Download THE ORIGIN OF ELECTRICITY

Lightning,which is the flow of electric charge in the atmosphere, is nature'smost spectacular display of electricity.
Here,multiple lightning strikes occur at night over an urban skyline. (RalphWetmore/Stone/Getty Images)
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 protonhas a mass of 1.673 X 10-27 kg, and a neutron has a slightly greater mass of
1675X 10-27kg. Surrounding the nucleus is a diffuse cloud of orbiting particles called
electrons,as Figure 18.1 suggests. An electron has a mass of 9.11 X 10-31 kg. Like mass,
electriccharge is an intrinsic property of protons and electrons, and only two types of
chargehave been discovered, positive and negative. A proton has a positive charge, and
anelectron has a
negative charge. A neutron has no net electric charge.
electron
proton
neutron
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.
537
538 | CHAPTER 18
ELECTRIC FORCES AND ELECTRIC
FIELDS
equals the
the charge on the proton exactly
Experiment reveals that the magnitude of
electron
the
and
the proton carries a charge +e,
magnitude of the charge on the electron;
is the
charge
the magnitude of an electric
carries a charge —e.The SI unit for measuring
experimentally to have the value
coulomb* (C), and e has been determined
e = 1.60 X 10-19C
electron and
of the charge on a proton or an
The symbol e represents only the magnitude
whether the charge is positive or negadoes not include the algebraic sign that indicates
equal numbers of protons and electrons.
tive. In nature, atoms are normally found withbecause the algebraic sum of the positive
Usually, then, an atom carries no net charge the electrons is zero. When an atom, or
charge of
charge of the nucleus and the negative
is said to be electrically neutral. The neutrons
object
any object, carries no net charge, the
particles.
in the nucleus are electrically neutral
an electron or on a proton are built up on
Charges of larger magnitude than that on
Thus, any charge of magnitude q is an
an object by adding or removing electrons.
is an integer. Because any electric charge q
integer multiple of e; that is, q = Ne, where N
indivisible charges of magnitude e, electric
occurs in integer multiples of elementary,
emphasizes the quantized nature of electric
charge is said to be quantized. Example I
charge.
Example 1 | A Lot of Electrons
How many electrons are there in one coulomb of negative charge?
electrons, since they carry
Reasoning The negativecharge is due to the presence of excess
is e = 1.60 x 10-19c,
magnitude
negative charge. Because an electron has a charge whose
on each electron.
e
charge
the
by
divided
the number of electrons is equal to the charge q
Solution The number N of electrons is
q
1.00 c
1.60 x 10-19c
Animal fur
Ebonite rod
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.
6.25 x 1018
CHARGED OBJECTS AND THE ELECTRIC FORCE
Electricity has many useful applications, and they are related to the fact that it is posSibleto 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 transfe1Tedto
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.
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.
*The definition of the coulomb depends on electric currents and magnetic fields, concepts that will be discussed
later. Therefore, we postpone its definition until Section 21.7.
18.2 | CHARGED OBJECTS AND THE ELECTRIC FORCE | 539
Electriccharges play a role in many situations other than rubbing two surfaces together.They are involved, for instance, in chemical reactions, electric circuits, and radecay.A great number of experimentshave verified that in any situation, the
dioactive
of electric charge is obeyed.
lawofconservation
LAWOF CONSERVATION OF ELECTRIC CHARGE
Duringany process, the net electric charge of an isolated system remains constant
(isconserved).
(a)
It is easy to demonstrate that two electrically charged objects exert a force on one another.Consider Figure 18.3a, which shows two small balls that have been oppositely
chargedand are light and free to move. The balls attract each other. On the other hand,
ballswith the same type of charge, either both positive or both negative, repel each other,
aspartsb and c of the drawing indicate. The behavior depicted in Figure 18.3 illustrates
thefollowingfundamental characteristic of electric charges:
Like charges repel and unlike charges attract each other.
ATA GLANCE
CONCEPTS
Like other forces that we have encountered, the electric
force(alsocalled the electrostatic force) can alter the motion of an object. It can do so by
contributingto the net external force EF that acts on the object. Newton's second law,
EF = ma, specifies the acceleration ä that arises because of the net external force. The
Concepts-at-a-Glancechart in Figure 18.4 is an expanded version of the charts in Figures
4.9,10.4,and 11.5, and emphasizes that any external electric force that acts on an object
mustbe included when determining the net external force to be used in the second law. <
Thephysicsof electronic ink. A new technology based on the electric force may revolutionizethe way books and other printed matter are made. This technology, called electronicink, allows letters and graphics on a page to be changed instantly, much like the
symbolsdisplayed on a computer monitor. Figure 18.5a illustrates the essential features
ofelectronicink. 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
dozenextremely tiny white beads that carry a slightly negative charge. The microcapsules
aresandwichedbetween two sheets, an opaque base layer and a transparent top layer, at
(b)
(c)
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.
