Skip to content Skip to navigation

Portal

You are here: Home » Content » Coulomb's Law

Navigation

Recently Viewed

This feature requires Javascript to be enabled.
 

Coulomb's Law

Module by: First Last. E-mail the author

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

  • Describe the electric force, both qualitatively and quantitatively
  • Calculate the force that charges exert on each other
  • Determine the direction of the electric force for different source charges
  • Correctly describe and apply the superposition principle for multiple source charges

Note: You are viewing an old style version of this document. The new style version is available here.

Experiments with electric charges have shown that if two objects each have electric charge, then they exert an electric force on each other. The magnitude of the force is linearly proportional to the net charge on each object and inversely proportional to the square of the distance between them. (Interestingly, the force does not depend on the mass of the objects.) The direction of the force vector is along the imaginary line joining the two objects and is dictated by the signs of the charges involved.

Let

  • q1,q2=q1,q2= the net electric charges of the two objects;
  • r12=r12= the vector displacement from q1q1 to q2q2.

The electric force FF on one of the charges is proportional to the magnitude of its own charge and the magnitude of the other charge, and is inversely proportional to the square of the distance between them:

Fq1q2r122.Fq1q2r122.

This proportionality becomes an equality with the introduction of a proportionality constant. For reasons that will become clear in a later chapter, the proportionality constant that we use is actually a collection of constants. (We discuss this constant shortly.)

Note: Coulomb’s Law:

The electric force (or Coulomb force) between two electrically charged particles is equal to

F12(r)=14πε0|q1q2|r122r^12F12(r)=14πε0|q1q2|r122r^12
(2)

We use absolute value signs around the product q1q2q1q2 because one of the charges may be negative, but the magnitude of the force is always positive. The unit vector r^r^ points directly from the charge q1q1 toward q2q2. If q1q1 and q2q2 have the same sign, the force vector on q2q2 points away from q1q1; if they have opposite signs, the force on q2q2 points toward q1q1 (Figure 1).

Figure 1: The electrostatic force FF between point charges q1q1 and q2q2 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 on q1q1 is equal in magnitude and opposite in direction to the force it exerts on q2q2. (a) Like charges; (b) unlike charges.
In part a, two charges q one and q two are shown separated by a distance r. Force vector arrow F one two points toward left and acts on q one. Force vector arrow F two one points toward right and acts on q two. Both forces act in opposite directions and are represented by arrows of same length. In part b, two charges q one and q two are shown at a distance r. Force vector arrow F one two points toward right and acts on q one. Force vector arrow F two one points toward left and acts on q two. Both forces act toward each other and are represented by arrows of same length.

It is important to note that the electric force is not constant; it is a function of the separation distance between the two charges. If either the test charge or the source charge (or both) move, then rr changes, and therefore so does the force. An immediate consequence of this is that direct application of Newton’s laws with this force can be mathematically difficult, depending on the specific problem at hand. It can (usually) be done, but we almost always look for easier methods of calculating whatever physical quantity we are interested in. (Conservation of energy is the most common choice.)

Finally, the new constant ε0ε0 in Coulomb’s law is called the permittivity of free space, or (better) the permittivity of vacuum. It has a very important physical meaning that we will discuss in a later chapter; for now, it is simply an empirical proportionality constant. Its numerical value (to three significant figures) turns out to be

ε0=8.85×10−12C2N·m2.ε0=8.85×10−12C2N·m2.

These units are required to give the force in Coulomb’s law the correct units of newtons. Note that in Coulomb’s law, the permittivity of vacuum is only part of the proportionality constant. For convenience, we often define a Coulomb’s constant:

ke=14πε0=8.99×109N·m2C2.ke=14πε0=8.99×109N·m2C2.

Example 1

The Force on the Electron in Hydrogen

A hydrogen atom consists of a single proton and a single electron. The proton has a charge of +e+e and the electron has ee. In the “ground state” of the atom, the electron orbits the proton at most probable distance of 5.29×10−11m5.29×10−11m (Figure 2). Calculate the electric force on the electron due to the proton.

Figure 2: A schematic depiction of a hydrogen atom, showing the force on the electron. This depiction is only to enable us to calculate the force; the hydrogen atom does not really look like this. Recall (Reference).
A positive charge is shown at the center of a sphere of radius r. An electron is depicted as a particle on the sphere. The force on the electron is along the radius, toward the nucleus.

Strategy

For the purposes of this example, we are treating the electron and proton as two point particles, each with an electric charge, and we are told the distance between them; we are asked to calculate the force on the electron. We thus use Coulomb’s law.

