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
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The learning objectives in this section will help your students master the following standards:
 (5) The student knows the nature of forces in the physical world. The student is expected to:
 (C)describe and calculate how the magnitude of the electrical force between two objects depends on their charges and the distance between them.
Section Key Terms
Coulomb’s law  inversesquare law 
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Teacher Support
This section presents Coulomb’s law and points out its similarities and differences with respect to Newton’s law of universal gravitation. The similarities include the inversesquare nature of the two laws and the analogous roles of mass and charge. The differences include the restriction of positive mass versus positive or negative charge.
[BL][OL]Discuss how Coulomb described this law long after Newton described the law of universal gravitation.
[AL]Ask why the law of force between electrostatic charge was discovered after that of gravity if gravity is weak compared to electrostatic forces.
More than 100 years before Thomson and Rutherford discovered the fundamental particles that carry positive and negative electric charges, the French scientist CharlesAugustin 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 glasswalled 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,
$$F\propto \frac{1}{{r}^{\mathrm{2,}}}$$
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
$$F\propto {q}_{\text{A}}{q}_{\text{B,}}$$
18.6
where ${q}_{\text{A}}$ is the charge on sphere A, and ${q}_{\text{B}}$ is the charge on sphere B.
Figure 18.15 A drawing of Coulomb’s torsion balance, which he used to measure the electrical force between charged spheres. (credit: CharlesAugustin de Coulomb)
Combining these two proportionalities, he proposed the following expression to describe the force between the charged spheres.
$$F=\frac{k{q}_{1}{q}_{2}}{{r}^{2}}$$
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 $k=8.99\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}{10}^{9}\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2.}}$
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 ${q}_{1}$ and ${q}_{2}$ are negative or if both are positive, the force between them is repulsive. This is shown in Figure 18.16(a). If ${q}_{1}$ is a negative charge and ${q}_{2}$ 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).
Figure 18.16 The magnitude of the electrostatic force F between point charges q_{1} and q_{2} 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 (F_{1},_{2}) on q_{1} is equal in magnitude and opposite in direction to the force (F_{2},_{1}) it exerts on q_{2}. (a) Like charges. (b) Unlike charges.
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Point out how the subscripts _{1}, _{2} means the force on object 1 due to object 2 (and vice versa).
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 ${q}_{1}$, 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 ${2}^{2}=4$ . 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 inversesquare law, which means the force depends on the square of the denominator. Another inversesquare law is Newton’s law of universal gravitation, which is $F=G{m}_{1}{m}_{2}/{r}^{2}$ . Although these laws are similar, they differ in two important respects: (i) The gravitational constant G is much, much smaller than k
( $G=6.67\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}{10}^{11}{\text{m}}^{3}\text{/kg}\cdot {\text{s}}^{2}$ ); 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.
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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 $20\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}{10}^{\mathrm{6}}\text{N}$ 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 ${q}_{1}$ and ${q}_{2}$ . 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 ${F}_{\text{f}}$ ) when the spheres are 3.0 cm apart, and the second is ${q}_{1}{q}_{2}$ .
Use the following notation: When the charges are 5.0 cm apart, the force is ${F}_{\text{i}}=20\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}{10}^{\mathrm{6}}\text{N}$ and ${r}_{\text{i}}=5.0\phantom{\rule{0ex}{0ex}}\text{cm}=0.050\phantom{\rule{0ex}{0ex}}\text{m}$, where the subscript i means initial. Once the charges are brought closer together, we know ${r}_{\text{f}}=3.0\phantom{\rule{0ex}{0ex}}\text{cm}=0.030\phantom{\rule{0ex}{0ex}}\text{m}$, where the subscript f means final.
