TITLE: COULOMB’S LAW negative, the forces are repulsive; when the charges have

OBJECTIVE: Calculate the electric field due to a opposite signs, the forces are attractive (Fig. 1.1b). The two
system of point charges using forces obey Newton’s third law; they are always equal in
Coulomb’s law and the superposition magnitude and opposite in direction, even when the charges
principle. are not equal in magnitude.
STEM_GP12EMIIIa-10
INTRODUCTION
We begin our study of electromagnetism in this chapter
by examining the nature of electric charge. We’ll find that
charge is quantized and obeys a conservation principle. When
charges are at rest in our frame of reference, they exert
electrostatic forces on each other. These forces are of
tremendous importance in chemistry and biology and have
many technological applications. Electrostatic forces are
governed by a simple relationship known as Coulomb’s law and
are most conveniently described by using the concept of
electric field.
Key Concepts
 Coulomb’s Law
 Point charges
 Electric Fields
DISCUSSION
 Coulomb’s Law
Charles Augustin de Coulomb (1736–1806) studied the
interaction forces of charged particles in detail in 1784. He used
a torsion balance (Fig. 1.1) similar to the one used 13 years later
by Cavendish to study the much weaker gravitational
interaction. For point charges, charged bodies that are very
small in comparison with the distance r between them,
Coulomb found that the electric force is proportional to 1/r 2
That is, when the distance doubles, the force decreases to one-
quarter of its initial value; when the distance is halved, the
force increases to four times its initial value.
The electric force between two point charges also
depends on the quantity of charge on each body, which we will
denote by q or Q. To explore this dependence, Coulomb divided
a charge into two equal parts by placing a small charged
spherical conductor into contact with an identical but (a) Measuring the electric force between point charges.
uncharged sphere; by symmetry, the charge is shared equally (b) The electric forces between point charges obey Newton’s
between the two spheres. (Note the essential role of the third law: F 1 on2=−F 2 on 1
principle of conservation of charge in this procedure.) Thus, he The proportionality of the electric force to 1/ r 2 has
could obtain one-half, one-quarter, and so on, of any initial been verified with great precision. There is no reason to suspect
charge. He found that the forces that two point charges q1 and that the exponent is different from precisely 2. Thus, the form
q2 exert on each other are proportional to each charge and of Eq. (above) is the same as that of the law of gravitation. But
therefore are proportional to the product q1 q2 of the two electric and gravitational interactions are two distinct classes of
charges. phenomena. Electric interactions depend on electric charges
Thus, Coulomb established what we now call and can be either attractive or repulsive, while gravitational
Coulomb’s law: interactions depend on mass and are always attractive (because
The magnitude of the electric force between two point there is no such thing as negative mass).
charges is directly proportional to the product of the charges  Fundamental Electric Constants
and inversely proportional to the square of the distance The value of the proportionality constant in Coulomb’s
between them. law depends on the system of units used. In our study of
In mathematical terms, the magnitude F of the force electricity and magnetism we will use SI units exclusively. The SI
that each of two point charges q1 and q2 a distance r apart electric units include most of the familiar units such as the volt,
exerts on the other can be expressed as, the ampere, the ohm, and the watt. (There is no British system
of electric units.) The SI unit of electric charge is called one
∥ q 1 q 2∥ coulomb (1 C). In SI units the constant k in Eq. (1.1) is
F=k 2
r
2
where k is a proportionality constant whose numerical m 9 2 2
k = -8.987551787 x 109 N. ≅ 8.988 x 10 N . m /C
value depends on the system of units used. The absolute value C
2

