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18
Electric Forces and Electric Fields
Lightning, which is the flow of electric charge in the atmosphere, is nature’s most spectacular display of electricity.
Here, multiple lightning strikes occur at night over an urban skyline. (Ralph Wetmore/Stone/Getty Images)
Copyright © 2007 John Wiley & Sons, Inc. All rights reserved.
18.1
The Origin of Electricity
The electrical nature of matter is inherent in atomic structure. An atom consists of a small,
relatively massive nucleus that contains particles called protons and neutrons. A proton has a
mass of
, and a neutron has a slightly greater mass of
.
Surrounding the nucleus is a diffuse cloud of orbiting particles called electrons, as Figure 18-1
. Like mass, electric charge is an intrinsic
suggests. An electron has a mass of
property of protons and electrons, and only two types of charge have been discovered, positive
and negative. A proton has a positive charge, and an electron has a negative charge. A neutron
has no net electric charge.
Figure 18-1 An atom contains a small, positively charged nucleus, about which
the negatively charged electrons move. The closed-loop paths shown here are
symbolic only. In reality, the electrons do not follow discrete paths, as Section
30.5 discusses.
Experiment reveals that the magnitude of the charge on the proton exactly equals the magnitude
of the charge on the electron; the proton carries a charge +e, and the electron carries a charge −e.
The SI unit for measuring the magnitude of an electric charge is the coulomb* (C), and e has
been determined experimentally to have the value
The symbol e represents only the magnitude of the charge on a proton or an electron and does not
include the algebraic sign that indicates whether the charge is positive or negative. In nature,
atoms are normally found with equal numbers of protons and electrons. Usually, then, an atom
carries no net charge because the algebraic sum of the positive charge of the nucleus and the
negative charge of the electrons is zero. When an atom, or any object, carries no net charge, the
object is said to be electrically neutral. The neutrons in the nucleus are electrically neutral
particles.
The charge on an electron or a proton is the smallest amount of free charge that has been
discovered. Charges of larger magnitude are built up on an object by adding or removing
, where N is
electrons. Thus, any charge of magnitude q is an integer multiple of e; that is,
an integer. Because any electric charge q occurs in integer multiples of elementary, indivisible
charges of magnitude e, electric charge is said to be quantized. Example 1 emphasizes the
quantized nature of electric charge.
Example 1 | A Lot of Electrons
How many electrons are there in one coulomb of negative charge?
Reasoning The negative charge is due to the presence of excess electrons, since they carry
negative charge. Because an electron has a charge whose magnitude is
, the
number of electrons is equal to the charge q divided by the charge e on each electron.
Solution The number N of electrons is
Copyright © 2007 John Wiley & Sons, Inc. All rights reserved.
Charged Objects and the Electric Force
It is easy to demonstrate that two electrically charged objects exert a force on one another. Consider Figure 183a, which shows two small balls that have been oppositely charged and are light and free to move. The balls
attract each other. On the other hand, balls with the same type of charge, either both positive or both negative,
During any process, the net electric charge of an isolated system remains constant (is conserved).
Law of Conservation of Electric Charge
Electric charges play a role in many situations other than rubbing two surfaces together. They are involved,
for instance, in chemical reactions, electric circuits, and radioactive decay. A great number of experiments
have verified that in any situation, the law of conservation of electric charge is obeyed.
When an ebonite rod is rubbed with animal fur, the rubbing process serves only to separate electrons and
protons already present in the materials. No electrons or protons are created or destroyed. Whenever an electron
is transferred to the rod, a proton is left behind on the fur. Since the charges on the electron and proton have
identical magnitudes but opposite signs, the algebraic sum of the two charges is zero, and the transfer does not
change the net charge of the fur/rod system. If each material contains an equal number of protons and electrons
to begin with, the net charge of the system is zero initially and remains zero at all times during the rubbing process.
Figure 18-2 When an ebonite rod is rubbed against animal fur, electrons from atoms of the fur are transferred to
the rod. This transfer gives the rod a negative charge (−) and leaves a positive charge (+) on the fur.
Electricity has many useful applications, and they are related to the fact that it is possible to transfer electric
charge from one object to another. Usually electrons are transferred, and the body that gains electrons acquires
an excess of negative charge. The body that loses electrons has an excess of positive charge. Such separation of
charge occurs often when two unlike materials are rubbed together. For example, when an ebonite (hard, black
rubber) rod is rubbed against animal fur, some of the electrons from atoms of the fur are transferred to the rod.
The ebonite becomes negatively charged, and the fur becomes positively charged, as Figure 18-2 indicates.
Similarly, if a glass rod is rubbed with a silk cloth, some of the electrons are removed from the atoms of the glass
and deposited on the silk, leaving the silk negatively charged and the glass positively charged. There are many
familiar examples of charge separation, as when you walk across a nylon rug or run a comb through dry hair. In
each case, objects become “electrified” as surfaces rub against one another.
18.2
CONCEPTS AT A GLANCE Like other forces that we have encountered, the electric force (also called
the electrostatic force) can alter the motion of an object. It can do so by contributing to the net external force
that acts on the object. Newton’s second law,
, specifies the acceleration
that arises
because of the net external force. The Concepts-at-a-Glance chart in Figure 18-4 is an expanded version of the
Figure 18-3 (a) A positive charge (+) and a negative charge (−) attract each other. (b) Two negative charges
repel each other. (c) Two positive charges repel each other.
repel each other, as parts b and c of the drawing indicate. The behavior depicted in Figure 18-3 illustrates the
following fundamental characteristic of electric charges:
The physics of electronic ink. A new technology based on the electric force may revolutionize the way books
and other printed matter are made. This technology, called electronic ink, allows letters and graphics on a page to
be changed instantly, much like the symbols displayed on a computer monitor. Figure 18-5a illustrates the
essential features of electronic ink. It consists of millions of clear microcapsules, each having the diameter of a
human hair and filled with a dark inky liquid. Inside each microcapsule are several dozen extremely tiny white
beads that carry a slightly negative charge. The microcapsules are sandwiched between two sheets, an opaque
base layer and a transparent top layer, at which the reader looks. When a positive charge is applied to a small region
of the base layer, as shown in part b of the drawing, the negatively charged white beads are drawn to it, leaving
dark ink at the top layer. Thus, a viewer sees only the dark liquid. When a negative charge is applied to a region of
the base layer, the negatively charged white beads are repelled from it and are forced to the top of the
microcapsules; now a viewer sees a white area due to the beads. Thus, electronic ink is based on the principle that
like charges repel and unlike charges attract each other; a positive charge causes one color to appear, and a
Figure 18-4 CONCEPTS AT A GLANCE External electrostatic forces that act on an object must be included
in the net external force when using Newton’s second law to determine acceleration. In a DNA analysis using
electrophoresis, an electrostatic force accelerates electrically charged DNA fragments to different terminal speeds
in a gel. The different fragments are represented by the glowing bands. (Eurelios/Phototake)
charts in Figures 4-9, 10-4, and 11-5, and emphasizes that any external electric force that acts on an object must
be included when determining the net external force to be used in the second law.
negative charge causes another color to appear. Each small region, whether dark or light, is known as a pixel (short
for “picture element”). Computer chips provide the instructions to produce the negative and positive charges on
the base layer of each pixel. Letters and graphics are produced by the patterns generated with the two colors.
Copyright © 2007 John Wiley & Sons, Inc. All rights reserved.
Figure 18-5 (a) Electronic ink consists of microcapsules filled with a dark, inky liquid and dozens of white beads.
(b) Dark and light pixels are formed when positive and negative charges are placed in the base layer by electronic
circuitry.
18.3
Conductors and Insulators
Electric charge can not only exist on an object, but it can also move through an object. However,
materials differ vastly in their abilities to allow electric charge to move or be conducted through
them. To help illustrate such differences in conductivity, Figure 18-6a recalls the conduction of heat
through a bar of material whose ends are maintained at different temperatures. As Section 13.2
discusses, metals conduct heat readily and, therefore, are known as thermal conductors. On the other
hand, substances that conduct heat poorly are referred to as thermal insulators.
Figure 18-6 (a) Heat is conducted from the hotter end of the metal bar to the
cooler end. (b) Electrons are conducted from the negatively charged end of the
metal bar to the positively charged end.
A situation analogous to the conduction of heat arises when a metal bar is placed between two
charged objects, as in Figure 18-6b. Electrons are conducted through the bar from the negatively
charged object toward the positively charged object. Substances that readily conduct electric charge
are called electrical conductors. Although there are exceptions, good thermal conductors are
generally good electrical conductors. Metals such as copper, aluminum, silver, and gold are
excellent electrical conductors and, therefore, are used in electrical wiring. Materials that conduct
electric charge poorly are known as electrical insulators. In many cases, thermal insulators are also
electrical insulators. Common electrical insulators are rubber, many plastics, and wood. Insulators,
such as the rubber or plastic that coats electrical wiring, prevent electric charge from going where it
is not wanted.
The difference between electrical conductors and insulators is related to atomic structure. As
electrons orbit the nucleus, those in the outer orbits experience a weaker force of attraction to the
nucleus than do those in the inner orbits. Consequently, the outermost electrons (also called the
valence electrons) can be dislodged more easily than the inner ones. In a good conductor, some
valence electrons become detached from a parent atom and wander more or less freely throughout
the material, belonging to no one atom in particular. The exact number of electrons detached from
each atom depends on the nature of the material, but is usually between one and three. When one
end of a conducting bar is placed in contact with a negatively charged object and the other end in
contact with a positively charged object, as in Figure 18-6b, the “free” electrons are able to move
readily away from the negative end and toward the positive end. The ready movement of electrons is
the hallmark of a good conductor. In an insulator the situation is different, for there are very few
electrons free to move throughout the material. Virtually every electron remains bound to its parent
atom. Without the “free” electrons, there is very little flow of charge when the material is placed
between two oppositely charged bodies, so the material is an electrical insulator.
Copyright © 2007 John Wiley & Sons, Inc. All rights reserved.
18.4
Charging by Contact and by Induction
When a negatively charged ebonite rod is rubbed on a metal object, such as the sphere in Figure
18-7a, some of the excess electrons from the rod are transferred to the object. Once the electrons
are on the metal sphere, where they can move readily, they repel one another and spread out over
the sphere’s surface. The insulated stand prevents them from flowing to the earth, where they
could spread out even more. When the rod is removed, as in part b of the picture, the sphere is left
with a negative charge distributed over its surface. In a similar manner, the sphere would be left
with a positive charge after being rubbed with a positively charged rod. In this case, electrons
from the sphere would be transferred to the rod. The process of giving one object a net electric
charge by placing it in contact with another object that is already charged is known as charging
by contact.
Figure 18-7 (a) Electrons are transferred by rubbing the negatively charged rod
on the metal sphere. (b) When the rod is removed, the electrons distribute
themselves over the surface of the sphere.
It is also possible to charge a conductor in a way that does not involve contact. In Figure 18-8, a
negatively charged rod is brought close to, but does not touch, a metal sphere. In the sphere, the
free electrons closest to the rod move to the other side, as part a of the drawing indicates. As a
result, the part of the sphere nearest the rod becomes positively charged and the part farthest away
becomes negatively charged. These positively and negatively charged regions have been
“induced” or “persuaded” to form because of the repulsive force between the negative rod and the
free electrons in the sphere. If the rod were removed, the free electrons would return to their
original places, and the charged regions would disappear.
Figure 18-8 (a) When a charged rod is brought near the metal sphere without
touching it, some of the positive and negative charges in the sphere are separated.
(b) Some of the electrons leave the sphere through the grounding wire, with the
result (c) that the sphere acquires a positive net charge.
Under most conditions the earth is a good electrical conductor. So when a metal wire is attached
between the sphere and the ground, as in Figure 18-8b, some of the free electrons leave the sphere
and distribute themselves over the much larger earth. If the grounding wire is then removed,
followed by the ebonite rod, the sphere is left with a positive net charge, as part c of the picture
shows. The process of giving one object a net electric charge without touching the object to a
second charged object is called charging by induction. The process could also be used to give the
sphere a negative net charge, if a positively charged rod were used. Then, electrons would be
drawn up from the ground through the grounding wire and onto the sphere.
If the sphere in Figure 18-8 were made from an insulating material like plastic, instead of metal,
the method of producing a net charge by induction would not work, because very little charge
would flow through the insulating material and down the grounding wire. However, the electric
force of the charged rod would have some effect, as Figure 18-9 illustrates. The electric force
would cause the positive and negative charges in the molecules of the insulating material to
separate slightly, with the negative charges being “pushed” away from the negative rod. Although
no net charge is created, the surface of the plastic does acquire a slight induced positive charge
and is attracted to the negative rod. For a similar reason, one piece of cloth can stick to another in
the phenomenon known as “static cling,” which occurs when an article of clothing has acquired
an electric charge while being tumbled about in a clothes dryer.
Figure 18-9 The negatively charged rod induces a slight positive surface charge
on the plastic.
Check Your Understanding 1
Two metal spheres are identical. They are electrically neutral and are touching. An electrically
charged ebonite rod is then brought near the spheres without touching them, as the drawing
shows. After a while, with the rod held in place, the spheres are separated, and the rod is then
and
of the spheres after they
removed. The following statements refer to the masses
are separated and the rod is removed. Which one or more of the statements is true? (a)
, (b)
if the rod is positive, (c)
if the rod is positive, (d)
if the rod is negative, (e)
if the rod is negative. [The answer(s) is (are)
given at the end of the book.]
