# Learning Objectives

### Learning Objectives

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

• State Coulomb's law in terms of how the electrostatic force changes with the distance between two objects
• Calculate the electrostatic force between two point charges, such as electrons or protons
• Compare the electrostatic force to the gravitational attraction for a proton and an electron; for a human and Earth

The information presented in this section supports the following AP® learning objectives and science practices:

• 3.A.3.3 The student is able to describe a force as an interaction between two objects and identify both objects for any force. (S.P. 1.4)
• 3.A.3.4 The student is able to make claims about the force on an object due to the presence of other objects with the same property: mass, electric charge. (S.P. 6.1, 6.4)
• 3.C.2.1 The student is able to use Coulomb's law qualitatively and quantitatively to make predictions about the interaction between two electric point charges—interactions between collections of electric point charges are not covered in Physics 1 and instead are restricted to Physics 2. (S.P. 2.2, 6.4)
• 3.C.2.2 The student is able to connect the concepts of gravitational force and electric force to compare similarities and differences between the forces. (S.P. 7.2)
Figure 1.18 This NASA image of Arp 87 shows the result of a strong gravitational attraction between two galaxies. In contrast, at the subatomic level, the electrostatic attraction between two objects, such as an electron and a proton, is far greater than their mutual attraction due to gravity. (NASA/HST)

Through the work of scientists in the late eighteenth century, the main features of the electrostatic force—the existence of two types of charge, the observation that like charges repel, unlike charges attract, and the decrease of force with distance—were eventually refined, and expressed as a mathematical formula. The mathematical formula for the electrostatic force is called Coulomb's law after the French physicist Charles Coulomb (1736–1806), who performed experiments and first proposed a formula to calculate it.

### Coulomb's Law

1.3 $F=k|q1q2|r2F=k|q1q2|r2 size 12{F=k { {q rSub { size 8{1} } q rSub { size 8{2} } } over {r rSup { size 8{2} } } } } {}$

Coulomb's law calculates the magnitude of the force $FF$ between two point charges, $q1q1 size 12{q rSub { size 8{1} } } {}$ and $q2q2 size 12{q rSub { size 8{2} } } {}$, separated by a distance $rr$. In SI units, the constant $kk$ is equal to

1.4 $k=8.988×109N⋅m2C2≈8.99×109N⋅m2C2.k=8.988×109N⋅m2C2≈8.99×109N⋅m2C2. size 12{k=8 "." "988" times "10" rSup { size 8{9} } { {N cdot m rSup { size 8{2} } } over {C rSup { size 8{2} } } } approx 9 "." "00" times "10" rSup { size 8{9} } { {N cdot m rSup { size 8{2} } } over {C rSup { size 8{2} } } } } {}$

The electrostatic force is a vector quantity and is expressed in units of newtons. The force is understood to be along the line joining the two charges (see Figure 1.19).

Although the formula for Coulomb's law is simple, it was no mean task to prove it. The experiments Coulomb did, with the primitive equipment then available, were difficult. Modern experiments have verified Coulomb's law to great precision. For example, it has been shown that the force is inversely proportional to distance between two objects squared $F∝1/r2F∝1/r2 size 12{ left (F prop {1} slash {r rSup { size 8{2} } } right )} {}$ to an accuracy of 1 part in $1016.1016. size 12{"10" rSup { size 8{"16"} } } {}$ No exceptions have ever been found, even at the small distances within the atom.

Figure 1.19 The magnitude of the electrostatic force $FF size 12{F} {}$ between point charges $q1q1 size 12{q rSub { size 8{1} } } {}$ and $q2q2 size 12{q rSub { size 8{2} } } {}$ separated by a distance $rr size 12{F} {}$ 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 on $q1q1 size 12{q rSub { size 8{1} } } {}$ is equal in magnitude and opposite in direction to the force it exerts on $q2q2 size 12{q rSub { size 8{2} } } {}$. (a) Like charges. (b) Unlike charges.

### Making Connections: Comparing Gravitational and Electrostatic Forces

Recall that the gravitational force (Newton's law of gravitation) quantifies force as $Fs=GmMr2Fs=GmMr2$.

The comparison between the two forces—gravitational and electrostatic—shows some similarities and differences. Gravitational force is proportional to the masses of interacting objects, and the electrostatic force is proportional to the magnitudes of the charges of interacting objects. Hence both forces are proportional to a property that represents the strength of interaction for a given field. In addition, both forces are inversely proportional to the square of the distances between them. It may seem that the two forces are related but that is not the case. In fact, there are huge variations in the magnitudes of the two forces as they depend on different parameters and different mechanisms. For electrons (or protons), electrostatic force is dominant and is much greater than the gravitational force. On the other hand, gravitational force is generally dominant for objects with large masses. Another major difference between the two forces is that gravitational force can only be attractive, whereas electrostatic could be attractive or repulsive, depending on the sign of charges; unlike charges attract and like charges repel.