bars are used in Eq. above because the charges and can be
either positive or negative, while the force magnitude is always The value of k is known to such a large number of
positive. significant figures because this value is closely related to the
The directions of the forces the two charges exert on speed of light in vacuum. Speed is defined to be exactly c=
each other are always along the line joining them. When the 2.99792458 x 108 m/s . The numerical value of is defined in
charges and have the same sign, either both positive or both terms of to be precisely,
 k= ( 10−7 N . s 2 /C 2 ) c 2
You should check this expression to confirm that has the right
units.
In principle we can measure the electric force between
two equal charges at a measured distance and use Coulomb’s
law to determine the charge. Thus, we could regard the value of
as an operational definition of the coulomb. For reasons of
experimental precision it is better to define the coulomb
instead in terms of a unit of electric current (charge per unit
time), the ampere, equal to 1 coulomb per second.
In SI units we usually write the constant as 1/4 πϵο
where (“epsilon-nought” or “epsilon-zero”) is another
constant. This appears to complicate matters, but it actually
simplifies many formulas that we will encounter in later EVALUATE: This astonishingly large number shows that
chapters. From now on, we will usually write Coulomb’s law as, the gravitational force in this situation is completely negligible
in comparison to the electric force. This is always true for
1 ∥ q 1 q 2∥ interactions of atomic and subnuclear particles. But within
F= 4 πϵο r
2 objects the size of a person or a planet, the positive and
negative charges are nearly equal in magnitude, and the net
(Coulomb’s law: force between two point charges) electric force is usually much smaller than the gravitational
force.
The constants in Eq. (above) are approximately,  Electric Field
To introduce this concept, let’s look at the mutual
12 2
ϵο=8.854 x 10 C / N . m and
2 repulsion of two positively charged bodies and (Fig. 1.2a).
Suppose B has charge qo and let F o be the electric force of A on
1 9
B. One way to think about this force is as an “action-at-
=k=8.988 x 10 N .m2 /C2 adistance” force—that is, as a force that acts across empty
4 πϵο
space without needing any matter (such as a push rod or a
The most fundamental unit of charge is the magnitude rope) to transmit it through the intervening space. (Gravity can
of the charge of an electron or a proton, which is denoted by. also be thought of as an “action-at-a-distance” force.) But a
The most precise value available as of the writing of this book is, more fruitful way to visualize the repulsion between and is as a
two-stage process. We first envision that body, as a result of the
e = 1.602176487 (40) x 10−19 C charge that it carries, somehow modifies the properties of the
space around it. Then body, as a result of the charge that it
carries, senses how space has been modified at its position. The
One coulomb represents the negative of the total
response of body B is to experience the force F o.
charge of about 6 x 1018electrons. For comparison, a copper
cube 1 cm on a side contains about 2.4 x 1024 electrons. About
9
10 electrons pass through the glowing filament of a flashlight
bulb every second.

 Sample Problem #1
Electric force versus gravitational force
An particle (the nucleus of a helium atom) has mass m=
6.64 x 10−27 kg and charge q= +2e = 3.2 x 10−19 Compare
the magnitude of the electric repulsion between two
(“alpha”) particles with that of the gravitational attraction
between them.
Our sketch for this problem.