Background: Mass and electric charge are both properties of electrons and protons. In a metal,
which move, electrons or protons? An object can be given an electric charge by contact with
another charged object or by induction.
For similar questions (including calculational counterparts), consult Self-Assessment Test
18.1, which is described at the end of Section 18.5.
Copyright © 2007 John Wiley & Sons, Inc. All rights reserved.
18.5
Coulomb’s Law
The Force that Point Charges Exert on each Other
The electrostatic force that stationary charged objects exert on each other depends on the amount of
charge on the objects and the distance between them. Experiments reveal that the greater the charge
and the closer together they are, the greater is the force. To set the stage for explaining these features
in more detail, Figure 18-10 shows two charged bodies. These objects are so small, compared to the
distance r between them, that they can be regarded as mathematical points. The “point charges” have
and
. If the charges have unlike signs, as in part a of the picture, each object is
magnitudes*
attracted to the other by a force that is directed along the line between them;
is the electric force
is the electric force exerted on object 2 by object 1. If, as in
exerted on object 1 by object 2 and
part b, the charges have the same sign (both positive or both negative), each object is repelled from the
other. The repulsive forces, like the attractive forces, act along the line between the charges. Whether
attractive or repulsive, the two forces are equal in magnitude but opposite in direction. These forces
always exist as a pair, each one acting on a different object, in accord with Newton’s action–reaction
law.
Figure 18-10 Each point charge exerts a force on the other. Regardless of whether
the forces are (a) attractive or (b) repulsive, they are directed along the line between
the charges and have equal magnitudes.
The French physicist Charles Augustin de Coulomb (1736–1806) carried out a number of experiments
to determine how the electric force that one point charge applies to another depends on the amount of
each charge and the separation between them. His result, now known as Coulomb’s law, is stated
below.
Coulomb’s Law
The magnitude F of the electrostatic force exerted by one point charge q1 on another point charge
q2 is directly proportional to the magnitudes
and
of the charges and inversely proportional
to the square of the distance r between them:
(18.1)
in SI units. The electrostatic
where k is a proportionality constant:
force is directed along the line joining the charges, and it is attractive if the charges have unlike
signs and repulsive if the charges have like signs.
It is common practice to express k in terms of another constant
called the permittivity of free space and has a value of
by writing
is
.
Equation 18.1 gives the magnitude of the electrostatic force that each point charge exerts on the other.
When using this equation, then, it is important to remember to substitute only the charge magnitudes
and
, as Example 2 illustrates.
(without algebraic signs) for
Example 2 | A Large Attractive Force
Two objects, whose charges are + 1.0 and −1.0 C, are separated by 1.0 km. Compared to 1.0 km, the
sizes of the objects are small. Find the magnitude of the attractive force that either charge exerts on the
other.
Reasoning Considering that the sizes of the objects are small compared to the separation distance, we
can treat the charges as point charges. Coulomb’s law may then be used to find the magnitude of the
attractive force, provided that only the magnitudes of the charges are used for the symbols
and
that appear in the law.
Solution The magnitude of the force is
(18.1)
Need more practice?
Interactive LearningWare18.1
An object of mass 215 kg is
located just above the surface of
the earth (
).
Suppose that the object and the
earth have an identical charge q.
Therefore, two forces–the
gravitational force and the
electrostatic force— act on the
object. Assuming that the earth’s
charge is located at the center of
the earth, determine q (magnitude
and sign) when the object is in
equilibrium.
Related Homework: Problem 61
The force calculated in Example 2 corresponds to about 2000 pounds and is so large because charges
of ± 1.0 C are enormous. Such large charges are encountered only in the most severe conditions, as in
a lightning bolt, where as much as 25 C can be transferred between the cloud and the ground. The
typical charges produced in the laboratory are much smaller and are measured conveniently in
microcoulombs
.
Coulomb’s law has a form that is remarkably similar to Newton’s law of gravitation
. The force in both laws depends on the inverse square (1/r2) of the distance
between the two objects and is directed along the line between them. In addition, the force is
proportional to the product of an intrinsic property of each of the objects, the magnitudes of the
and
in Coulomb’s law and the masses m1 and m2 in the gravitation law. But there is a
charges
major difference between the two laws. The electrostatic force can be either repulsive or attractive,
depending on whether or not the charges have the same sign; in contrast, the gravitational force is
always an attractive force.
Section 5.5 discusses how the gravitational attraction between the earth and a satellite provides the
centripetal force that keeps a satellite in orbit. Example 3 illustrates that the electrostatic force of
attraction plays a similar role in a famous model of the atom created by the Danish physicist Niels
Bohr (1885 – 1962).
ANALYZING
PROBLEMS
MULTIPLE-CONCEPT
Example 3 | A Model of the Hydrogen Atom
In the Bohr model of the hydrogen atom, the electron (charge = −e) is in a circular orbit about the
nuclear proton (charge = + e) at a radius of
, as Figure 18-11 shows. The mass of the
electron is
. Determine the speed of the electron.
Figure 18-11 In the Bohr model of the hydrogen atom, the electron (− e) orbits
the proton (+ e) at a distance of r = 5.29 × 10−11 m. The velocity of the electron is
.
Reasoning Recall from Section 5.3 that a net force is required to keep an object such as an electron
moving on a circular path. This net force is called the centripetal force and always points toward the
center of the circle. It has a magnitude given by
, where m and v are, respectively, the
mass and speed of the electron and r is the radius of the orbit. This equation can be solved for the
speed of the electron. Since the mass and orbital radius are known, we can calculate the electron’s
speed provided that a value for the centripetal force can be found. For the electron in the hydrogen
atom, the centripetal force is provided almost exclusively by the electrostatic force that the proton
exerts on the electron. This attractive force points toward the center of the circle and its magnitude
is given by Coulomb’s law. The electron is also pulled toward the proton by the gravitational force.
However, the gravitational force is negligible in comparison to the electrostatic force.
Knowns and Unknowns The data for this problem are:
Description
Symbol
Electron charge
−e
Electron mass
m
Proton charge
+e
Radius of orbit
r
Value
Unknown Variable
Orbital speed of electron
v
?
Modeling the Problem
Step 1 Centripetal Force
An electron of mass m that moves with a constant speed v on a circular path of radius r
experiences a net force, called the centripetal force. The magnitude Fc of this force is given by
(Equation 5.3). By solving this equation for the speed, we obtain Equation 1 at
the right. The mass and radius in this expression are known. However, the magnitude of the
centripetal force is not known, so we will evaluate it in Step 2.
(1)
Step 2 Coulomb’s Law
As the electron orbits the proton in the hydrogen atom, it is attracted to the proton by the
electrostatic force. The magnitude F of the electrostatic force is given by Coulomb’s law as
(Equation 18.1), where
and
are the magnitudes of the charges, r is
the orbital radius, and
. Since the centripetal force is provided
almost entirely by the electrostatic force, it follows that
. Furthermore,
. With these substitutions, Equation 18.1 becomes
and
All the variables on the right side of this expression are known, so we substitute it into Equation
1, as indicated in the right column.
Solution Algebraically combining the results of the modeling steps, we have
The speed of the orbiting electron is
Related Homework: Problems 16 and 20
Since the electrostatic force depends on the inverse square of the distance between the charges, it
becomes larger for smaller distances, such as those involved when a strip of adhesive tape is stuck to a
smooth surface. Electrons shift over the small distances between the tape and the surface. As a result,
the materials become oppositely charged. Since the distance between the charges is relatively small,
the electrostatic force of attraction is large enough to contribute to the adhesive bond. Figure 18-12
shows an image of the sticky surface of a piece of tape after it has been pulled off a metal surface. The
image was obtained using an atomic-force microscope and reveals the tiny pits left behind when
microscopic portions of the adhesive remain stuck to the metal because of the strong adhesive bonding
forces.
Figure 18-12 After a strip of adhesive tape has been pulled off a metal surface, there
are tiny pits (approximately one ten-millionth of a meter in diameter) in the stick
surface of the tape, as this image shows. It was obtained using an atomic-force
microscope. (Courtesy Louis Scudiero and J. Thomas Dickinson, Washington State
University.)
The physics of adhesion.
The Force on a Point Charge Due to Two or More Other
Point Charges
Up to now, we have been discussing the electrostatic force on a point charge (magnitude
) due to
another point charge (magnitude
). Suppose that a third point charge (magnitude
) is also
present. What would be the net force on q1 due to both q2 and q3?. It is convenient to deal with such a
problem in parts. First, find the magnitude and direction of the force exerted on q1 by q2 (ignoring q3).
Then, determine the force exerted on q1 by q3 (ignoring q2). The net force on q1 is the vector sum of
these forces. Examples 4 and 5 illustrate this approach when the charges lie along a straight line and
on a plane, respectively.
Example 4 | Three Charges on a Line
Figure 18-13a shows three point charges that lie along the x axis in a vacuum. Determine the
magnitude and direction of the net electrostatic force on q1.
Figure 18-13 (a) Three charges lying along the x axis. (b) The force exerted on q1
by q2 is
, while the force exerted on q1 by q3 is
.
Reasoning Part b of the drawing shows a free-body diagram of the forces that act on q1. Since q1 and
q2 have opposite signs, they attract one another. Thus, the force exerted on q1 by q2 is
points to the left. Similarly, the force exerted on q1 by q3 is
, and it
and is also an attractive force. It
points to the right in Figure 18-13b. The magnitudes of these forces can be obtained from Coulomb’s
law. The net force is the vector sum of
and
.
Solution The magnitudes of the forces are
Since
force
points in the negative x direction, and
points in the positive x direction, the net
is
The plus sign in the answer indicates that the net force points to the right in the drawing.
Problem solving insight When
using Coulomb’s law
, remember
and
that the symbols
stand for the charge
magnitudes. Do not substitute
negative numbers for these
symbols.
Example 5 | Three Charges in a Plane
Figure 18-14a shows three point charges that lie in the x, y plane in a vacuum. Find the magnitude and
direction of the net electrostatic force on q1.
Figure 18-14 (a) Three charges lying in a plane. (b) The net force acting on q1 is
=
+
. The angle that
Reasoning The force exerted on q1 by q2 is
makes with the + x axis is .
and is an attractive force because the two charges
have opposite signs. It points along the line between the charges. The force exerted on q1 by q3 is
and is also an attractive force. It points along the line between q1 and q3. Coulomb’s law
specifies the magnitudes of these forces. Since the forces point in different directions (see Figure 1814b), we will use vector components to find the net force.
Need more practice?
Interactive LearningWare18.2
A positive charge is placed on the x
,
axis
and another positive charge is placed
on the y axis
.A
third charge
is placed at the coordinate origin. If the
charge at the origin were free to move,
what would be the magnitude of its
acceleration?
Related Homework: Problem 17
Solution The magnitudes of the forces are
The net force
components of
is the vector sum of
and
that lie in the x and y directions are
, as part b of the drawing shows. The
and
, respectively. Our approach to
is the same as that used in Chapters 1 and 4. The forces
finding
x and y components. Then, the x components are combined to give
. Once
combined to give
determined using trigonometry.
Force
and
and
are resolved into
, and the y components are
are known, the magnitude and direction of
x component
y component
+ 18 N
0N
can be
The magnitude F and the angle of the net force are
Problem solving insight The
electrostatic force is a vector
and has a direction as well as a
magnitude. When adding
electrostatic forces, take into
account the directions of all
forces, using vector
components as needed.
Check Your Understanding 2
The drawing shows three point charges arranged in different ways. The charges are + q, − q, and
− q each has the same magnitude, one being positive and the other two negative. In each part of the
drawing the distance d is the same. Rank the arrangements in descending order (largest first)
according to the magnitude of the net electrostatic force that acts on the positive charge. (The
answer is given at the end of the book.)
Background: The electrostatic force that one point charge exerts on another is described by
Coulomb’s law. Like any force, this force is a vector quantity, and the rules of vector addition
apply.
For similar questions (including calculational counterparts), consult Self-Assessment Test 18.1,
which is described next.
Test your understanding of the material in Sections 18.1, 18.2, 18.3, 18.4 and 18.5 :
· The Origin of Electricity
· Charged Objects and the Electric Force
· Conductors and Insulators
· Charging by Contact and by Induction
· Coulomb’s Law
Copyright © 2007 John Wiley & Sons, Inc. All rights reserved.
18.6
The Electric Field
Definition
As we know, a charge can experience an electrostatic force due to the presence of other charges. For
, which is the vector sum of
instance, the positive charge q0 in Figure 18-15 experiences a force
the forces exerted by the charges on the rod and the two spheres. It is useful to think of q0 as a test
charge for determining the extent to which the surrounding charges generate a force. However, in
using a test charge, we must be careful to select one with a very small magnitude, so that it does not
alter the locations of the other charges. The next example illustrates how the concept of a test charge
is applied.
Figure 18-15 A positive charge q0 experiences an electrostatic force
the surrounding charges on the ebonite rod and the two spheres.
Example 6 | A Test Charge
due to
The positive test charge shown in Figure 18-15 is
and experiences a force
in the direction shown in the drawing. (a) Find the force per coulomb that the test
charge experiences. (b) Using the result of part (a), predict the force that a charge of
would experience if it replaced q0.
Reasoning The charges in the environment apply a force
coulomb experienced by the test charge is
to the test charge q0. The force per
. If q0 is replaced by a new charge q, then the
force on this new charge is the force per coulomb times q.
Solution
a. The force per coulomb of charge is
b. The result from part (a) indicates that the surrounding charges can exert 2.0 newtons of force
would experience a force whose
per coulomb of charge. Thus, a charge of
magnitude is
The direction of this force would be the same as that experienced by the test charge, since
both have the same positive sign.