Solution

Our two charges and the distance between them are,

q1=+e=+1.602×10−19Cq2=e=−1.602×10−19Cr=5.29×10−11m.q1=+e=+1.602×10−19Cq2=e=−1.602×10−19Cr=5.29×10−11m.

The magnitude of the force on the electron is

F=14πϵ0|e|2r2=14π(8.85×10−12C2N·m2)(1.602×10−19C)2(5.29×10−11m)2=8.25×10−8N.F=14πϵ0|e|2r2=14π(8.85×10−12C2N·m2)(1.602×10−19C)2(5.29×10−11m)2=8.25×10−8N.

As for the direction, since the charges on the two particles are opposite, the force is attractive; the force on the electron points radially directly toward the proton, everywhere in the electron’s orbit. The force is thus expressed as

F=(8.25×10−8N)r^.F=(8.25×10−8N)r^.

Significance

This is a three-dimensional system, so the electron (and therefore the force on it) can be anywhere in an imaginary spherical shell around the proton. In this “classical” model of the hydrogen atom, the electrostatic force on the electron points in the inward centripetal direction, thus maintaining the electron’s orbit. But note that the quantum mechanical model of hydrogen (discussed in Quantum Mechanics) is utterly different.

Note:

Exercise 1

Check Your Understanding What would be different if the electron also had a positive charge?

Solution

The force would point outward.

Multiple Source Charges

The analysis that we have done for two particles can be extended to an arbitrary number of particles; we simply repeat the analysis, two charges at a time. Specifically, we ask the question: Given N charges (which we refer to as source charge), what is the net electric force that they exert on some other point charge (which we call the test charge)? Note that we use these terms because we can think of the test charge being used to test the strength of the force provided by the source charges.

Like all forces that we have seen up to now, the net electric force on our test charge is simply the vector sum of each individual electric force exerted on it by each of the individual test charges. Thus, we can calculate the net force on the test charge Q by calculating the force on it from each source charge, taken one at a time, and then adding all those forces together (as vectors). This ability to simply add up individual forces in this way is referred to as the principle of superposition, and is one of the more important features of the electric force. In mathematical form, this becomes

Note:

F(r)=14πε0Qi=1Nqiri2r^i.F(r)=14πε0Qi=1Nqiri2r^i.

In this expression, Q represents the charge of the particle that is experiencing the electric force FF, and is located at rr from the origin; the qi’sqi’s are the N source charges, and the vectors ri=rir^iri=rir^i are the displacements from the position of the ith charge to the position of Q. Each of the N unit vectors points directly from its associated source charge toward the test charge. All of this is depicted in Figure 3. Please note that there is no physical difference between Q and qiqi; the difference in labels is merely to allow clear discussion, with Q being the charge we are determining the force on.

Figure 3: The eight source charges each apply a force on the single test charge Q. Each force can be calculated independently of the other seven forces. This is the essence of the superposition principle.
Eight source charges are shown as small spheres distributed within an x y z coordinate system. The sources are labeled q sub 1, q sub 2, and so on. Sources 1, 2, 4, 7 and 8 are shaded red and sources 3, 5, and 6 are shaded blue. A test charge is also shown, shaded in green and labeled as plus Q. The r vectors from each source to the test charge Q are shown as arrows with tails at the sources and heads at the test charge. The vector from q sub 1 to the test charge is labeled as r sub 1. The vector from q sub 2 to the test charge is labeled as r sub 2, and so on for all eight vectors.

(Note that the force vector FiFi does not necessarily point in the same direction as the unit vector r^ir^i; it may point in the opposite direction, r^ir^i. The signs of the source charge and test charge determine the direction of the force on the test charge.)

There is a complication, however. Just as the source charges each exert a force on the test charge, so too (by Newton’s third law) does the test charge exert an equal and opposite force on each of the source charges. As a consequence, each source charge would change position. However, by Equation 8, the force on the test charge is a function of position; thus, as the positions of the source charges change, the net force on the test charge necessarily changes, which changes the force, which again changes the positions. Thus, the entire mathematical analysis quickly becomes intractable. Later, we will learn techniques for handling this situation, but for now, we make the simplifying assumption that the source charges are fixed in place somehow, so that their positions are constant in time. (The test charge is allowed to move.) With this restriction in place, the analysis of charges is known as electrostatics, where “statics” refers to the constant (that is, static) positions of the source charges and the force is referred to as an electrostatic force.

Example 2

The Net Force from Two Source Charges

Three different, small charged objects are placed as shown in Figure 4. The charges q1q1 and q3q3 are fixed in place; q2q2 is free to move. Given q1=2eq1=2e, q2=−3eq2=−3e, and q3=−5eq3=−5e, and that d=2.0×10−7md=2.0×10−7m, what is the net force on the middle charge q2q2?