Solution
Coulomb’s law applied to the spheres in their initial positions gives
$${F}_{\text{i}}=\frac{k{q}_{1}{q}_{2}}{{r}_{\text{i}}^{2.}}\text{.}$$
18.8
Coulomb’s law applied to the spheres in their final positions gives
$${F}_{\text{f}}=\frac{k{q}_{1}{q}_{2}}{{r}_{\text{f}}^{2.}}\text{.}$$
18.9
Dividing the second equation by the first and solving for the final force ${F}_{\text{f}}$ leads to
$$\begin{array}{ccc}\hfill \frac{{F}_{\text{f}}}{{F}_{\text{i}}}& \hfill =\hfill & \frac{k{q}_{1}{q}_{2}/{r}_{\text{f}}^{2}}{k{q}_{1}{q}_{2}/{r}_{\text{i}}^{2}}\text{.}\hfill \\ \hfill & \hfill =\hfill & \frac{{r}_{\text{i}}^{2}}{{r}_{\text{f}}^{2}}\hfill \\ \hfill {F}_{\text{f}}& \hfill =\hfill & {F}_{\text{i}}\frac{{r}_{\text{i}}^{2}}{{r}_{\text{f}}^{2}}\text{.}\hfill \end{array}$$
18.10
Inserting the known quantities yields
$$\begin{array}{ccc}{F}_{\text{f}}& =& {F}_{\text{i}}\frac{{r}_{\text{i}}^{2}}{{r}_{\text{f}}^{2}}\\ \hfill & \hfill =\hfill & \left(20\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}{10}^{\mathrm{6}}\text{N}\right)\frac{{\left(0.050\phantom{\rule{0ex}{0ex}}\text{m}\right)}^{2}}{{\left(0.030\phantom{\rule{0ex}{0ex}}\text{m}\right)}^{2}}\hfill \\ \hfill & \hfill =\hfill & 56\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}{10}^{\mathrm{6}}\text{N}\hfill \\ \hfill & \hfill =\hfill & 5.6\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}{10}^{\mathrm{5}}\text{N.}\hfill \end{array}$$
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 $\left{q}_{1}{q}_{2}\right$ . By using the first equation, we find
$$\begin{array}{ccc}\hfill {F}_{\text{f}}& =& \frac{k{q}_{1}{q}_{2}}{{r}_{\text{i}}^{2}}\hfill \\ \hfill {q}_{1}{q}_{2}& \hfill =\hfill & \frac{{F}_{\text{i}}{r}_{\text{i}}^{2}}{k}\hfill \\ \hfill & \hfill =\hfill & \frac{\left(20\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}{10}^{\mathrm{6}}\text{N}\right){\left(0.050\phantom{\rule{0ex}{0ex}}\text{m}\right)}^{2}}{8.99\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}{10}^{9}\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}}\hfill \\ \hfill & \hfill =\hfill & 5.6\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}{10}^{\mathrm{18}}{\text{C}}^{2}\hfill \end{array}$$
18.12
Note how the units cancel in the secondtolast 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
$${q}_{1}={q}_{2}=\pm \sqrt{5.6\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}{10}^{\mathrm{18}}{\text{C}}^{2}}=\pm 2.4\phantom{\rule{0ex}{0ex}}\text{nC.}$$
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 $F=5.5\phantom{\rule{0ex}{0ex}}\text{mN}$ on its partner. If each ink drop carries a charge ${q}_{\text{ink}\text{drop}}=\mathrm{1}\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}{10}^{\mathrm{10}}\text{C}$, 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: $F=5.5\phantom{\rule{0ex}{0ex}}\text{mN}=5.5\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}{10}^{\mathrm{3}}\text{N.}$
Solution
The charges in Coulomb’s law are ${q}_{1}={q}_{2}={q}_{\text{ink}\text{drop}}^{}\text{,}$ so the numerator in Coulomb’s law takes the form ${q}_{1}{q}_{2}={q}_{\text{ink}\text{drop}}^{2}$ . Inserting this into Coulomb’s law and solving for the distance r gives
$$\begin{array}{ccc}\hfill F& \hfill =\hfill & \frac{k{q}_{\text{ink}\text{drop}}^{2}}{{r}^{2}}\hfill \\ \hfill r& \hfill =\hfill & \pm \sqrt{\frac{k{q}_{\text{ink}\text{drop}}^{2}}{F}}\hfill \\ & =& \pm \sqrt{\frac{\left(8.99\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}{10}^{9}\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right){\left(\mathrm{1}\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}{10}^{\mathrm{10}}\text{C}\right)}^{2}}{5.5\times {10}^{\mathrm{3}}\text{N}}}\\ \hfill & \hfill =\hfill & \pm 1.3\times {10}^{\mathrm{4}}\text{m}\hfill \end{array}$$
18.14
or 130 microns (about onetenth of a millimeter).
Discussion
The plusminus 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 × 10^{14} N, and the force is attractive
 6.74 × 10^{23} N, and the force is attractive
 −ŷ, 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 inversesquare 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