The electric force per coulomb,
, calculated in Example 6(a) is one illustration of an idea
that is very important in the study of electricity. The idea is called the electric field. Equation 18.2
presents the definition of the electric field.
Definition of the Electric Field
The electric field
that exists at a point is the electrostatic force
test charge* q0 placed at that point divided by the charge itself:
experienced by a small
(18.2)
The electric field is a vector, and its direction is the same as the direction of the force
positive test charge.
on a
SI Unit of Electric Field: newton per coulomb (N/C)
Equation 18.2 indicates that the unit for the electric field is that of force divided by charge, which is
a newton/coulomb (N/C) in SI units.
It is the surrounding charges that create an electric field at a given point. Any positive or negative
charge placed at the point interacts with the field and, as a result, experiences a force, as the next
example indicates.
Example 7 | An Electric Field Leads to a Force
In Figure 18-16a the charges on the two metal spheres and the ebonite rod create an electric field
at the spot indicated. This field has a magnitude of 2.0 N/C and is directed as in the drawing.
Determine the force on a charge placed at that spot, if the charge has a value of (a)
and (b)
.
Figure 18-16 The electric field
that exists at a given spot can exert a variety
of forces. The force exerted depends on the magnitude and sign of the charge
placed at that spot. (a) The force on a positive charge points in the same direction as
while (b) the force on a negative charge points opposite to
.
Reasoning The electric field at a given spot can exert a variety of forces, depending on the
magnitude and sign of the charge placed there. The charge is assumed to be small enough that it
does not alter the locations of the surrounding charges that create the field.
Solution
a. The magnitude of the force is the product of the magnitudes of q0 and
:
(18.2)
Since q0 is positive, the force points in the same direction as the electric field, as part a of the
drawing indicates.
b. In this case, the magnitude of the force is
(18.2)
The force on the negative charge points in the direction opposite to the force on the positive
charge—that is, opposite to the electric field (see part b of the drawing).
At a particular point in space, each of the surrounding charges contributes to the net electric field
that exists there. To determine the net field, it is necessary to obtain the various contributions
separately and then find the vector sum of them all. Such an approach is an illustration of the
principle of linear superposition, as applied to electric fields. (This principle is introduced in Section
17.1, in connection with waves.) Example 8 emphasizes the vector nature of the electric field, and
Example 9 illustrates that a charged particle accelerates in an electric field.
Example 8 | Electric Fields Add as Vectors Do
Figure 18-17 shows two charged objects, A and B. Each contributes as follows to the net electric
field at point P:
directed to the right, and
directed
downward. Thus,
and
are perpendicular. What is the net electric field at P?
Figure 18-17 The electric field contributions
and
, which come from
at
the two charge distributions, are added vectorially to obtain the net field
point P.
Reasoning The net electric field
is the vector sum of
and
:
. As
illustrated in Figure 18-17,
and
are perpendicular, so
is the diagonal of the rectangle
and
shown in the drawing. Thus, we can use the Pythagorean theorem to find the magnitude of
trigonometry to find the directional angle .
Solution The magnitude of the net electric field is
The direction of
is given by the angle in the drawing:
ANALYZING
PROBLEMS
MULTIPLE-CONCEPT
Example 9 | A Proton Accelerating in an Electric Field
In a vacuum, a proton
is moving parallel to a uniform
electric field that is directed along the + x axis (see Figure 18-18). The proton starts with a
velocity of
and accelerates in the same direction as the electric field, which has
. Find the velocity of the proton when its displacement is + 2.0 mm
a value of
from the starting point.
Figure 18-18 A proton, moving to the right, accelerates in the presence of the
electric field
.
Reasoning Since we know the initial velocity and displacement of the proton, we can determine
its final velocity from an equation of kinematics, provided the proton’s acceleration can be found.
The acceleration is given by Newton’s second law as the net force acting on the proton divided by
its mass. The net force is the electrostatic force, since the proton is moving in an electric field.
The electrostatic force depends on the proton’s charge and the electric field, both of which are
known.
Knowns and Unknowns The data for this problem are listed as follows:
Description
Symbol
Value
Proton charge
+e
Proton mass
m
Initial velocity of proton
v0
Electric field
E
Displacement of proton
x
+ 2.0 mm
v
?
x
x
Unknown Variable
Final velocity of proton
x
Comment
Modeling the Problem
Step 1 Kinematics
To obtain the final velocity v of the proton we employ Equation 3.6a from the equations of
x
kinematics:
. We have chosen this equation because two of the variables, the
initial velocity v0 and the displacement x, are known. Taking the square root of each side of
x
this relation and choosing the + sign, since the proton is moving in the + x direction (see
Figure 18-18), we arrive at Equation 1 in the right column. Although the acceleration a is
x
not known, we will obtain an expression for it in Step 2.
(1)
Step 2 Newton’s Second Law
Newton’s second law, as given in Equation 4.2a, states that the acceleration a of the proton
x
acting on it divided by the proton’s mass
is equal to the net force
the electrostatic force F acts on the proton, so it is the net force. Setting
x
. Only
in
Newton’s second law gives
This expression can be substituted into Equation 1, as indicated at the right. The electrostatic
force is not known, so we proceed to Step 3 to evaluate it using the concept of the electric
field.
Step 3 The Electric Field
Since the proton is moving in a uniform electric field E , it experiences an electrostatic force
x
F given by
x
(Equation 18.2), where q0 is the charge. Setting q = e for the proton,
0
we have
All the variables on the right side of this equation are known, so we substitute it into Equation
2, as shown in the right column.
Solution Algebraically combining the results of the three steps, we have
The final velocity of the proton is
where the + sign denotes that the final velocity points along the + x axis.
Related Homework: Problems 38, 40 and 66
Point Charges
A more complete understanding of the electric field concept can be gained by considering the field
created by a point charge, as in the following example.
Example 10 | The Electric Field of a Point Charge
There is an isolated point charge of
in a vacuum at the left in Figure 18-19a. Using a
, determine the electric field at point P, which is 0.20 m away.
test charge of
Figure 18-19 (a) At location P, a positive test charge q0 experiences a repulsive
force
due to the positive point charge q. (b) At P the electric field
is
directed to the right. (c) If the charge q were negative rather than positive, the
electric field would have the same magnitude as in (b) but would point to the left.
Reasoning Following the definition of the electric field, we place the test charge q0 at point P,
determine the force acting on the test charge, and then divide the force by the test charge.
Solution Coulomb’s law (Equation 18.1), gives the magnitude of the force:
Equation 18.2 gives the magnitude of the electric field:
The electric field
points in the same direction as the force
on the positive test charge. Since
the test charge experiences a force of repulsion directed to the right, the electric field vector also
points to the right, as Figure 18-19b shows.
The electric field produced by a point charge q can be obtained in general terms from Coulomb’s
law. First, note that the magnitude of the force exerted by the charge q on a test charge q0 is
. Then, divide this value by
to obtain the magnitude of the field. Since
eliminated algebraically from the result, the electric field does not depend on the test charge:
is
(18.3)
As in Coulomb’s law, the symbol |q| denotes the magnitude of q in Equation 18.3, without regard to
whether q is positive or negative. If q is positive, then
is directed away from q, as in Figure 1819b. On the other hand, if q is negative, then
is directed toward q, since a negative charge
attracts a positive test charge. For instance, Figure 18-19c shows the electric field that would exist at
P if there were a charge of − q instead of + q at the left of the drawing. Example 11 reemphasizes
the fact that all the surrounding charges make a contribution to the electric field that exists at a given
place.
Example 11 | The Electric Fields from Separate Charges May
Cancel
Two positive point charges,
and
, are separated in a vacuum by a
distance of 3.0 m, as Figure 18-20 illustrates. Find the spot on the line between the charges where
the net electric field is zero.
Figure 18-20 The two point charges q1 and q2 create electric fields
and
that cancel at a location P on the line between the charges.
Reasoning Between the charges the two field contributions have opposite directions, and the net
electric field is zero at the place where the magnitude of
equals that of
. However, since
q2 is smaller than q this location must be closer to q2, in order that the field of the smaller charge
1
can balance the field of the larger charge. In the drawing, the cancellation spot is labeled P, and its
distance from q1 is d.
Solution At P,
, and using the expression
Rearranging this expression shows that
of this equation reveals that
, we have
. Taking the square root of each side
The plus and minus signs on the right occur because either the positive or negative root can be
taken. Therefore, there are two possible values for d: + 2.0 m and + 6.0 m. The value + 6.0 m
and
corresponds to a location off to the right of both charges, where the magnitudes of
are equal, but where the directions are the same. Thus,
and
do not cancel at this spot. The
other value for d corresponds to the location shown in the drawing and is the zero-field location:
.
Problem solving insight
Equation 18.3 gives only the
magnitude of the electric field
produced by a point charge.
Therefore, do not use negative
numbers for the symbol |q| in
this equation.
When point charges are arranged in a symmetrical fashion, it is often possible to deduce useful
information about the magnitude and direction of the electric field by taking advantage of the
symmetry. Conceptual Example 12 illustrates the use of this technique.
Conceptual Example 12 | Symmetry and the Electric Field
Figure 18-21 shows point charges fixed to the corners of a rectangle in two different ways. The
charges all have the same magnitudes, but they have different signs. Consider the net electric field at
the center C of the rectangle in each case. Which field is larger?
Figure 18-21 Charges of identical magnitude, but different signs, are placed at the
corners of a rectangle. The charges give rise to different electric fields at the center
C of the rectangle, depending on the signs of the charges.
Reasoning and Solution In Figure 18-21a, the charges at corners 1 and 3 are both + q. The positive
charge at corner 1 produces an electric field at C that points toward corner 3. In contrast, the positive
charge at corner 3 produces an electric field at C that points toward corner 1. Thus, the two fields
have opposite directions. The magnitudes of the fields are identical because the charges have the
same magnitude and are equally far from the center. Therefore, the fields from the two positive
charges cancel.
Now, let’s look at the electric field produced by the charges on corners 2 and 4 in Figure 18-21a.
The electric field due to the negative charge at corner 2 points toward corner 2, and the field due to
the positive charge at corner 4 points the same way. Furthermore, the magnitudes of these fields are
equal because each charge has the same magnitude and is located at the same distance from the
shown
center of the rectangle. As a result, the two fields combine to give the net electric field
in the drawing.
In Figure 18-21b, the charges on corners 2 and 4 are identical to those in part a of the drawing, so
. The charges on corners 1 and 3 are identical to
this pair gives rise to the electric field labeled
those on corners 2 and 4, so they give rise to an electric field labeled
, which has the same
. The net electric field
is the vector sum of
and
and is also
magnitude as
shown in the drawing. Clearly, this sum is greater than
alone. Therefore, the net field in part
b is larger than that in part a.
Related Homework: Conceptual Question 12, Problem 34
Need more practice?
Interactive LearningWare18.3
A constant electric field exists in
a region of space. The field has a
magnitude of 1600 N/C and
points due north. A point charge
of
is then placed
in this electric field. (a) What is
the magnitude and direction of
the net electric field at a spot 13
cm due east of the charge?
Specify the direction as an angle
relative to due east. (b) If a
charge were placed
at this spot, what would be the
electrostatic force (magnitude
and direction) exerted on it?
Related Homework: Problem 59
The Parallel Plate Capacitor
Equation 18.3, which gives the electric field of a point charge, is a very useful result. With the aid of
integral calculus, this equation can be applied in a variety of situations where point charges are
distributed over one or more surfaces. One such example that has considerable practical importance
is the parallel plate capacitor. As Figure 18-22 shows, this device consists of two parallel metal
plates, each with area A. A charge + q is spread uniformly over one plate, while a charge − q is
spread uniformly over the other plate. In the region between the plates and away from the edges, the
electric field points from the positive plate toward the negative plate and is perpendicular to both. It
can be shown (see Example 16 in Section 18.9) that the electric field has a magnitude of
(18.4)
where is the permittivity of free space. In this expression the Greek symbol sigma ( ) denotes the
charge per unit area (
) and is sometimes called the charge density. Except in the region
near the edges, the field has the same value at all places between the plates. The field does not
depend on the distance from the charges, in distinct contrast to the field created by an isolated point
charge.
Figure 18-22 A parallel plate capacitor.
Check Your Understanding 3
A positive point charge + q is fixed in position at the center of a square, as the drawing shows. A
second point charge is fixed to either corner B, corner C, or corner D. The net electric field at
corner A is zero. (a) At which corner is the second charge located? (b) Is the second charge
positive or negative? (c) Does the second charge have a greater, a smaller, or the same magnitude
as the charge at the center? (The answers are given at the end of the book.)
Background: The electric field is a vector quantity. Its magnitude and direction depend on the
magnitudes and algebraic signs of the charges that create it.
For similar questions (including calculational counterparts), consult Self-Assessment Test
18.2, which is described at the end of Section 18.9.
Copyright © 2007 John Wiley & Sons, Inc. All rights reserved.
18.7
Electric Field Lines
As we have seen, electric charges create an electric field in the space surrounding them. It is useful to
have a kind of “map” that gives the direction and indicates the strength of the field at various places.
The great English physicist Michael Faraday (1791–1867) proposed an idea that provides such a
“map,” the idea of electric field lines. Since the electric field is the electric force per unit charge, the
electric field lines are sometimes called lines of force.
To introduce the electric field line concept, Figure 18-23a shows a positive point charge + q. At the
locations numbered 1–8, a positive test charge would experience a repulsive force, as the arrows in the
drawing indicate. Therefore, the electric field created by the charge + q is directed radially outward.