Figure 4: Source charges q1q1 and q3q3 each apply a force on q2q2.
Three charges are shown in an x y coordinate system. Charge q sub 1 is at x=0, y=d. Charge q sub 2 is at x=2 d, y=0. Charge q sub 3 is at the origin. Force F 1 2 is exerted on charge q sub 2 and points up. Force F 2 3 is exerted on charge q sub 2 and points to the left. Force F is exerted on charge q sub 2 and points at an angle theta above the minus x direction.

Strategy

We use Coulomb’s law again. The way the question is phrased indicates that q2q2 is our test charge, so that q1q1 and q3q3 are source charges. The principle of superposition says that the force on q2q2 from each of the other charges is unaffected by the presence of the other charge. Therefore, we write down the force on q2q2 from each and add them together as vectors.

Solution

We have two source charges (q1(q1 and q3),q3), a test charge (q2),(q2), distances (r21(r21 and r23),r23), and we are asked to find a force. This calls for Coulomb’s law and superposition of forces. There are two forces:

F=F21+F23=14πε0[q2q1r212j^+(q2q3r232i^)].F=F21+F23=14πε0[q2q1r212j^+(q2q3r232i^)].

We can’t add these forces directly because they don’t point in the same direction: F12F12 points only in the −x-direction, while F13F13 points only in the +y-direction. The net force is obtained from applying the Pythagorean theorem to its x- and y-components:

F=Fx2+Fy2F=Fx2+Fy2

where

Fx=F23=14πε0q2q3r232=(8.99×109N·m2C2)(4.806×10−19C)(8.01×10−19C)(4.00×10−7m)2=2.16×10−14NFx=F23=14πε0q2q3r232=(8.99×109N·m2C2)(4.806×10−19C)(8.01×10−19C)(4.00×10−7m)2=2.16×10−14N

and

Fy=F21=14πε0q2q1r212=(8.99×109N·m2C2)(4.806×10−19C)(3.204×10−19C)(2.00×10−7m)2=3.46×10−14N.Fy=F21=14πε0q2q1r212=(8.99×109N·m2C2)(4.806×10−19C)(3.204×10−19C)(2.00×10−7m)2=3.46×10−14N.

We find that

F=Fx2+Fy2=4.08×10−14NF=Fx2+Fy2=4.08×10−14N

at an angle of

ϕ=tan−1(FyFx)=tan−1(3.46×10−14N−2.16×10−14N)=−58°,ϕ=tan−1(FyFx)=tan−1(3.46×10−14N−2.16×10−14N)=−58°,

that is, 58°58° above the −x-axis, as shown in the diagram.

Significance

Notice that when we substituted the numerical values of the charges, we did not include the negative sign of either q2q2 or q3q3. Recall that negative signs on vector quantities indicate a reversal of direction of the vector in question. But for electric forces, the direction of the force is determined by the types (signs) of both interacting charges; we determine the force directions by considering whether the signs of the two charges are the same or are opposite. If you also include negative signs from negative charges when you substitute numbers, you run the risk of mathematically reversing the direction of the force you are calculating. Thus, the safest thing to do is to calculate just the magnitude of the force, using the absolute values of the charges, and determine the directions physically.

It’s also worth noting that the only new concept in this example is how to calculate the electric forces; everything else (getting the net force from its components, breaking the forces into their components, finding the direction of the net force) is the same as force problems you have done earlier.

Note:

Exercise 2

Check Your Understanding What would be different if q1q1 were negative?

Solution

The net force would point 58°58° below the −x-axis.

Summary

  • Coulomb’s law gives the magnitude of the force between point charges. It is
    F12(r)=14πε0q1q2r122r^12F12(r)=14πε0q1q2r122r^12

    where q2q2 and q2q2 are two point charges separated by a distance r. This Coulomb force is extremely basic, since most charges are due to point-like particles. It is responsible for all electrostatic effects and underlies most macroscopic forces.

Conceptual Questions

Exercise 3

Would defining the charge on an electron to be positive have any effect on Coulomb’s law?

Exercise 4

An atomic nucleus contains positively charged protons and uncharged neutrons. Since nuclei do stay together, what must we conclude about the forces between these nuclear particles?

Solution

The force holding the nucleus together must be greater than the electrostatic repulsive force on the protons.

Exercise 5

Is the force between two fixed charges influenced by the presence of other charges?

