unknown. We thus have two equations and two unknowns, which we can solve. The first unknown is the force (which we call $F_{f} F_{f}$ ) when the spheres are 3.0 cm apart, and the second is $q_{1} q_{2} q_{1} q_{2}$ .

Use the following notation: When the charges are 5.0 cm apart, the force is $F_{i} = 20 \times 10^{−6} N F_{i} = 20$ and $r_{i} = 5.0 cm = 0.050 m r_{i} = 5.0$ , where the subscript i means initial. Once the charges are brought closer together, we know $r_{f} = 3.0 cm = 0.030 m r_{f} = 3.0$ , where the subscript f means final.

Show solution

Solution

Coulomb’s law applied to the spheres in their initial positions gives

F_{i} = \frac{k q_{1} q_{2}}{r_{i}^{2.}} . F_{i} = \frac{k q_{1} q_{2}}{r_{i}^{2.}} .

18.8

Coulomb’s law applied to the spheres in their final positions gives

F_{f} = \frac{k q_{1} q_{2}}{r_{f}^{2.}} . F_{f} = \frac{k q_{1} q_{2}}{r_{f}^{2.}} .

18.9

Dividing the second equation by the first and solving for the final force $F_{f} F_{f}$ leads to

\begin{matrix} \frac{F_{f}}{F_{i}} & = & \frac{k q_{1} q_{2} / r_{f}^{2}}{k q_{1} q_{2} / r_{i}^{2}} . \\ = & \frac{r_{i}^{2}}{r_{f}^{2}} \\ F_{f} & = & F_{i} \frac{r_{i}^{2}}{r_{f}^{2}} . \end{matrix} \begin{matrix} \frac{F_{f}}{F_{i}} & = & \frac{k q_{1} q_{2} / r_{f}^{2}}{k q_{1} q_{2} / r_{i}^{2}} . \\ = & \frac{r_{i}^{2}}{r_{f}^{2}} \\ F_{f} & = & F_{i} \frac{r_{i}^{2}}{r_{f}^{2}} . \end{matrix}

18.10

Inserting the known quantities yields

\begin{matrix} F_{f} & = & F_{i} \frac{r_{i}^{2}}{r_{f}^{2}} \\ = & (20 \times 10^{−6} N) \frac{{(0.050 m)}^{2}}{{(0.030 m)}^{2}} \\ = & 56 \times 10^{−6} N \\ = & 5.6 \times 10^{−5} N. \end{matrix} \begin{matrix} F_{f} & = & F_{i} \frac{r_{i}^{2}}{r_{f}^{2}} \\ = & (20) \frac{{(0.050)}^{2}}{{(0.030)}^{2}} \\ = & 56 \\ = & 5.6 \end{matrix}

18.11

The force acts along the line joining the centers of the spheres. Because the same type of charge is on each sphere, the force is repulsive.

Discussion

As expected, the force between the charges is greater when they are 3.0 cm apart than when they are 5.0 cm apart. Note that although it is a good habit to convert cm to m (because the constant k is in SI units), it is not necessary in this problem, because the distances cancel out.

We can also solve for the second unknown $| q_{1} q_{2} | | q_{1} q_{2} |$ . By using the first equation, we find

\begin{matrix} F_{f} & = & \frac{k q_{1} q_{2}}{r_{i}^{2}} \\ q_{1} q_{2} & = & \frac{F_{i} r_{i}^{2}}{k} \\ = & \frac{(20 \times 10^{−6} N) {(0.050 m)}^{2}}{8.99 \times 10^{9} N \cdot m^{2} {/C}^{2}} \\ = & 5.6 \times 10^{−18} C^{2} \end{matrix} \begin{matrix} F_{f} & = & \frac{k q_{1} q_{2}}{r_{i}^{2}} \\ q_{1} q_{2} & = & \frac{F_{i} r_{i}^{2}}{k} \\ = & \frac{(20) {(0.050)}^{2}}{8.99} \\ = & 5.6 \end{matrix}

18.12

Note how the units cancel in the second-to-last line. Had we not converted cm to m, this would not occur, and the result would be incorrect. Finally, because the charge on each sphere is the same, we can further deduce that

q_{1} = q_{2} = \pm \sqrt{5.6 \times 10^{−18} C^{2}} = \pm 2.4 nC. q_{1} = q_{2} = \pm \sqrt{5.6} = \pm 2.4

18.13

Worked Example

Using Coulomb’s law to find the distance between charged objects

An engineer measures the force between two ink drops by measuring their acceleration and their diameter. She finds that each member of a pair of ink drops exerts a repulsive force of $F = 5.5 mN F = 5.5$ on its partner. If each ink drop carries a charge $q_{ink drop} = −1 \times 10^{−10} C q_{ink drop} = −1$ , how far apart are the ink drops?

Strategy

We know the force and the charge on each ink drop, so we can solve Coulomb’s law for the distance r between the ink drops. Do not forget to convert the force into SI units: $F = 5.5 mN = 5.5 \times 10^{−3} N. F = 5.5$

Show solution

Solution

The charges in Coulomb’s law are $q_{1} = q_{2} = q_{ink drop}^{}, q_{1} = q_{2} = q_{ink drop}^{},$ so the numerator in Coulomb’s law takes the form $q_{1} q_{2} = q_{ink drop}^{2} q_{1} q_{2} = q_{ink drop}^{2}$ . Inserting this into Coulomb’s law and solving for the distance r gives

\begin{matrix} F & = & \frac{k q_{ink drop}^{2}}{r^{2}} \\ r & = & \pm \sqrt{\frac{k q_{ink drop}^{2}}{F}} \\ = & \pm \sqrt{\frac{(8.99 \times 10^{9} N \cdot m^{2} {/C}^{2}) {(−1 \times 10^{−10} C)}^{2}}{5.5 \times 10^{−3} N}} \\ = & \pm 1.3 \times 10^{−4} m \end{matrix} \begin{matrix} F & = & \frac{k q_{ink drop}^{2}}{r^{2}} \\ r & = & \pm \sqrt{\frac{k q_{ink drop}^{2}}{F}} \\ = & \pm \sqrt{\frac{(8.99) {(−1)}^{2}}{5.5 \times 10^{−3} N}} \\ = & \pm 1.3 \times 10^{−4} m \end{matrix}

18.14

or 130 microns (about one-tenth of a millimeter).

Discussion

The plus-minus sign means that we do not know which ink drop is to the right and which is to the left, but that is not important, because both ink drops are the same.

Practice Problems

11.

A charge of −4 × 10⁻⁹ C is a distance of 3 cm from a charge of 3 × 10⁻⁹ C . What is the magnitude and direction of the force between them?

1.2 × 10⁻⁴ N, and the force is attractive
1.2 × 10¹⁴ N, and the force is attractive
6.74 × 10²³ N, and the force is attractive
3.6 × 10⁻⁶ N, and the force is attractive

12.

Two charges are repelled by a force of 2.0 N. If the distance between them triples, what is the force between the charges?

0.22 N
0.67 N
2.0 N
18.0 N

Check Your Understanding

13 .

How are electrostatic force and charge related?

The force is proportional to the product of two charges.
The force is inversely proportional to the product of two charges.
The force is proportional to any one of the charges between which the force is acting.
The force is inversely proportional to any one of the charges between which the force is acting.

14.

Why is Coulomb’s law called an inverse-square law?

because the force is proportional to the inverse of the distance squared between charges
because the force is proportional to the product of two charges
because the force is proportional to the inverse of the product of two charges
because the force is proportional to the distance squared between charges