The electric field lines are lines drawn to show this direction, as part b of the drawing illustrates. They
begin on the charge + q and point radially outward. Figure 18-24 shows the field lines in the vicinity
of a negative charge − q. In this case they are directed radially inward because the force on a positive
test charge is one of attraction, indicating that the electric field points inward. In general, electric field
lines are always directed away from positive charges and toward negative charges.
Figure 18-23 (a) At any of the eight marked spots around a positive point charge
+ q, a positive test charge would experience a repulsive force directed radially
outward. (b) The electric field lines are directed radially outward from a positive point
charge + q.
Figure 18-24 The electric field lines are directed radially inward toward a negative
point charge − q.
The electric field lines in Figures 18-23 and 18-24 are drawn in only two dimensions, as a matter of
convenience. Field lines radiate from the charges in three dimensions, and an infinite number of lines
could be drawn. However, for clarity only a small number is ever included in pictures. The number is
chosen to be proportional to the magnitude of the charge; thus, five times as many lines would emerge
from a + 5q charge as from a + q charge.
The pattern of electric field lines also provides information about the magnitude or strength of the
field. Notice in Figures 18-23 and 18-24 that near the charges, where the electric field is stronger, the
lines are closer together. At distances far from the charges, where the electric field is weaker, the lines
are more spread out. It is true in general that the electric field is stronger in regions where the field
lines are closer together. In fact, no matter how many charges are present, the number of lines per unit
area passing perpendicularly through a surface is proportional to the magnitude of the electric field.
In regions where the electric field lines are equally spaced, there is the same number of lines per unit
area everywhere, and the electric field has the same strength at all points. For example, Figure 18-25
shows that the field lines between the plates of a parallel plate capacitor are parallel and equally
spaced, except near the edges where they bulge outward. The equally spaced, parallel lines indicate
that the electric field has the same magnitude and direction at all points in the central region of the
capacitor.
Figure 18-25 In the central region of a parallel plate capacitor, the electric field lines
are parallel and evenly spaced, indicating that the electric field there has the same
magnitude and direction at all points.
Often, electric field lines are curved, as in the case of an electric dipole. An electric dipole consists of
two separated point charges that have the same magnitude but opposite signs. The electric field of a
dipole is proportional to the product of the magnitude of one of the charges and the distance between
the charges. This product is called the dipole moment. Many molecules, such as H2O and HCl, have
dipole moments. Figure 18-26 depicts the field lines in the vicinity of a dipole. For a curved field line,
the electric field vector at a point is tangent to the line at that point (see points 1, 2, and 3 in the
drawing). The pattern of the lines for the dipole indicates that the electric field is greatest in the region
between and immediately surrounding the two charges, since the lines are closest together there.
Figure 18-26 The electric field lines of an electric dipole are curved and extend from
the positive to the negative charge. At any point, such as 1, 2, or 3, the field created
by the dipole is tangent to the line through the point.
Notice in Figure 18-26 that any given field line starts on the positive charge and ends on the negative
charge. In general, electric field lines always begin on a positive charge and end on a negative
charge and do not start or stop in midspace. Furthermore, the number of lines leaving a positive
charge or entering a negative charge is proportional to the magnitude of the charge. This means, for
example, that if 100 lines are drawn leaving a
charge, then 75 lines would have to end on a
charge and 25 lines on a
charge. Thus, 100 lines leave the charge of
and end
, So the lines begin and end on equal amounts of total charge.
on a total charge of
The electric field lines are also curved in the vicinity of two identical charges. Figure 18-27 shows the
pattern associated with two positive point charges and reveals that there is an absence of lines in the
region between the charges. The absence of lines indicates that the electric field is relatively weak
between the charges.
Figure 18-27 The electric field lines for two identical positive point charges. If the
charges were both negative, the directions of the lines would be reversed.
Some of the important properties of electric field lines are considered in Conceptual Example 13.
Conceptual Example 13 | Drawing Electric Field Lines
Figure 18-28a shows three negative point charges (− q, − q, and − 2q) and one positive point charge
(+ 4q), along with some electric field lines drawn between the charges. There are three things wrong
with this drawing. What are they?
Figure 18-28 (a) Incorrectly and (b) correctly drawn electric field lines.
Reasoning and Solution One aspect of Figure 18-28a that is incorrect is that electric field lines cross
at point P. Field lines can never cross, and here’s why. An electric charge placed at P experiences a
single net force due to the presence of the other charges in its environment. Therefore, there is only
one value for the electric field (which is the force per unit charge) at that point. If two field lines
intersected, there would be two electric fields at the point of intersection, one associated with each
line. Since there can be only one value of the electric field at any point, there can be only one electric
field line passing through that point.
Another mistake in Figure 18-28a is the number of electric field lines that end on the negative charges.
Remember that the number of field lines leaving a positive charge or entering a negative charge is
proportional to the magnitude of the charge. The − 2q charge has half the magnitude of the + 4q
charge. Therefore, since 8 lines leave the + 4q charge, 4 of them (one-half of them) must enter the − 2q
charge. Of the remaining 4 lines that leave the positive charge, 2 enter each of the − q charges,
according to a similar line of reasoning.
The third error in Figure 18-28a is the way in which the electric field lines are drawn between the + 4q
charge and the − q charge at the left of the drawing. As drawn, the lines are parallel and evenly spaced.
This would indicate that the electric field everywhere in this region has a constant magnitude and
direction, as is the case in the central region of a parallel plate capacitor. But the electric field between
the + 4q and − q charges is not constant everywhere. It certainly is stronger in places close to the + 4q
or − q charge than it is midway between them. The field lines, therefore, should be drawn with a
curved nature, similar (but not identical) to those that surround a dipole. Figure 18-28b shows more
nearly correct representations of the field lines for the four charges.
Related Homework: Problems 25 and 26
Copyright © 2007 John Wiley & Sons, Inc. All rights reserved.
18.8
The Electric Field Inside a Conductor: Shielding
In conducting materials such as copper, electric charges move readily in response to the forces that electric
fields exert. This property of conducting materials has a major effect on the electric field that can exist within
and around them. Suppose that a piece of copper carries a number of excess electrons somewhere within it,
as in Figure 18-29a. Each electron would experience a force of repulsion because of the electric field of its
neighbors. And, since copper is a conductor, the excess electrons move readily in response to that force. In
fact, as a consequence of the 1/r2 dependence on distance in Coulomb’s law, they rush to the surface of the
copper. Once static equilibrium is established with all of the excess charge on the surface, no further
movement of charge occurs, as part b of the drawing indicates. Similarly, excess positive charge also moves
to the surface of a conductor. In general, at equilibrium under electrostatic conditions, any excess charge
resides on the surface of a conductor.
Figure 18-29 (a) Excess charge within a conductor (copper) moves quickly (b) to the
surface.
Now consider the interior of the copper in Figure 18-29b. The interior is electrically neutral, although there
are still free electrons that can move under the influence of an electric field. The absence of a net movement
of these free electrons indicates that there is no net electric field present within the conductor. In fact, the
excess charges arrange themselves on the conductor surface precisely in the manner needed to make the
electric field zero within the material. Thus, at equilibrium under electrostatic conditions, the electric field
is zero at any point within a conducting material. This fact has some fascinating implications.
Figure 18-30a shows an uncharged, solid, cylindrical conductor at equilibrium in the central region of a
parallel plate capacitor. Induced charges on the surface of the cylinder alter the electric field lines of the
capacitor. Since an electric field cannot exist within the conductor under these conditions, the electric field
lines do not penetrate the cylinder. Instead, they end or begin on the induced charges. Consequently, a test
charge placed inside the conductor would feel no force due to the presence of the charges on the capacitor. In
other words, the conductor shields any charge within it from electric fields created outside the conductor.
The shielding results from the induced charges on the conductor surface.
Figure 18-30 (a) A cylindrical conductor (shown as an end view) is placed between the
oppositely charged plates of a capacitor. The electric field lines do not penetrate the
conductor. The blowup shows that, just outside the conductor, the electric field lines are
perpendicular to its surface. (b) The electric field is zero in a cavity within the conductor.
Since the electric field is zero inside the conductor, nothing is disturbed if a cavity is cut from the interior of
the material, as in part b of the drawing. Thus, the interior of the cavity is also shielded from external electric
fields, a fact that has important applications, particularly for shielding electronic circuits. “Stray” electric
fields are produced by various electrical appliances (e.g., hair dryers, blenders, and vacuum cleaners), and
these fields can interfere with the operation of sensitive electronic circuits, such as those in stereo amplifiers,
televisions, and computers. To eliminate such interference, circuits are often enclosed within metal boxes
that provide shielding from external fields.
The physics of shielding
electronic circuits.
The blowup in Figure 18-30a shows another aspect of how conductors alter the electric field lines created by
external charges. The lines are altered because the electric field just outside the surface of a conductor is
perpendicular to the surface at equilibrium under electrostatic conditions. If the field were not
perpendicular, there would be a component of the field parallel to the surface. Since the free electrons on the
surface of the conductor can move, they would do so under the force exerted by that parallel component. In
reality, however, no electron flow occurs at equilibrium. Therefore, there can be no parallel component, and
the electric field is perpendicular to the surface.
The preceding discussion deals with features of the electric field within and around a conductor at
equilibrium under electrostatic conditions. These features are related to the fact that conductors contain
mobile free electrons and do not apply to insulators, which contain very few free electrons. Conceptual
Example 14 further explores the behavior of a conducting material in the presence of an electric field.
Conceptual Example 14 | A Conductor in an Electric Field
A charge + q is suspended at the center of a hollow, electrically neutral, spherical conductor, as Figure 18-31
illustrates. Show that this charge induces (a) a charge of − q on the interior surface and (b) a charge of + q on
the exterior surface of the conductor.
Figure 18-31 A positive charge + q is suspended at the center of a hollow spherical
conductor that is electrically neutral. Induced charges appear on the inner and outer surfaces
of the conductor. The electric field within the conductor itself is zero.
Reasoning and Solution
a. Electric field lines emanate from the positive charge + q. Since the electric field inside the metal
conductor must be zero at equilibrium under electrostatic conditions, each field line ends when it
reaches the conductor, as the picture shows. Since field lines terminate only on negative charges, there
must be an induced negative charge on the interior surface of the conductor. Furthermore, the lines
begin and end on equal amounts of charge, so the magnitude of the total induced charge is the same as
the magnitude of the charge at the center. Thus, the total induced charge on the interior surface is − q.
b. Before the charge + q is introduced, the conductor is electrically neutral. Therefore, it carries no net
charge. We have also seen that there can be no excess charge within the metal. Thus, since an induced
charge of − q appears on the interior surface, a charge of + q must be induced on the outer surface.
The positive charge on the outer surface generates field lines that radiate outward (see the drawing) as
if they originated from the central charge and the conductor were absent. The conductor does not
shield the outside from the field produced by the charge on the inside.
Related Homework: Problem 57
Copyright © 2007 John Wiley & Sons, Inc. All rights reserved.
Gauss’ Law
We begin by developing a version of Gauss’ law that applies only to a point charge, which we assume to be
positive. The electric field lines for a positive point charge radiate outward in all directions from the charge, as
,
Figure 18-23b indicates. The magnitude E of the electric field at a distance r from the charge is
according to Equation 18.3, in which we have replaced the symbol |q| with the symbol q since we are assuming that
Figure 18-32 CONCEPTS AT A GLANCE The electric field and the surface through which it passes are
brought together to define the concept of electric flux. It is in terms of electric flux that Gauss’ law is formulated.
When the electric field is produced by a distribution of charges, such as that on the silvery metal sphere in the
photograph, Gauss’ law relates the electric flux to the charge. (© SSPL/ The Image Works)
CONCEPTS AT A GLANCE In presenting Gauss’ law, it will be necessary to introduce a new idea
called electric flux, and the Concepts-at-a-Glance chart in Figure 18-32 outlines the approach we will follow. The
idea of flux involves both the electric field and the surface through which it passes. By bringing together the
electric field and the surface through which it passes, we will be able to define electric flux and then present
Gauss’ law.
Section 18.6 discusses how a point charge creates an electric field in the space around the charge. There are also
many situations in which an electric field is produced by charges that are spread out over a region, rather than by
a single point charge. Such an extended collection of charges is called a “charge distribution.” For example,
the electric field within the parallel plate capacitor in Figure 18-22 is produced by positive charges spread
uniformly over one plate and an equal number of negative charges spread over the other plate. As we will see,
Gauss’ law describes the relationship between a charge distribution and the electric field it produces. This law
was formulated by the German mathematician and physicist Carl Friedrich Gauss (1777–1855).
18.9
, where
.
is
Figure 18-34 shows a charge distribution whose net charge is labeled Q. The charge distribution is surrounded by
a Gaussian surface—that is, an imaginary closed surface. The surface can have any arbitrary shape (it need not
be spherical), but it must be closed (an open surface would be like that of half an eggshell). The direction of
the electric field is not necessarily perpendicular to the Gaussian surface. Furthermore, the magnitude of the
electric field need not be constant on the surface, but can vary from point to point.
Equation 18.5 is the result we have been seeking, for it is the form of Gauss’ law that applies to a point charge.
depends only on the charge q within
This result indicates that, aside from the constant , the electric flux
the Gaussian surface and is independent of the radius r of the surface. We will now see how to generalize
Equation 18.5 to account for distributions of charges and Gaussian surfaces with arbitrary shapes.