Problems

Exercise 6

Two point particles with charges +3μC+3μC and +5μC+5μC are held in place by 3-N forces on each charge in appropriate directions. (a) Draw a free-body diagram for each particle. (b) Find the distance between the charges.

Exercise 7

Two charges +3μC+3μC and +12μC+12μC are fixed 1 m apart, with the second one to the right. Find the magnitude and direction of the net force on a −2-nC charge when placed at the following locations: (a) halfway between the two (b) half a meter to the left of the +3μC+3μC charge (c) half a meter above the +12μC+12μC charge in a direction perpendicular to the line joining the two fixed charges

Solution

a. charge 1 is 3μC3μC; charge 2 is 12μC12μC, F31=2.16×10−4NF31=2.16×10−4N to the left,
F32=8.63×10−4NF32=8.63×10−4N to the right,
Fnet=6.47×10−4NFnet=6.47×10−4N to the right;
b. F31=2.16×10−4NF31=2.16×10−4N to the right,
F32=9.59×10−5NF32=9.59×10−5N to the right,
Fnet=3.12×10−4NFnet=3.12×10−4N to the right,
Three charges are shown. Charge 1 is a 3 micro Coulomb charge at the bottom left. Charge 2 is a 12 micro Coulomb charge at the bottom right, 1 meter to the right of charge 1. Charge 3 is a minus 2 nano Coulomb charge 0.5 meters above charge 2. The charges define a right triangle, with charge 2 at the right angle. The angle at the vertex with charge one is theta. The forces on charge three are shown. F 3 1 points down and to the left, toward charge 1. Force F 3 2 points vertically down.;
c. F31x=−2.76×10−5Ni^F31x=−2.76×10−5Ni^,
F31y=−1.38×10−5Nj^F31y=−1.38×10−5Nj^,
F32y=−8.63×10−4Nj^F32y=−8.63×10−4Nj^
Fnet=−2.76×10−5Ni^8.77×10−4Nj^Fnet=−2.76×10−5Ni^8.77×10−4Nj^

Exercise 8

In a salt crystal, the distance between adjacent sodium and chloride ions is 2.82×10−10m.2.82×10−10m. What is the force of attraction between the two singly charged ions?

Exercise 9

Protons in an atomic nucleus are typically 10−15m10−15m apart. What is the electric force of repulsion between nuclear protons?

Solution

F=230.7NF=230.7N

Exercise 10

Suppose Earth and the Moon each carried a net negative charge −Q. Approximate both bodies as point masses and point charges.

(a) What value of Q is required to balance the gravitational attraction between Earth and the Moon?

(b) Does the distance between Earth and the Moon affect your answer? Explain.

(c) How many electrons would be needed to produce this charge?

Exercise 11

Point charges q1=50μCq1=50μC and q2=−25μCq2=−25μC are placed 1.0 m apart. What is the force on a third charge q3=20μCq3=20μC placed midway between q1q1 and q2q2?

Solution

F=53.94NF=53.94N

Exercise 12

Where must q3q3 of the preceding problem be placed so that the net force on it is zero?

Exercise 13

Two small balls, each of mass 5.0 g, are attached to silk threads 50 cm long, which are in turn tied to the same point on the ceiling, as shown below. When the balls are given the same charge Q, the threads hang at 5.0°5.0° to the vertical, as shown below. What is the magnitude of Q? What are the signs of the two charges?

Two small balls are attached to threads which are in turn tied to the same point on the ceiling. The threads hang at an angle of 5.0 degrees to either side of the vertical. Each ball has a charge Q.

Solution

The tension is T=0.049NT=0.049N. The horizontal component of the tension is 0.0043N0.0043N
d=0.088m,q=6.1×10−8Cd=0.088m,q=6.1×10−8C.
The charges can be positive or negative, but both have to be the same sign.

Exercise 14

Point charges Q1=2.0μCQ1=2.0μC and Q2=4.0μCQ2=4.0μC are located at r1=(4.0i^2.0j^+5.0k^)mr1=(4.0i^2.0j^+5.0k^)m and r2=(8.0i^+5.0j^9.0k^)mr2=(8.0i^+5.0j^9.0k^)m. What is the force of Q2Q2 on Q1Q1?

Exercise 15

The net excess charge on two small spheres (small enough to be treated as point charges) is Q. Show that the force of repulsion between the spheres is greatest when each sphere has an excess charge Q/2. Assume that the distance between the spheres is so large compared with their radii that the spheres can be treated as point charges.