***The electric force on a charged body is exerted by

the electric field created by other charged bodies.
To find out experimentally whether there is an electric
field at a particular point, we place a small charged body, which
we call a test charge, at the point. If the test charge experiences
an electric force, then there is an electric field at that point. This
field is produced by charges other than qo.
Force is a vector quantity, so electric field is also a
vector quantity. (Note the use of vector signs as well as
boldface letters and plus, minus, and equals signs in the
following discussion.) We define the electric field E at a point as 2. Two small spheres spaced apart have equal charge. How
the electric force F experienced by a test charge q at the point, many excess electrons must be present on each sphere if
divided by the charge q. That is, the electric field at a certain the magnitude of the force of repulsion between them is
point is equal to the electric force per unit charge experienced 4.57 x 10−21 N ?
by a charge at that point:
 Electric Field
F0 3. A proton is placed in a uniform electric field of 2.75x
E= q0
3
10 N /C Calculate: (a) the magnitude of the electric force
felt by the proton; (b) the proton’s acceleration; (c) the
(definition of electric field as electric force per unit charge) proton’s speed after 1.00 μs in the field, assuming it starts
In SI units, in which the unit of force is 1 N and the unit from rest.
of charge is 1 C, the unit of electric field magnitude is 1 newton
per coulomb ( 1 N/C ). 4. A particle has charge (a) Find the magnitude and
direction of the electric field due to this particle at a point
0.250 m directly above it. (b) At what distance from this
 Electric Field of a Point Charge particle does its electric field have a magnitude of 12.0 N/C?
If the source distribution is a point charge , it is easy to  Assessment
find the electric field that it produces. We call the location of Solve the following problems carefully. Draw the
the charge the source point, and we call the point where we direction of the magnitude using Free Body Diagram
are determining the field the field point. (FBD).
It is also useful to introduce a unit vector that points
along the line from source point to field point (Fig. below). This 1. A proton is traveling horizontally to the right at 4.50 x
6
unit vector r is equal to the displacement vector r from the 10 m/s (a) Find the magnitude and direction of the
source point to the field point, divided by the distance r = ∕ r ∕ weakest electric field that can bring the proton uniformly
between these two points; that is, r = r/r If we place a small to rest over a distance of 3.20 cm (b) How much time
test charge at the field point , at a distance from the source does it take the proton to stop after entering the field?
point, the magnitude F of the force is given by Coulomb’s law, (c) What minimum field (magnitude and direction) would
be needed to stop an electron under the conditions of
1 q1q2 part (a)?
F= 4 πϵο 2
r
2. An electron is released from rest in a uniform electric
The magnitude E of the electric field at P is, field. The electron accelerates vertically upward,
traveling 4.50m in the first 3.00 μs after it is released. (a)
1 q What are the magnitude and direction of the electric
E= field? (b) Are we justified in ignoring the effects of
4 πϵο r 2
gravity? Justify your answer quantitatively.
 Sample Problem #2
3. An average human weighs about 650 N. If two such
Electric-field magnitude for a point charge
generic humans each carried 1.0 coulomb of excess
What is the magnitude of the electric field E at a field
charge, one positive and one negative, how far apart
point 2.0 m from a point charge q = 4.0 nC?
would they have to be for the electric attraction
between them to equal their 650-N weight?
IDENTIFY and SET UP: This problem concerns the electric
field due to a point charge. We are given the magnitude of
4. Two small aluminum spheres, each having mass are
the charge and the distance from the charge to the field
separated by 80 cm (a) How many electrons does each
point, to calculate the field magnitude E.
sphere contain? (The atomic mass of aluminum is 26.982
g/mol and its atomic number is 13.) (b) How many
EXECUTE:
electrons would have to be removed from one sphere
−9 and added to the other to cause an attractive force
1 ∥ q∥ 2 4.0 x 10 C
E=
9 2 between the spheres of magnitude 1.00 x 104 (roughly 1
4 πεο r 2 = (9.0x10 N.m /C ) (2.0 m)2 = 9.0 N/C ton)? Assume that the spheres may be treated as point
charges. (c) What fraction of all the electrons in each
EVALUATE: Our result means that if we placed a 1.0-C sphere does this represent?
charge at a point 2.0 m from q, it would experience a 9.0-N
force. The force on a 2.0-C charge at that point would be 5. Two positive point charges q are placed on the -axis, one
(2.0 C) (9.0 N/C) = 18 N, and so on. at x=a and one at x= -a (a) Find the magnitude and
direction of the electric field at x=0 (b) Derive an
 Activity # 1 expression for the electric field at points on the -axis.
Solve the following problem carefully. Use your result to graph the -component of the electric
field as a function of x, for values of x between -4a and
 Coulomb’s Law +4a.
1. Excess electrons are placed on a small lead sphere with
mass so that its net charge is -3.20 x 10 -9 C (a) Find the
number of excess electrons on the sphere. (b) How many References
excess electrons are there per lead atom? The atomic
number of lead is 82, and its atomic mass is 207 g/mol.
Shipman, J. Wilson J., & Higgins, C. (2013). An Introduction to
Physical Science (13th ed). 20 Channel Center Street,
Boston, MA 02210 USA.

Tillery, B. W. (2009). Physical Sciences (9th ed.) Arizona State

University. McGraw-Hill Companies, Inc., 1221 Avenue of
the Americas, New York, NY 100020.