Figure 18-33 A positive point charge is located at the center of an imaginary spherical surface of radius r. Such
a surface is one example of a Gaussian surface. Here the electric field is perpendicular to the surface and has the
same magnitude everywhere on it.
The left side of Equation 18.5 is the product of the magnitude E of the electric field at any point on the
Gaussian surface and the area A of the surface. In Gauss’ law this product is especially important and is called
the electric flux,
:
. (It will be necessary to modify this definition of flux when we consider the
general case of a Gaussian surface with an arbitrary shape.)
We now place this point charge at the center of an imaginary spherical surface of radius r, as Figure 18-33
shows. Such a hypothetical closed surface is called a Gaussian surface, although in general it need not be
spherical. The surface area A of a sphere is
, and the magnitude of the electric field can be written in terms
of this area as
, or
the permittivity of free space. With this substitution, the magnitude of the electric field becomes
the charge is positive. As mentioned in Section 18.5, the constant k can be expressed as
(18.5)
Gauss’ Law
where, as usual, the symbol means “the sum of.” Gauss’ law relates the electric flux
Q enclosed by the arbitrarily shaped Gaussian surface.
to the net charge
To determine the electric flux through such a surface, we divide the surface into many tiny sections with areas
, and so on. Each section is so small that it is essentially flat and the electric field
is a constant (both
in magnitude and direction) over it. For reference, a dashed line called the “normal” is drawn perpendicular to
each section on the outside of the surface. To determine the electric flux for each of the sections, we use only
that is perpendicular to the surface—that is, the component of the electric field that
the component of
passes through the surface. From the drawing it can be seen that this component has a magnitude of E
, where
)
is the angle between the electric field and the normal. The electric flux through any one section is then (E
. The electric flux
that passes through the entire Gaussian surface is the sum of all of these individual
, or
fluxes:
Figure 18-34 The charge distribution Q is surrounded by an arbitrarily shaped Gaussian surface. The electric flux
through any tiny segment of the surface is the product of
and the area
of the segment:
. The angle is the angle between the electric field and the normal to the surface.
(18.6)
,
Reasoning Because the charge is distributed uniformly over the spherical shell, the electric field is symmetrical.
This means that the electric field is directed radially outward in all directions, and its magnitude is the same at
all points that are equidistant from the shell. All such points lie on a sphere, so the symmetry is called
spherical symmetry. With this symmetry in mind, we will use a spherical Gaussian surface to evaluate the electric
flux
. We will then use Gauss’ law to determine the magnitude of the electric field.
(b) inside the shell, respectively.
Figure 18-35 A uniform distribution of positive charge resides on a thin spherical shell of radius R. The
spherical Gaussian surfaces (a) S and (b) S1 are used in Example 15 to evaluate the electric flux (a) outside and
Figures 18-35a and b show a thin spherical shell of radius R. A positive charge q is spread uniformly over the
shell. Find the magnitude of the electric field at any point (a) outside the shell and (b) inside the shell.
Example 15 | The Electric Field of a Charged Thin Spherical Shell
Although we arrived at Gauss’ law by assuming the net charge Q was positive, Equation 18.7 also applies when Q
is also negative. Gauss’ law is often used to find the magnitude of
is negative. In this case, the electric flux
the electric field produced by a distribution of charges. The law is most useful when the distribution is uniform
and symmetrical. In the next two examples we will see how to apply Gauss’ law in such situations.
SI Unit of Electric Flux: N·m2/C
The electric flux
through a Gaussian surface is equal to the net charge Q enclosed by the surface divided by
the permittivity of free space:
(18.7)
. In accord with Gauss’ law,
Example 16 | The Electric Field Inside a Parallel Plate Capacitor
Gauss’ law allows us to deduce that there is no electric field inside a uniform spherical shell of charge. An
electric field exists only on the outside.
, where Q is the net charge inside the Gaussian surface. But now Q = 0 C,
the electric flux must be equal to
or
since all the charge lies on the shell that is outside the surface S1. Consequently, we have
the electric flux through the Gaussian surface is
the electric field (if it exists) must also have spherical symmetry. Therefore, using reasoning like that in part (a),
b. To find the magnitude of the electric field inside the charged shell, we select a spherical Gaussian surface that
lies inside the shell and is concentric with it. See the surface labeled S1 in Figure 18-35b. Inside the charged shell,
This is a surprising result, for it is the same as that for a point charge (see Equation 18.3 with
). Thus,
the electric field outside a uniformly charged spherical shell is the same as if all the charge q were concentrated as
a point charge at the center of the shell.
. Thus, we can solve for E and find that
. Since the only charge within the Gaussian surface is the charge q on the shell, it follows
is just the sum of the tiny areas that make up the Gaussian surface. Since the area of a spherical
, we have
. Setting the electric flux equal to
, as specified by Gauss’ law,
that the net charge within the Gaussian surface is
yields
The term
surface is
a. To find the magnitude of the electric field outside the charged shell, we evaluate the electric flux
by using a spherical Gaussian surface of radius r (r > R) that is concentric with the shell. See
is everywhere perpendicular to the
the surface labeled S in Figure 18-35a. Since the electric field
Gaussian surface,
and
. In addition, E has the same value at all points on the surface, since they
are equidistant from the charged shell. Being constant over the surface, E can be factored outside the summation,
with the result that
Solution
Surface 1 —the flat left end of the cylinder—is embedded inside the positive metal plate. As discussed in Section
18.8, the electric field is zero everywhere inside a conductor that is in equilibrium under electrostatic conditions.
Since
, the electric flux through this surface is also zero:
Solution Figure 18-36b shows that we have placed our Gaussian cylinder so that its left end is inside the positive
metal plate, and the right end is in the space between the plates. To determine the electric flux through this
Gaussian surface, we evaluate the flux through each of the three parts—labeled 1, 2, and 3 in the drawing—that
make up the total surface of the cylinder and then add up the fluxes.
Figure 18-36 (a) A side view of a parallel plate capacitor, showing some of the electric field lines. (b) The
Gaussian surface is a cylinder oriented so its axis is perpendicular to the positive plate and its left end is inside
the plate.
Reasoning Figure 18-36a shows the electric field inside a parallel plate capacitor. Because the positive and
negative charges are distributed uniformly over the surfaces of the plates, symmetry requires that the electric field
be perpendicular to the plates. We will take advantage of this symmetry by choosing our Gaussian surface to be
a small cylinder whose axis is perpendicular to the plates (see part b of the figure). With this choice, we will be able
to evaluate the electric flux and then, with the aid of Gauss’ law, determine E.
According to Equation 18.4, the electric field inside a parallel plate capacitor, and away from the edges, is constant
, where is the charge density (the charge per unit area) on a plate. Use Gauss’
and has a magnitude of
law to prove this result.
For similar questions (including calculational counterparts), consult Self-Assessment Test 18.2, which is
described next.
Background: Electric flux is defined in terms of the electric field and the surface area through which the field
passes. Gauss’ law is formulated in terms of electric flux and relates the flux to the net electric charge.
The drawing shows an arrangement of three charges. In parts (a) and (b) different Gaussian surfaces are
shown. Through which surface, if either, does the greater electric flux pass? (The answer is given at the end of
the book.)
Check Your Understanding 4
, where Q is the net charge inside the
According to Gauss’ law, we set the electric flux equal to
. But Q/A is the charge per unit area, , on the plate. Therefore, we arrive at the
Gaussian cylinder:
. Notice that the distance of the right end of
value of the electric field inside a parallel plate capacitor:
the Gaussian cylinder from the positive plate does not appear in this result, indicating that the electric field has
the same value everywhere between the plates.
The electric flux through the entire Gaussian cylinder is the sum of the three fluxes determined above:
Surface 3—the flat right end of the cylinder—is perpendicular to the electric field between the plates,
. The electric field is constant over this surface, so E can be taken outside the summation
so
is the area of surface 3, we find that the electric flux through this surface is
in Equation 18.6. Noting that
Surface 2—the curved wall of the cylinder—is everywhere parallel to the electric field between the plates, so
. Therefore, the electric flux through this surface is also zero:
that
Copyright © 2007 John Wiley & Sons, Inc. All rights reserved.
· Gauss’ Law
· Electric Field Lines
· The Electric Field
Test your understanding of the material in Sections 18.6, 18.7, 18.8 and 18.9 :
18.10
*Copiers and Computer Printers
The electrostatic force that charged particles exert on one another plays the central role in an office copier.
The copying process is called xerography, from the Greek xeros view and graphos, meaning “dry writing.” The
heart of a copier is the xerographic drum, an aluminum cylinder coated with a layer of selenium (see Figure 1837a). Aluminum is an excellent electrical conductor. Selenium, on the other hand, is a photoconductor: it is
an insulator in the dark but becomes a conductor when exposed to light. Consequently, a positive charge deposited
on the selenium surface will remain there, provided the selenium is kept in the dark. When the drum is exposed
to light, however, electrons from the aluminum pass through the conducting selenium and neutralize the
positive charge.
Figure 18-37 (a) This cutaway shows the essential elements of a copying machine. (b) The five steps in
the xerographic process.
The physics of xerography.
The photoconductive property of selenium is critical to the xerographic process, as Figure 18-37b illustrates. First,
an electrode called a corotron gives the entire selenium surface a positive charge in the dark. Second, a series of
lenses and mirrors focuses an image of a document onto the revolving drum. The dark and light areas of the
document produce corresponding areas on the drum. The dark areas retain their positive charge, but the light
areas become conducting and lose their positive charge, ending up neutralized. Thus, a positive-charge image of
the document remains on the selenium surface. In the third step, a special dry black powder, called the toner, is given
a negative charge and then spread onto the drum, where it adheres selectively to the positively charged areas.
The fourth step involves transferring the toner onto a blank piece of paper. However, the attraction of the
positive-charge image holds the toner to the drum. To transfer the toner, the paper is given a greater positive
charge than that of the image, with the aid of another corotron. Last, the paper and adhering toner pass through
heated pressure rollers, which melt the toner into the fibers of the paper and produce the finished copy.
A laser printer is used with computers to provide high-quality copies of text and graphics. It is similar in operation
to the xerographic machine, except that the information to be reproduced is not on paper. Instead, the information
is transferred from the computer’s memory to the printer, and laser light is used to copy it onto the seleniumaluminum drum. A laser beam, focused to a fine point, is scanned rapidly from side to side across the rotating drum,
as Figure 18-38 indicates. While the light remains on, the positive charge on the drum is neutralized. As the laser
beam moves, the computer turns the beam off at the right moments during each scan to produce the desired
positive-charge image, which is the letter “A” in the picture.
Figure 18-38 As the laser beam scans back and forth across the surface of the xerographic drum, a positivecharge image of the letter “A” is created.
The physics of a laser printer.
An inkjet printer is another type of printer that uses electric charges in its operation. While shuttling back and
forth across the paper, the inkjet printhead ejects a thin stream of ink. Figure 18-39 illustrates the elements of one
type of printhead. The ink is forced out of a small nozzle and breaks up into extremely small droplets, with
diameters that can be as small as 9 × 10−6 m. About 150 000 droplets leave the nozzle each second and travel with
a speed of approximately 18 m/s toward the paper. During their flight, the droplets pass through two
electrical components, an electrode and the deflection plates (a parallel plate capacitor). When the printhead
moves over regions of the paper that are not to be inked, the charging control is turned on and an electric field
is established between the printhead and the electrode. As the drops pass through the electric field, they acquire a
net charge by the process of induction. The deflection plates divert the charged droplets into a gutter and thus
prevent them from reaching the paper. Whenever ink is to be placed on the paper, the charging control, responding
to instructions from the computer, turns off the electric field. The uncharged droplets fly straight through the
deflection plates and strike the paper.
Figure 18-39 An inkjet printhead ejects a steady flow of ink droplets. Charged droplets are deflected into a
gutter by the deflection plates, while uncharged droplets fly straight onto the paper. Letters formed by an inkjet
printer look normal, except when greatly enlarged and the patterns from the drops become apparent.
The physics of an inkjet printer.
Copyright © 2007 John Wiley & Sons, Inc. All rights reserved.
18.11
Concepts & Calculations
In this chapter we have studied electric forces and electric fields. We conclude now by presenting
some examples that review important features of these concepts. The three-part format of the
examples stresses the role of conceptual understanding in problem solving. First, the problem
statement is given. Then, there is a concept question-and-answer section, followed by the solution
section. The purpose of the concept question-and-answer section is to provide help in
understanding the solution and to illustrate how a review of the concepts can help in anticipating
some of the characteristics of the numerical answers.
Concepts & Calculations Example 17 | The Vector Nature of
Electric Forces
The charges on three identical metal spheres are
,
, and
. The spheres
are brought together so they simultaneously touch each other. They are then separated and placed
on the x and y axes, as in Figure 18-40a. What is the net force (magnitude and direction) exerted
on the sphere at the origin? Treat the spheres as if they were particles.
Figure 18-40 (a) Three equal charges lie on the x and y axes. (b) The net force
.
exerted on q1 by the other two charges is
Concept Questions and Answers Is the net charge on the system comprising the three spheres
the same before and after touching?
Answer Yes. The conservation of electric charge states that during any process the net electric
charge of an isolated system remains constant (is conserved). Therefore, the net charge on the
three spheres before they touch (
) is the same as the
net charge after they touch.
After the spheres touch and are separated, do they have identical charges?
Answer Yes. Since the spheres are identical, the net charge
distributes itself
equally over the three spheres. After they are separated, each has one-third of the net charge:
.