CONCEPTSAT A GLANCE
External Forces
1. Gravitational Force
(Section4.7)
2. Normal Force
(Section 4.8)
3. Frictional Forces
(Section 4.9)
4. Tension Force
(Section 4.10)
5. Restoring Force of
a Spring
(Section 10.l)
6. CollisionalForces
(Section 11.2)
7. Electrostatic Force
Figure18.4 CONCEPTS AT A GLANCE External electrostatic forces that act on an object must
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
DNAfragmentsto different terminal speeds in a gel. The different fragments are represented by the
glowing
bands. (Eurelios/Phototake)
Newton's Second Law
540 | CHAPTER 18
ELECTRIC FORCES AND ELECTRIC FIELDS
See your sales
person for the
right fit
Base layer
Beads
Microcapsules
(a)
Dark liquid
Base layer
Dark pixel
Light pixel
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.
(b)
which the reader looks. When a positive charge is applied to a small region of thebase
layer, as shown in part b of the drawing, the negatively charged white beads are drawnto
it, leaving dark ink at the top layer. Thus, a viewer sees only the dark liquid. When a neg-
ative charge is applied to a region of the base layer, the negatively charged whitebeads
are repelled from it and are forced to the top of the microcapsules; now a viewerseesa
white area due to the beads. Thus, electronic ink is based on the principle thatlike
charges repel and unlike charges attract each other; a positive charge causes one colorto
appear, and a negative charge causes another color to appear. Each
small region, whether
dark or light, is known as a pixel (short for "picture element").
Computerchipsprovide
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•
CONDUCTORS AND INSULATORS
Electric charge can not only exist on an object, but it
can also move throughan
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 readilyand'
therefore, are known as thermal conductors. On the other
hand, substances that conduct
heat poorly are referred to as thermal insulators.
1 541
18.4 | CHARGING BY CONTACT AND BY INDUCTION
Hotter
object
Cooler
object
Heat
Negatively
charged
object
(a)
+ Positively
charged
object
(b)
to the conduction of heat arises when a metal bar is placed beA situationanalogous
objects, as in Figure 18.6b. Electrons are conducted through the bar from
twocharged
tween
charged object toward the positively charged object. Substances that readily
thenegatively
electrical conductors. Although there are exceptions, good
electriccharge are called
conduct
good electrical conductors. Metals such as copper, alugenerally
conductors are
thermal
electrical conductors and, therefore, are used in electrisilver,and gold are excellent
minum,
charge poorly are known as electrical insulators.
electric
conduct
that
Materials
calWiling.
electrical
also
insulators. Common electrical insulators
are
insulators
Inmanycases,thermal
such
Insulators,
as the rubber or plastic that coats elecwood.
and
arerubber,manyplastics,
from
going
where
it is not wanted.
wiring,prevent electric charge
trical
atomic strucThedifferencebetween electrical conductors and insulators is related to
force of atthe
outer
weaker
in
orbits
a
those
experience
nucleus,
the
orbit
Aselectrons
ture.
elecoutermost
the
inner
orbits.
Consequently,
in
the
do
those
than
to the nucleus
traction
In
ones.
the
inner
be
dislodged
more
easily
than
can
electrons)
(alsocalled the valence
Irons
wander
and
atom
become
detached
from
a
parent
agoodconductor,some valence electrons
or less freely throughout the material, belonging to no one atom in particular. The exmore
actnumberof electrons detached from each atom depends on the nature of the material, but
isusuallybetweenone and three. When one end of a conducting bar is placed in contact
witha negativelycharged object and the other end in contact with a positively charged object,asin Figure 18.6b, the "free" electrons are able to move readily away from the negative
endandtowardthe positive end. The ready movement of electrons is the hallmark of a good
In an insulator the situation is different, for there are very few electrons free to
conductor.
move
throughoutthe material. Virtually every electron remains bound to its parent atom.
Witlmtthe "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.
from
Figure 18.6 (a) Heat is conducted
the
to
bar
the hotter end of the metal
conducted
cooler end. (b) Electrons are
of the
end
from the negatively charged
end.
charged
metal bar to the positively
Ebonite rod
CHARGING BY CONTACT AND BY INDUCTION
Whena 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.
Oncethe electrons are on the metal sphere, where they can move readily, they repeloneanother and spread out over the sphere's surface. The insulated stand prevents
themfromflowing 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
overits surface.In a similar manner, the sphere would be left with a positive charge after
being
rubbedwith a positively charged rod. In this case, electrons from the sphere would
betransferredto the rod. The process of giving one object a net electric charge by placing
itincontactwith
another object that is already charged is known as charging by contact.
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
Metal sphere
Insulated stand
(a)
(b)
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 distributethemselves over
the surface of the sphere.
Ebonite rod
Metal sphere
Grounding
wire
Insulated stand
-z Connection
to ground
(a)
(b)
(c)
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.