Solution

Let the charge on one of the spheres be rQ, where r is a fraction between 0 and 1. In the numerator of Coulomb’s law, the term involving the charges is rQ(1r)Q.rQ(1r)Q. This is equal to (rr2)Q2(rr2)Q2. Finding the maximum of this term gives 12r=0r=1212r=0r=12

Exercise 16

Two small, identical conducting spheres repel each other with a force of 0.050 N when they are 0.25 m apart. After a conducting wire is connected between the spheres and then removed, they repel each other with a force of 0.060 N. What is the original charge on each sphere?

Exercise 17

A charge q=2.0μCq=2.0μC is placed at the point P shown below. What is the force on q?

Two charges are shown, placed on a horizontal line and separated by 2.0 meters. The charge on the left is a positive 1.0 micro Coulomb charge. The charge on the right is a negative 2.0 micro Coulomb charge. Point P is 1.0 to the right of the negative charge.

Solution

Define right to be the positive direction and hence left is the negative direction, then F=−0.05NF=−0.05N

Exercise 18

What is the net electric force on the charge located at the lower right-hand corner of the triangle shown here?

Charges are shown at the vertices of an equilateral triangle with sides length a. The bottom of the triangle is on the x axis of an x y coordinate system, and the bottom left vertex is at the origin. The charge at the origin is positive q. The charge at the bottom right hand corner is also positive q. The charge at the top vertex is negative two q.

Exercise 19

Two fixed particles, each of charge 5.0×10−6C,5.0×10−6C, are 24 cm apart. What force do they exert on a third particle of charge −2.5×10−6C−2.5×10−6C that is 13 cm from each of them?

Solution

The particles form triangle of sides 13, 13, and 24 cm. The x-components cancel, whereas there is a contribution to the y-component from both charges 24 cm apart. The y-axis passing through the third charge bisects the 24-cm line, creating two right triangles of sides 5, 12, and 13 cm.
Fy=2.56NFy=2.56N in the negative y-direction since the force is attractive. The net force from both charges is Fnet=−5.12Nj^Fnet=−5.12Nj^.

Exercise 20

The charges q1=2.0×10−7C,q2=−4.0×10−7C,q1=2.0×10−7C,q2=−4.0×10−7C, and q3=−1.0×10−7Cq3=−1.0×10−7C are placed at the corners of the triangle shown below. What is the force on q1?q1?

Charges are shown at the vertices of a right triangle. The bottom of the triangle is length 4 meters, the vertical side on the left is length 3 meters, and the hypotenuse is length 5 meters. The charge at the top is q sub one and positive, the charge at the bottom left is q sub 3 and negative and the charge at the bottom right is q sub 2 and negative.

Exercise 21

What is the force on the charge q at the lower-right-hand corner of the square shown here?

Charges are shown at the corners of a square with sides length a. All of the charges are positive and all are magnitude q.

Solution

The diagonal is 2a2a and the components of the force due to the diagonal charge has a factor cosθ=12cosθ=12;
Fnet=[kq2a2+kq22a212]i^[kq2a2+kq22a212]j^Fnet=[kq2a2+kq22a212]i^[kq2a2+kq22a212]j^

Exercise 22

Point charges q1=10μCq1=10μC and q2=−30μCq2=−30μC are fixed at r1=(3.0i^4.0j^)mr1=(3.0i^4.0j^)m and r2=(9.0i^+6.0j^)m.r2=(9.0i^+6.0j^)m. What is the force of q2onq1q2onq1?

Glossary

Coulomb force:
another term for the electrostatic force
Coulomb’s law:
mathematical equation calculating the electrostatic force vector between two charged particles
electrostatic force:
amount and direction of attraction or repulsion between two charged bodies; the assumption is that the source charges remain motionless
electrostatics:
study of charged objects which are not in motion
permittivity of vacuum:
also called the permittivity of free space, and constant describing the strength of the electric force in a vacuum
principle of superposition:
useful fact that we can simply add up all of the forces due to charges acting on an object

Content actions

Download module as:

Add module to:

My Favorites (?)

'My Favorites' is a special kind of lens which you can use to bookmark modules and collections. 'My Favorites' can only be seen by you, and collections saved in 'My Favorites' can remember the last module you were on. You need an account to use 'My Favorites'.

| A lens I own (?)

Definition of a lens

Lenses

A lens is a custom view of the content in the repository. You can think of it as a fancy kind of list that will let you see content through the eyes of organizations and people you trust.

What is in a lens?

Lens makers point to materials (modules and collections), creating a guide that includes their own comments and descriptive tags about the content.

Who can create a lens?

Any individual member, a community, or a respected organization.

What are tags? tag icon

Tags are descriptors added by lens makers to help label content, attaching a vocabulary that is meaningful in the context of the lens.

| External bookmarks