Do q2 and q3 exert forces of equal magnitude on q1?
Answer Yes. The charges q2 and q3 have equal magnitudes and are the same distance from q1.
According to Coulomb’s law, then, they exert forces of equal magnitude on q1.
Is the magnitude of the net force exerted on q1 equal to 2F, where F is the magnitude of the force
that either q2 or q3 exerts on q1?
Answer No. Although the two forces that act on q1 have equal magnitudes, they have different
directions. The forces are repulsive forces, since all of the charges in part a of the drawing are
exerted on q1 by q2 and the force
identical. Figure 18-40b shows the force
exerted on q1 by q3. To obtain the net force
, we must take these directions into account by
using vector addition.
Solution The magnitude F12 of the force exerted on q1 by q2 is given by Coulomb’s law, Equation
18.1, as
Problem solving insight Often
charge magnitudes are
specified in microcoulombs
. When using Coulomb’s
law, be sure to convert
microcoulombs into coulombs
before
substituting for the charge
and
.
magnitudes
Note that we have used the magnitudes of q1 and q2 in Coulomb’s law. As mentioned previously,
the magnitude of the force
exerted on q1 by q3 has the same value as F12, and so
. Since the forces
and
are perpendicular to each other (see Figure
18-40b), we may use the Pythagorean theorem to find the magnitude F of the net force:
The angle that the net force makes with the −x axis (see part b of the drawing) is
Concepts & Calculations Example 18 | Becoming Familiar
with Electric Fields
Two point charges are lying on the y axis in Figure 18-41a:
and
. They are equidistant from the point P, which lies on the x axis. (a) What is the
and mass m = 1.20 g is
net electric field at P? (b) A small object of charge
placed at P. When it is released, what is its acceleration?
Figure 18-41 (a) Two charges q1 and q2 produce an electric field at the point P.
(b) The electric fields
and
add to give the net electric field
.
Concept Questions and Answers There is no charge at P in part (a). Is there an electric field at
P?
Answer Yes. An electric field is produced by the charges q1 and q2, and it exists throughout the
entire region that surrounds them. If a test charge were placed at this point, it would experience
a force due to the electric field. The force would be the product of the charge and the electric
field.
The charge q1 produces an electric field at the point P. What is the direction of this field?
Answer The electric field created by a charge always points away from a positive charge and
points toward it (see
toward a negative charge. Since q1 is negative, the electric field
Figure 18-41b).
What is the direction of the electric field produced by q2 at P?
Answer Since q2 is positive, the electric field
that it produces points away from q2, as
shown in the drawing.
Is the magnitude of the net electric field equal to E1 + E2, where E1 and E2 are the magnitudes of
the electric fields produced by q1 and q2?
Answer No, because the electric fields have different directions. We must add the individual
fields as vectors to obtain the net electric field. Only then can we determine its magnitude.
Solution
a. The magnitude of the electric fields that q1 and q2 produce at P are given by Equation 18.3,
where the distances are specified in the drawing:
The x and y components of these fields and the total field
table:
are given in the following
Electric
field
x component
= 0 N/C
The net electric field
y component
= + 7.56 × 104 N/C
has only a component along the + y axis. So,
of an object placed
b. According to Newton’s second law, Equation 4.2, the acceleration
is
at this point is equal to the net force acting on it divided by its mass. The net force
, as indicated by Equation
the product of the charge and the net electric field,
18.2. Thus, the acceleration is
At the end of the problem set for this chapter, you will find homework problems that contain both
conceptual and quantitative parts. These problems are grouped under the heading Concepts &
Calculations.
Copyright © 2007 John Wiley & Sons, Inc. All rights reserved.
Concept Summary
If you need more help with a concept, use the text examples and media noted next to the
discussion or equation. Go to for:
Interactive LearningWare (ILW)—examples solved in a five-step interactive format.
Concept Simulations (CS)—animated text figures or animations of important concepts.
Interactive Solutions (IS)—models for certain types of problems in the chapter homework. The
calculations are carried out interactively.
Self-Assessment Tests—qualitative and quantitative questions with extensive feedback.
Topic
Discussion
The coulomb (C)
18.1 THE ORIGIN OF ELECTRICITY There are
two kinds of electric charge: positive and negative. The
SI unit of electric charge is the coulomb (C). The
magnitude of the charge on an electron or a proton is
Magnitude of
charge on electron
or proton
Since the symbol e denotes a magnitude, it has no
algebraic sign. Thus, the electron carries a charge of
− e, and the proton carries a charge of + e.
Charge is quantized The charge on any object, whether positive or negative,
is quantized, in the sense that the charge consists of an
integer number of protons or electrons.
Law of
conservation of
electric charge
18.2 CHARGED OBJECTS AND THE ELECTRIC
FORCE The law of conservation of electric charge
states that the net electric charge of an isolated system
remains constant during any process.
Electric repulsion
and attraction
Like charges repel and unlike charges attract each other.
Conductor
18.3 CONDUCTORS AND INSULATORS An
electrical conductor is a material, such as copper, that
conducts electric charge readily.
Learning Aids
Insulator
An electrical insulator is a material, such as rubber, that
conducts electric charge poorly.
Charging by contact 18.4 CHARGING BY CONTACT AND BY
INDUCTION Charging by contact is the process of
giving one object a net electric charge by placing it in
contact with an object that is already charged.
Charging by
induction
Charging by induction is the process of giving an object
a net electric charge without touching it to a charged
object.
Point charge
18.5 COULOMB’s LAW A point charge is a charge
that occupies so little space that it can be regarded as a
mathematical point.
Coulomb’s law gives the magnitude F of the electric
force that two point charges q1 and q2 exert on each
other:
Coulomb’s law
(18.1)
where
and
are the magnitudes of the charges
and have no algebraic sign. The term k is a constant and
. The force
has the value
specified by Equation 18.1 acts along the line between
the two charges.
Permittivity of free
space
The permittivity of free space
relation
is defined by the
Use Self-Assessment Test 18.1 to evaluate your understanding of Sections 18.1, 18.2,
18.3, 18.4 and 18.5 .
18.6 THE ELECTRIC FIELD The electric field
at a given spot is a vector and is the electrostatic force
experienced by a very small test charge q0 placed
at that spot divided by the charge itself:
Electric field
(18.2)
The direction of the electric field is the same as the
direction of the force on a positive test charge. The SI
unit for the electric field is the newton per coulomb (N/
C). The source of the electric field at any spot is the
charged objects surrounding that spot.
The magnitude of the electric field created by a point
charge q is
Electric field of a
point charge
(18.3)
where |q| is the magnitude of the charge and has no
algebraic sign and r is the distance from the charge. The
points away from a positive charge
electric field
and toward a negative charge.
For a parallel plate capacitor that has a charge per unit
area of on each plate, the magnitude of the electric
field between the plates is
Electric field of a
parallel plate
capacitor
Electric field lines
Direction of electric
field Strength of
electric field
(18.4)
18.7 ELECTRIC FIELD LINES Electric field lines
are lines that can be thought of as a “map,” insofar as
the lines provide information about the direction and
strength of the electric field. The lines are directed
away from positive charges and toward negative
charges. The direction of the lines gives the direction of
the electric field, since the electric field vector at a
point is tangent to the line at that point. The electric
field is strongest in regions where the number of lines
per unit area passing perpendicularly through a surface
is the greatest—that is, where the lines are packed
together most tightly.
Excess charge
carried by a
conductor at
equilibrium
18.8 THE ELECTRIC FIELD INSIDE A
CONDUCTOR: SHIELDING Excess negative or
positive charge resides on the surface of a conductor at
equilibrium under electrostatic conditions. In such a
situation, the electric field at any point within the
conducting material is zero, and the electric field just
outside the surface of the conductor is perpendicular to
the surface.
through a
18.9 Gauss’ LAW The electric flux
surface is related to the magnitude E of the electric
field, the area A of the surface, and the angle that
specifies the direction of the field relative to the normal
to the surface:
Electric flux
(18.6)
Gauss’ law states that the electric flux through a closed
surface (a Gaussian surface) is equal to the net charge
Q enclosed by the surface divided by , the
permittivity of free space:
Gauss’ law
(18.7)
Use Self-Assessment Test 18.2 to evaluate your understanding of Sections 18.6, 18.7,
18.8 and 18.9 .
Copyright © 2007 John Wiley & Sons, Inc. All rights reserved.
Conceptual Questions
ssm Solution is in the Student Solutions Manual
1 In Figure 18-8 the grounding wire is removed first, followed by the rod, and the sphere is
left with a positive charge. If the rod were removed first, followed by the grounding wire,
would the sphere be left with a charge? Account for your answer.
2 A metallic object is given a positive charge by the process of induction, as illustrated in
Figure 18-8. (a) Does the mass of the object increase, decrease, or remain the same? Why?
(b) What happens to the mass of the object if it is given a negative charge by induction?
Explain.
3 A rod made from insulating material carries a net charge, while a copper sphere is neutral.
The rod and the sphere do not touch. Is it possible for the rod and the sphere to (a) attract
one another and (b) repel one another? Explain.
4 On a dry day, just after washing your hair to remove natural oils and drying it thoroughly,
run a plastic comb through it. Small bits of paper will be attracted to the comb. Explain
why.
5 Blow up a balloon and rub it against your shirt a number of times. In so doing you give the
balloon a net electric charge. Now touch the balloon to the ceiling. On being released, the
balloon will remain stuck to the ceiling. Why?
6 A proton and an electron are held in place on the x axis. The proton is at x = −d, while the
electron is at x = + d. They are released simultaneously, and the only force that affects
their motions is the electrostatic force of attraction that each applies to the other. Which
particle reaches the origin first? Give your reasoning.
7 A particle is attached to a spring and is pushed so that the spring is compressed more and
more. As a result, the spring exerts a greater and greater force on the particle. Similarly, a
charged particle experiences a greater and greater force when pushed closer and closer to
another particle that is fixed in position and has a charge of the same polarity. In spite of
the similarity, the charged particle will not exhibit simple harmonic motion on being
released, as will the particle on the spring. Explain why not.
8 ssm Identical point charges are fixed to opposite corners of a square. Where does a third
point charge experience the greater force, at one of the empty corners or at the center of
the square? Account for your answer.
9 On a thin, nonconducting rod, positive charges are spread evenly, so that there is the same
amount of charge per unit length at every point. On another identical rod, positive charges
are spread evenly over only the left half, and the same amount of negative charges are
spread evenly over the right half. For each rod, deduce the direction of the electric field at
a point that is located directly above the midpoint of the rod. Give your reasoning.
10 There is an electric field at point P. A very small charge is placed at this point and
experiences a force. Another very small charge is then placed at this point and
experiences a force that differs in both magnitude and direction from that experienced by
the first charge. How can these two different forces result from the single electric field
that exists at point P?
11 Three point charges are fixed to the corners of a square, one to a corner, in such a way
that the net electric field at the empty corner is zero. Do these charges all have (a) the
same sign and (b) the same magnitude (but, possibly, different signs)? Justify your
answers.
12 Review Conceptual Example 12 as an aid in answering this question. Suppose in Figure
18-21 that charges + q are placed on corners 1 and 3 of the rectangle, and charges − q are
placed on corners 2 and 4. What is the net electric field at the center C of the rectangle?
13 In Figure 18-26 there is no place on the line through the charges where the electric field
is zero, neither to the left of the positive charge, nor between the charges, nor to the right
of the negative charge. Now, suppose the magnitude of the negative charge were greater
than the magnitude of the positive charge. Is there any place on the line through the
charges where the electric field is zero? Justify your answer.
14 ssm Drawings I and II show two examples of electric field lines. Decide which of the
following statements are true and which are false, defending your choice in each case. (a)
In both I and II the electric field is the same everywhere. (b) As you move from left to
right in each case, the electric field becomes stronger. (c) The electric field in I is the
same everywhere but becomes stronger in II as you move from left to right. (d) The
electric fields in both I and II could be created by negative charges located somewhere
on the left and positive charges somewhere on the right. (e) Both I and II arise from a
single positive point charge located somewhere on the left.
15 A positively charged particle is moving horizontally when it enters the region between
the plates of a capacitor, as the drawing illustrates. (a) Draw the trajectory that the
particle follows in moving through the capacitor. (b) When the particle is within the
capacitor, which of the following four vectors, if any, are parallel to the electric field
inside the capacitor: the particle’s displacement, its velocity, its linear momentum, its
acceleration? For each vector explain why the vector is, or is not, parallel to the electric
field of the capacitor.
16 Refer to Figure 18-27. Imagine a plane that is perpendicular to a line between the
charges, midway between them, and is half into and half out of the paper. The electric
flux through this plane is zero. Explain why.
17 ssm Two charges, + q and − q, are inside a Gaussian surface. Since the net charge inside
the Gaussian surface is zero, Gauss’ law states that the electric flux through the surface is
. Does the fact that
imply that the electric field
at
also zero; that is,
any point on the Gaussian surface is also zero? Justify your answer.
Hint:
(Imagine a Gaussian surface that encloses the two charges in Figure 18-26.)
18 The drawing shows three charges, labeled q1, q2, and q3. A Gaussian surface is drawn
around q1 and q2. (a) Which charges determine the electric flux through the Gaussian
surface? (b) Which charges produce the electric field at the point P? Justify your
answers.
19 A charge + q is placed inside a spherical Gaussian surface. The charge is not located at
the center of the sphere. (a) Can Gauss’ law tell us exactly where the charge is located
inside the sphere? Justify your answer. (b) Can Gauss’ law tell us about the magnitude of
the electric flux through the Gaussian surface? Why?