### Example 1.1How Strong is the Coulomb Force Relative to the Gravitational Force?

Compare the electrostatic force between an electron and proton separated by $0.530×10−10m0.530×10−10m size 12{0 "." "530" times "10" rSup { size 8{ - "10"} } m} {}$ with the gravitational force between them. This distance is their average separation in a hydrogen atom.

Strategy

To compare the two forces, we first compute the electrostatic force using Coulomb's law, $F=k|q1q2|r2F=k|q1q2|r2 size 12{F=k { {q rSub { size 8{1} } q rSub { size 8{2} } } over {r rSup { size 8{2} } } } } {}$. We then calculate the gravitational force using Newton's universal law of gravitation. Finally, we take a ratio to see how the forces compare in magnitude.

Solution

Entering the given and known information about the charges and separation of the electron and proton into the expression of Coulomb's law yields

1.5 $F=k|q1q2|r2F=k|q1q2|r2 size 12{F=k { {q rSub { size 8{1 } } q rSub { size 8{2} } } over {r rSup { size 8{2} } } } } {}$
1.6 =8.99× 109N⋅ m2/C2×(1.60× 10–19C)(1.60× 10–19 C)(0.530× 10–10m)2.=8.99× 109N⋅ m2/C2×(1.60× 10–19C)(1.60× 10–19 C)(0.530× 10–10m)2.alignl { stack { size 12{" "= left (9 "." "00 " times " 10" rSup { size 8{9} } N cdot " m" rSup { size 8{2} } /C rSup { size 8{2} } right ) times { { $$"-1" "." "60 " times " 10" rSup { size 8{"-19"} } C$$ $$1 "." "60" times " 10" rSup { size 8{"-19 "} } C$$ } over { $$0 "." "530 " times " 10" rSup { size 8{"-10"} } m$$ rSup { size 8{2} } } } } {} # {} } } {}

Thus the Coulomb force is

1.7 $F=8.19× 10–8N.F=8.19× 10–8N. size 12{F=" -8" "." "20 " times " 10" rSup { size 8{"-8"} } N} {}$

The charges are opposite in sign, so this is an attractive force. This is a very large force for an electron—it would cause an acceleration of $8.99×1022m/s28.99×1022m/s2 size 12{9 "." "00" times "10" rSup { size 8{"22"} } {m} slash {s rSup { size 8{2} } } } {}$, verification is left as an end-of-section problem.The gravitational force is given by Newton's law of gravitation as

1.8 $FG= G mMr2,FG= G mMr2, size 12{F rSub { size 8{G} } =" G " { {"mM"} over {r rSup { size 8{2} } } } } {}$

where $G=6.67×10−11N⋅m2/kg2G=6.67×10−11N⋅m2/kg2 size 12{G=6 "." "67" times "10" rSup { size 8{ - "11"} } {N cdot m rSup { size 8{2} } } slash { ital "kg" rSup { size 8{2} } } } {}$. Here $mm$ and $MM$ represent the electron and proton masses, which can be found in the appendices. Entering values for the knowns yields

1.9 $FG=(6.67× 10–11N⋅m2/kg2)×(9.11× 10–31kg)(1.67×10–27kg)(0.530× 10–10m)2=3.61× 10–47N.FG=(6.67× 10–11N⋅m2/kg2)×(9.11× 10–31kg)(1.67×10–27kg)(0.530× 10–10m)2=3.61× 10–47N.$

This is also an attractive force, although it is traditionally shown as positive since gravitational force is always attractive. The ratio of the magnitude of the electrostatic force to gravitational force in this case is, thus

1.10 $FFG= 2.27 × 1039.FFG= 2.27 × 1039. size 12{ { {F} over {F rSub { size 8{G} } } } =" 2" "." "27 " times " 10" rSup { size 8{"39"} } } {}$

Discussion

This is a remarkably large ratio! Note that this will be the ratio of electrostatic force to gravitational force for an electron and a proton at any distance, taking the ratio before entering numerical values shows that the distance cancels. This ratio gives some indication of just how much larger the Coulomb force is than the gravitational force between two of the most common particles in nature.

As the example implies, gravitational force is completely negligible on a small scale, where the interactions of individual charged particles are important. On a large scale, such as between Earth and a person, the reverse is true. Most objects are nearly electrically neutral, and so attractive and repulsive Coulomb forces nearly cancel. Gravitational force on a large scale dominates interactions between large objects because it is always attractive, while Coulomb forces tend to cancel.