Copyright © 2007 John Wiley & Sons, Inc. All rights reserved.
. What is the net charge after
electrons have been placed on it?
of water? (b) What is the net charge of all these electrons?
Section 18.5 Coulomb’s Law
7 ssm Two charges attract each other with a force of 1.5 N. What will be the force if the distance between them
is reduced to one-ninth of its original value?
electrons are there in one liter
. The objects are
4 Four identical metallic objects carry the following charges: + 1.6, + 6.2, − 4.8, and
brought simultaneously into contact, so that each touches the others. Then they are separated, (a) What is the
final charge on each object? (b) How many electrons (or protons) make up the final charge on each object?
5 ssm Consider three identical metal spheres, A, B, and C. Sphere A carries a charge of + 5q. Sphere B carries a
charge of − q. Sphere C carries no net charge. Spheres A and B are touched together and then separated. Sphere C
is then touched to sphere A and separated from it. Last, sphere C is touched to sphere B and separated from it. (a)
How much charge ends up on sphere C? What is the total charge on the three spheres (b) before they are allowed
to touch each other and (c) after they have touched?
* 6 Water has a mass per mole of 18.0 g/mol, and each water molecule (H2O) has 10 electrons. (a) How many
3 A metal sphere has a charge of
1 ssm How many electrons must be removed from an electrically neutral silver dollar to give it a charge of
, while a rod carries a charge of
How many electrons must
2 A plate carries a charge of
be transferred from the plate to the rod, so that both objects have the same charge?
Section 18.4 Charging by Contact and by Induction
Section 18.3 Conductors and Insulators,
Section 18.2 Charged Objects and the Electric Force,
Section 18.1 The Origin of Electricity,
This icon represents a biomedical application.
ssm Solution is in the Student Solutions Manual.
are presented in a guided tutorial format that provides enhanced interactivity. See Preface for additional details.
Note to Instructors: Most of the homework problems in this chapter are available for assignment via an
online homework management program such as WileyPLUS or WebAssign, and those marked with the icon
Problems that are not marked with a star are considered the easiest to solve. Problems that are marked with a
single star (*) are more difficult, while those marked with a double star (**) are the most difficult.
Problems
?
. Find (a) the charge on each particle and (b) the mass of particle 2.
and
. They are separated by
14 Two tiny conducting spheres are identical and carry charges of
a distance of 2.50 cm. (a) What is the magnitude of the force that each sphere experiences, and is the force attractive
or repulsive? (b) The spheres are brought into contact and then separated to a distance of 2.50 cm. Determine
the magnitude of the force that each sphere now experiences, and state whether the force is attractive or repulsive.
15 Interactive Solution 18.15 at provides a model for solving this type of problem. Two small objects, A and B, are
fixed in place and separated by 3.00 cm in a vacuum. Object A has a charge of
, and object B has a
charge of
. How many electrons must be removed from A and put onto B to make the electrostatic
force that acts on each object an attractive force whose magnitude is 68.0 N?
* 16 Multiple-Concept Example 3 provides some pertinent background for this problem. Suppose a single electron
orbits about a nucleus containing two protons (+ 2e), as would be the case for a helium atom from which one of
. Determine the
the two naturally occurring electrons is removed. The radius of the orbit is
magnitude of the electron’s centripetal acceleration.
of
13 ssm www Two particles, with identical positive charges and a separation of
, are released from
rest. Immediately after the release, particle 1 has an acceleration
whose magnitude is
,
while particle 2 has an acceleration
whose magnitude is
. Particle 1 has a mass
9 ssm www Two very small spheres are initially neutral and separated by a distance of 0.50 m. Suppose that
electrons are removed from one sphere and placed on the other. (a) What is the magnitude of
the electrostatic force that acts on each sphere? (b) Is the force attractive or repulsive? Why?
10 A charge + q is located at the origin, while an identical charge is located on the x axis at x = + 0.50 m. A third
charge of + 2 q is located on the x axis at such a place that the net electrostatic force on the charge at the
origin doubles, its direction remaining unchanged. Where should the third charge be located?
11 ssm Consult Concept Simulation 18.1 at for insight into this problem. Three charges are fixed to an x, y
coordinate system. A charge of
is on the y axis at y = + 3.0 m. A charge of
is at the origin. Last,
is on the x axis at x = + 3.0 m. Determine the magnitude and direction of the net electrostatic
a charge of
force on the charge at x = + 3.0 m. Specify the direction relative to the − x axis.
is fixed at the center of a compass. Two additional charges are fixed on the circle of
12 A charge of
the compass (radius = 0.100 m). The charges on the circle are
at the position due north and
at the position due east. What is the magnitude and direction of the net electrostatic force acting on the charge at
the center? Specify the direction relative to due east.
8 Two spherical objects are separated by a distance of
. The objects are initially electrically neutral
and are very small compared to the distance between them. Each object acquires the same negative charge due to
the addition of electrons. As a result, each object experiences an electrostatic force that has a magnitude
of
. How many electrons did it take to produce the charge on one of the objects?
* 20 Multiple-Concept Example 3 illustrates several of the concepts used in this problem. A single electron orbits a
lithium nucleus that contains three protons (+ 3e). The radius of the orbit is
. Determine the
kinetic energy of the electron.
* 21 An electrically neutral model airplane is flying in a horizontal circle on a 3.0-m guideline, which is nearly parallel
to the ground. The line breaks when the kinetic energy of the plane is 50.0 J. Reconsider the same situation, except
that now there is a point charge of + q on the plane and a point charge of − q at the other end of the guideline. In
this case, the line breaks when the kinetic energy of the plane is 51.8 J. Find the magnitude of the charges.
* 19 ssm Two spheres are mounted on identical horizontal springs and reston a frictionless table, as in the drawing.
When the spheres are uncharged, the spacing between them is 0.0500 m, and the springs are unstrained. When
, the spacing doubles. Assuming that the spheres have a negligible
each sphere has a charge of
diameter, determine the spring constant of the springs.
points vertically downward and has a magnitude of 405 N. Determine the magnitudes and algebraic signs of
the charges qA and qB.
* 18 The drawing shows an equilateral triangle, each side of which has a length of 2.00 cm. Point charges are fixed to
charge experiences a net force due to the charges qA and qB. This net force
each corner, as shown. The
1.50 g and it were free to move, what would be its acceleration?
* 17 Interactive LearningWare 18.2 at provides one approach to solving problems such as this one. The drawing
shows three point charges fixed in place. The charge at the coordinate origin has a value of
; the
other two charges have identical magnitudes, but opposite signs:
and
. (a)
Determine the net force (magnitude and direction) exerted on q1 by the other two charges. (b) If q1 had a mass of
28 Four point charges have the same magnitude of
and are fixed to the corners of a square that is 4.0
cm on a side. Three of the charges are positive and one is negative. Determine the magnitude of the net electric
field that exists at the center of the square.
, are fixed in place and separated by 3.0 m. (a) At what spot along a line through
29 Two charges, −16 and
the charges is the net electric field zero? Locate this spot relative to the positive charge. (Hint: The spot does
placed at this spot?
not necessarily lie between the two charges.) (b) What would be the force on a charge of
Section 18.8 The Electric Field Inside a Conductor: Shielding
25 Concept Simulation 18.2 at provides background concerning the electric field lines that are the focus of this
problem. Review the important features of electric field lines discussed in Conceptual Example 13. Three
point charges (+ q, + 2q, and −3q) are at the corners of an equilateral triangle. Sketch in six electric field lines
between the three charges
26 Review Conceptual Example 13 as an aid in working this problem. Charges of −4q are fixed to diagonally
opposite corners of a square. A charge of + 5q is fixed to one of the remaining corners, and a charge of + 3q is fixed
to the last corner. Assuming that ten electric field lines emerge from the + 5q charge, sketch the field lines in
the vicinity of the four charges.
. What electric field (magnitude and direction)
27 ssm www A tiny ball (mass = 0.012 kg) carries a charge of
is needed to cause the ball to float above the ground?
Section 18.7 Electric Field Lines,
Section 18.6 The Electric Field,
. Two are positive and two are negative. The charges are
** 24 There are four charges, each with a magnitude of
fixed to the corners of a 0.30-m square, one to a corner, in such a way that the net force on any charge is
directed toward the center of the square. Find the magnitude of the net electrostatic force experienced by any charge.
is held 0.150 m away from the sphere and directly to the right of it, so
negligible mass. A charge of
the wire makes an angle with the vertical (see the drawing). Find (a) the angle and (b) the tension in the wire.
** 22 Two identical small insulating balls are suspended by separate 0.25-m threads that are attached to a common point
on the ceiling. Each ball has a mass of 8.0 × 10−4 kg. Initially the balls are uncharged and hang straight down.
They are then given identical positive charges and, as a result, spread apart with an angle of 36° between the
threads. Determine (a) the charge on each ball and (b) the tension in the threads.
** 23 ssm A small spherical insulator of mass
and charge
is hung by a thin wire of
The membrane surrounding a living cell consists of an inner and an outer wall that are separated by a
is at x1 = + 9.0 cm. Find the net electric field (magnitude and direction) at (a) x = 0 cm and
that is perpendicular to the
. It is placed at a certain spot where there is an electric field. When
in the direction of the + x axis. Determine
released, the object experiences an acceleration of
the magnitude and direction of the electric field.
* 38 Multiple-Concept Example 9 illustrates some of the concepts pertinent to this problem. An electron is released
from rest at the negative plate of a parallel plate capacitor. The charge per unit area on each plate
, and the plates are separated by a distance of
. How fast is the electron
is
moving just before it reaches the positive plate?
and a charge of
the plate radius for the first capacitor.
* 37 Interactive Solution 18.37 at provides a model for problems of this kind. A small object has a mass of
r between the two charges?
* 36 Two parallel plate capacitors have circular plates. The magnitude of the charge on these plates is the same.
, while the field within
However, the electric field between the plates of the first capacitor is
. Determine the ratio r2/r1 of the plate radius for the second capacitor to
the second capacitor is
is
. The charge per unit area on each plate has a magnitude of
. The force
on q1 due to q2 equals the force on q1 due to the electric field of the parallel plate capacitor. What is the distance
rod. With respect to the z axis, find the magnitude of the net torque applied to the rod.
33 ssm A small drop of water is suspended motionless in air by a uniform electric field that is directed upward and has
a magnitude of 8480 N/C. The mass of the water drop is
. (a) Is the excess charge on the water
drop positive or negative? Why? (b) How many excess electrons or protons reside on the drop?
34 Review Conceptual Example 12 before attempting to work this problem. The magnitude of each of the charges
. The lengths of the sides of the rectangles are 3.00 cm and 5.00 cm. Find
in Figure 18-21 is
the magnitude of the electric field at the center of the rectangle in Figures 18-21a and b.
* 35 Two charges are placed between the plates of a parallel plate capacitor. One charge is + q1 and the other
x, y plane there is a constant external electric field
(b) x = + 6.0 cm.
32 A long, thin rod (length = 4.0 m) lies along the x axis, with its midpoint at the origin. In a vacuum, a
point charge is fixed to one end of the rod, and a
point charge is fixed to the other end. Everywhere in the
other
small space. Assume that the membrane acts like a parallel plate capacitor in which the effective charge density on
. (a) What is the magnitude of the electric field
the inner and outer walls has a magnitude of
within the cell membrane? (b) Find the magnitude of the electric force that would be exerted on a potassium ion (K
+; charge = + e) placed inside the membrane.
is at x1 = + 3.0 cm and the
31 ssm Two charges are placed on the x axis. One of the charges
30
?
** 43 A small plastic ball with a mass of
and with a charge of
is suspended from an
insulating thread and hangs between the plates of a capacitor (see the drawing). The ball is in equilibrium, with
the thread making an angle of 30.0° with respect to the vertical. The area of each plate is 0.0150 m2. What is
the magnitude of the charge on each plate?
** 42 The drawing shows an electron entering the lower left side of a parallel plate capacitor and exiting at the upper
. The capacitor is 2.00 cm long, and its plates
right side. The initial speed of the electron is
are separated by 0.150 cm. Assume that the electric field between the plates is uniform everywhere and find
its magnitude.
a circle. At the center of the circle they produce a net electric field that is directed upward along the vertical
axis. Determine the ratio
/
of the charge magnitudes.
* 41 Interactive Solution 18.41 at for help with this problem. The drawing shows two positive charges q1 and q2 fixed to
of + 480 N/C. What is the displacement of the particle after a time of
* 40 In Multiple-Concept Example 9 you can see the concepts that are important in this problem. A particle of
charge
and mass
is released from rest in a region where there is a constant electric field
at the (empty) upper right corner is zero. Find the magnitudes of q1 and q2. Express your an swers in terms of q.
* 39 ssm www A rectangle has a length of 2d and a height of d. Each of the following three charges is located at a corner
of the rectangle: + q1 (upper left corner), + q2 (lower right corner), and −q (lower left corner). The net electric field
. Find the angle
in the drawing.
P
What is Q?
49 A vertical wall
in a house faces due east. A uniform electric field has a magnitude of 150 N/C.
This field is parallel to the ground and points 35° north of east. What is the electric flux through the wall?
50 A charge Q is located inside a rectangular box. The electric flux through each of the six surfaces of the box is:
48 A rectangular surface
is oriented in a uniform electric field of 580 N/C. What is the
maximum possible electric flux through the surface?
Section 18.9 Gauss’ Law
46 A spherical surface completely surrounds a collection of charges. Find the electric flux through the surface if
the collection consists of (a) a single
charge, (b) a single
charge, and (c) both of
the charges in (a) and (b).
47 ssm The drawing shows an edge-on view of two planar surfaces that intersect and are mutually perpendicular.
Surface 1 has an area of 1.7 m2, while surface 2 has an area of 3.2 m2. The electric field
in the drawing is
uniform and has a magnitude of 250 N/C. Find the electric flux through (a) surface 1 and (b) surface 2.
is
M
** 44 Two particles are in a uniform electric field whose value is + 2500 N/C. The mass and charge of particle 1
and
, while the corresponding values for particle 2 are
are
and
. Initially the particles are at rest. The particles are both located on the
same electric field line, but are separated from each other by a distance d. When released, they accelerate, but
always remain at this same distance from each other. Find d.
** 45 ssm Two point charges of the same magnitude but opposite signs are fixed to either end of the base of an
isosceles triangle, as the drawing shows. The electric field at the midpoint M between the charges has a magnitude
E . The field directly above the midpoint at point P has a magnitude E . The ratio of these two field magnitudes
.
. (Hint: For a Gaussian surface, use a sphere of radius r
sphere at a radius r has a magnitude of
Copyright © 2007 John Wiley & Sons, Inc. All rights reserved.
centered within the solid sphere. Note that the net charge within any volume is the charge density times the volume.)
* 52 Refer to Concept Simulation 18.3 at for a perspective that is useful in solving this problem. Two spherical shells
have a common center. A
charge is spread uniformly over the inner shell, which has a radius
of 0.050 m. A
charge is spread uniformly over the outer shell, which has a radius of 0.15 m. Find
the magnitude and direction of the electric field at a distance (measured from the common center) of (a) 0.20 m,
(b) 0.10 m, and (c) 0.025 m.
* 53 ssm A cube is located with one corner at the origin of an x, y, z coordinate system. One of the cube’s faces lies in the
x, y plane, another in the y, z plane, and another in the x, z plane. In other words, the cube is in the first octant of
the coordinate system. The edges of the cube are 0.20 m long. A uniform electric field is parallel to the x, y plane
and points in the direction of the + y axis. The magnitude of the field is 1500 N/C. (a) Find the electric flux
through each of the six faces of the cube. (b) Add the six values obtained in part (a) to show that the electric
flux through the cubical surface is zero, as Gauss’ law predicts, since there is no net charge within the cube.
** 54 A long, thin, straight wire of length L has a positive charge Q distributed uniformly along it. Use Gauss’ law to
show that the electric field created by this wire at a radial distance r has a magnitude of
,
where
. (Hint: For a Gaussian surface, use a cylinder aligned with its axis along the wire and note that
the cylinder has a flat surface at either end, as well as a curved surface.)
. Use Gauss’ law to show that the electric field at a point within the
charge per unit volume, therefore, is
* 51 Interactive Solution 18.51 at offers help with this problem in an interactive environment. A solid
nonconducting sphere has a positive charge q spread uniformly throughout its volume. The charge density or
Additional Problems
55 An electric field of 260 000 N/C points due west at a certain spot. What are the
at this spot?
magnitude and direction of the force that acts on a charge of
56 The force of repulsion that two like charges exert on each other is 3.5 N. What will the
force be if the distance between the charges is increased to five times its original value?
57 Conceptual Example 14 deals with the hollow spherical conductor in Figure 18-31. The
conductor is initially electrically neutral, and then a charge + q is placed at the center of
the hollow space. Suppose the conductor initially has a net charge of + 2q instead of
being neutral. What is the total charge on the interior and on the exterior surface when
the + q charge is placed at the center?
58 At a distance r1 from a point charge, the magnitude of the electric field created by the
charge is 248 N/C. At a distance r2 from the charge, the field has a magnitude of 132 N/
C. Find the ratio r2/r1.
59 ssm Background pertinent to this problem is available in Interactive LearningWare
18.3 at . A
point charge is placed in an external uniform electric field of
. At what distance from the charge is the net electric field zero?
60 In a vacuum, two particles have charges of q1 and q2, where
. They are
separated by a distance of 0.26 m, and particle 1 experiences an attractive force of 3.4 N.
What is q2 (magnitude and sign)?
61 ssm Interactive LearningWare 18.1 at offers some perspective on this problem. Two
tiny spheres have the same mass and carry charges of the same magnitude. The mass of
each sphere is
. The gravitational force that each sphere exerts on the other
is balanced by the electric force. (a) What algebraic signs can the charges have? (b)
Determine the charge magnitude.
is located in an electric field. The x and y components of the
62 A charge of
and
, respectively. (a)
electric field are
What is the magnitude of the force on the charge? (b) Determine the angle that the force
makes with the + x axis.
at x1 = + 4.0 cm, and
* 63 Two charges are located along the x axis:
at x2 = − 4.0 cm. Two other charges are located on the y axis:
at y3 = + 5.0 cm, and
electric field (magnitude and direction) at the origin.
at y4 = + 7.0 cm. Find the net
* 64 Four point charges have equal magnitudes. Three are positive, and one is negative, as the
drawing shows. They are fixed in place on the same straight line, and adjacent charges
are equally separated by a distance d. Consider the net electrostatic force acting on each
charge. Calculate the ratio of the largest to the smallest net force.
* 65 A proton is moving parallel to a uniform electric field. The electric field accelerates the
proton and thereby increases its linear momentum to
from
in a time of
. What is the magnitude of the electric
field?
* 66 Multiple-Concept Example 9 illustrates the concepts employed in this problem. A small
object, which has a charge
and mass
, is placed in a
constant electric field. Starting from rest, the object accelerates to a speed of
in a time of 0.96 s. Determine the magnitude of the electric field.
* 67 ssm In the rectangle in the drawing, a charge is to be placed at the empty corner to make
the net force on the charge at corner A point along the vertical direction. What charge
(magnitude and algebraic sign) must be placed at the empty corner?
** 68 Two objects are identical and small enough that their sizes can be ignored relative to the
distance between them, which is 0.200 m. In a vacuum, each object carries a different
charge, and they attract each other with a force of 1.20 N. The objects are brought into
contact, so the net charge is shared equally, and then they are returned to their initial
positions. Now it is found that the objects repel one another with a force whose
magnitude is equal to that of the initial attractive force. What is the initial charge on each
object? Note that there are two answers.
Copyright © 2007 John Wiley & Sons, Inc. All rights reserved.
Concepts & Calculations
Note: Each of these problems consists of Concept Questions followed by a related quantitative
Problem. The Concept Questions involve little or no mathematics. They focus on the concepts
with which the problems deal. Recognizing the concepts is the essential initial step in any
problem-solving technique.
69
Concept Questions Two identical metal spheres have charges of q1 and q2. They
are brought together so they touch, and then they are separated. (a) How is the net charge
on the two spheres before they touch related to the net charge after they touch? (b) After
they touch and are separated, is the charge on each sphere the same? Why?
Problem Four identical metal spheres have charges of
,
,
, and
. (a) Two of the spheres are brought together so
they touch and then they are separated. Which spheres are they, if the final charge on
each of the two is
? (b) In a similar manner, which three spheres are brought
together and then separated, if the final charge on each of the three is
(c) How
many electrons would have to be added to one of the spheres in part (b) to make it
electrically neutral?
70
Concept Questions The drawings show three charges that have the same
magnitude, but different signs. In all cases the distance d between charges 1 and 2 and
between 2 and 3 is the same. (a) Draw the electrical force that each charge exerts on
charge 2. Each force should be drawn in the correct direction, and its magnitude should
be correct relative to that of the other force. (b) Rank the magnitudes of the net electrical
force on charge 2, largest first. Explain.
Problem The magnitude of the charges |q| = 8.6 μC, and the distance between them is
3.8 mm. Determine the magnitude of the net force on charge 2 for each of the three
drawings. Verify that your answers are consistent with your answers to the Concept
Questions.
71
Concept Questions Suppose you want to neutralize the V gravitational attraction
between the earth and the moon by placing equal amounts of charge on each. (a) Should
the charges be both positive, both negative, or one positive and the other negative? Why?
(b) Do you need to know the distance between the earth and the moon to find the
magnitude of the charge? Why or why not?
and
,
Problem The masses of the earth and moon are
respectively. Identical amounts of charge are placed on each body, such that the net force
(gravitational plus electrical) on each is zero. What is the magnitude of the charge?
72
Concept Questions Suppose you want to determine the electric field in a certain
region of space. You have a small object of known charge and an instrument that
measures the magnitude and direction of the force exerted on the object by the electric
field. How would you determine the magnitude and direction of the electric field if the
object were (a) positively charged and (b) negatively charged?
Problem (a) The object has a charge of
and the instrument indicates that the
electric force exerted on it is
, due east. What is the magnitude and direction of
the electric field? (b) What is the magnitude and direction of the electric field if the
object has a charge of
and the instrument indicates that the force is
due west?
73
Concept Question The drawing shows two situations in which charges are placed
on the x and y axes. They are all located at the same distance from the origin O. Without
doing any calculations, does the net electric field at the origin in part (a) have a
magnitude that is greater than, less than, or equal to the magnitude of the field at the
origin in part (b)? Justify your answer.
Problem The distance between each of the charges and the origin is 6.1 cm. For each of
the situations shown in the drawing, determine the magnitude of the net electric field at
the origin O. Check to see that your results are consistent with your answer to the
Concept Question.
74
Concept Questions A proton and an electron are moving due east in a constant
electric field that also points due east. (a) Does each experience an electric force of the
same magnitude and direction? (b) What is the direction of the proton’s acceleration, and
what is the direction of the electron’s acceleration? (c) Is the magnitude of the proton’s
acceleration greater than, less than, or the same as that of the electron’s acceleration?
Explain your answers.
Problem The electric field points due east and has a magnitude of
.
Determine the magnitude of the acceleration of the proton and the electron. Check that
your answers are consistent with part (c) of the Concept Questions.
75
Concept Questions Three point charges have equal magnitudes, two being positive
and one negative. These charges are fixed to the corners of an equilateral triangle, as the
drawing shows. (a) The charge at any one corner experiences forces from the charges at
the other corners. Do the individual forces exerted by the charges have the same or
different magnitudes? (b) At which one or more corners does (do) the charge(s)
experience a net force that has the greatest magnitude? (c) At which one or more corners
does (do) the charge(s) experience a net force that has the smallest magnitude?
Problem The magnitude of each of the charges is
, and the lengths of the sides of
the triangle are 3.0 cm. Calculate the magnitude of the net force that each charge
experiences. Be sure that your answers are consistent with your answers to the Concept
Questions.
76
Concept Questions The drawing shows a positive point charge + q1, a second
point charge q2 that may be positive or negative, and a spot labeled P, all on the same
straight line. The distance d between the two charges is the same as the distance between
q1 and the point P. With q2 present, the magnitude of the net electric field at P is twice
what it is when q1 is present alone. (a) When the second charge is positive, is its
magnitude smaller than, equal to, or greater than the magnitude of q1? Explain your
reasoning. (b) When the second charge is negative, is its magnitude smaller than, equal
to, or greater than that in question (a)? Account for your answer.
Problem Given that
, determine q2 when it is (a) positive and (b)
negative. Verify that your answers are consistent with your answers to the Concept
Questions.
Copyright © 2007 John Wiley & Sons, Inc. All rights reserved.
CHAPTER 18 ELECTRIC FORCES AND
ELECTRIC FIELDS
CONCEPTUAL QUESTIONS
____________________________________________________________________________________________
8.
2
REASONING AND SOLUTION
Identical point 1
charges are fixed to opposite corners of a square, as
shown in the figure at the right.
There exists an electric field at each point in space in
the vicinity of this configuration. The electric field at a
given point is the resultant of the electric field at that
C
point due to the charge at corner 1 and the charge at
corner 3.
3
The magnitude of the electric field due to a single 4
point charge q is given by Equation 18.3: E k q / r 2 ,
where r is the distance from the charge to the point in question.
Since the point charges at corners 1 and 3 are identical, they each have the same value of
q. Furthermore, they are equidistant from the center point C. Therefore, at the center of the
square, each charge gives rise to an electric field that is equal in magnitude and opposite in
direction. The resultant electric field at the center of the square is, therefore, zero. Note that
this result is independent of the polarity of the charges.
The distance from either empty corner, 2 or 4, to either of the charges is the same. Since
the charges are equidistant from either empty corner, each charge gives rise to an electric
field that is equal in magnitude. Since the direction of the electric field due to a point
charge is radial (radially inward for negative charges and radially outward for positive
charges), we see that the electric fields due to each of the two point charges will be mutually
perpendicular. Their resultant can be found by using the Pythagorean theorem.
The magnitude of the force experienced by a third point charge placed in this system is,
from Equation 18.2: F q0 E , where q0 is the magnitude of the charge and E is the
magnitude of the electric field at the location of the charge. Since the electric field is zero at
the center of the square, a third point charge will experience no force there. Thus, a third
point charge will experience the greater force at one of the empty corners of the square.
____________________________________________________________________________________________
14. REASONING AND SOLUTION In general, electric field lines are always directed away
from positive charges and toward negative charges. At a point in space, the direction of the
electric field is tangent to the electric field line that passes through that point. The
magnitude of the electric field is strongest in regions where the field lines are closest
together. Furthermore, the text states that electric field lines created by a positive point
charge are directed radially outward, while those created by a negative point charge are
directed radially inward.
a. In both drawings I and II the electric field is the same everywhere. This statement is
false. By inspection, we see that the field lines in II get closer together as we proceed from