542 | CHAPTER 18 I ELECTRIC FORCES AND ELECTRIC FIELDS
Ebomte
rod
Positive surface
charge
Plastic
Figure 18.9 The negatively charged rod
induces a slight positive surface charge
on the plastic.
sphere. In the sphere, the free electrons closest to the rod move to the other side, aspana
of the drawing indicates. As a result, the part of the sphere nearest the rod becomesposi.
tively 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.Ifthe
rod were removed, the free electrons would return to their original places, andthe
charged regions would disappear.
Under most conditions the earth is a good electrical conductor. So when a metalwire
is attached between the sphere and the ground, as in Figure 18.8b, some of the freeelec.
trons leave the sphere and distribute themselves over the much larger earth. If the ground.
ing 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 electm
charge without touching the object to a second charged object is called charging byin.
duction. The process could also be used to give the sphere a negative net charge,if a posi.
tively 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,in.
stead of metal, the method of producing a net charge by induction would not work,be.
cause very little charge would flow through the insulating material and down the ground.
ing wire. However, the electric force of the charged rod would have some effect,
Figure 18.9 illustrates. The electric force would cause the positive and negative chargesin
the molecules of the insulating material to separate slightly, with the negative chargesbe.
ing "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.
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,andthe
rod is then removed. The following statements refer to the masses rnAand rnBof the
spheres after they are separated and the rod is removed. Which one or more of thestatements is true? (a)
rnB,(b) 'DA>
if the rod is positive, (c) rnA< rnBif the rod is posi•
tive, (d) rnA>
if the rod is negative, (e) tm <
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.Ina
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.
COULOMB'S LAW
THE FORCE THAT POINT CHARGES EXERT ON EACH OTHER
The electrostatic force that stationary charged objects exert on each other depenb
on the amount of charge on the objects and the distance between them.
'
reveal that the greater the charge and the closer together they are, the greater is theforce
18.5 I COULOMB'S LAW | 543
for explaining these features in more detail, Figure
18.10 shows two
Toset the stage
objects are so small, compared to the
distance r between them, that
bodies. These
If the charges have unlike signs, as in part a of the picture, each object is atat
the other by a force th is directed along the line between them; +F is the
to
tracted
on object I by
and
object 2 and —F is the electric force
exerted
exerted on obelectricforce
in
as
part
b,
the
charges have the same sign (both positive
If,
l.
object
or both
ject2 by
each object is repelled from the other. The repulsive forces, like the attractive
negative),
the line between the charges. Whether attractive or repulsive, the two
forces,act along magnitude but
opposite in direction. These forces always exist as a
in
forcesare equal
on a different object, in accord with Newton's action—reactionlaw.
pair,eachone acting
Charles Augustin de Coulomb (1736—1806)carried out a numphysicist
TheFrench
determine
how the electric force that one point charge applies to anto
berofexperiments
amount of each charge and the separation between them. His result,
otherdependson the
law, is stated below.
Coulomb's
nowknownas
(a)
(b)
Figure18.10 Each point charge exerts a
force on the other Regardlessof
whether the forces are (a) attractive or
(b) repulsive, they are directed along
the line between the charges and have
equal magnitudes.
COULOMB'SLAW
Themagnitude F of the electrostatic force exerted by one point charge qi on another
pointcharge is directly proportional to the magnitudes I and Iq21of the charges
andinverselyproportional to the square of the distance r between them:
Iqlllq21
(18.1)
2
wherek is a proportionality constant: k = 8.99 X 109N • m2/C2in SI units. The electrostaticforce is directed along the line joining the charges, and it is attractive if the
chargeshave unlike signs and repulsive if the charges have like signs.
It is common practice to express k in terms of another constant €0, by writing k —
€0is called the permittivity of free space and has a value of €0= 1/(47k) =
8.85X 10-12 C2/(N • m2). Equation 18.1 gives the magnitude of the electrostatic force
thateachpoint charge exerts on the other. When using this equation, then, it is important
torememberto substitute only the charge magnitudes (without algebraic signs) for Iql I
andIq21,
as Example 2 illustrates.
V
Needmorepractice?
Example 2 | A Large Attractive Force
Interactive LearningWare 18.1 An
Twoobjects,whose charges are +1.0 and —1.0C, are separated by 1.0 km. Compared to
1.0km,the sizes of the objects are small. Find the magnitude of the attractive force that either
chargeexertson the other.
ReasoningConsidering 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 magnitudesof the charges are used
object of mass 215 kg is located
just above the surface of the earth
(radius = 6.38 x 106 m). Suppose
that the object and the earth have
a. identical charge q. Therefore,
two forces—the gravitational
force and the electrostatic force—
SolutionThe magnitude of
the force is
act on the object. Assuming that
the earth's charge is located at the
center of the earth, determine q
forthesymbols Iqi
I and Iq21that appear in the law.
Iql llq21
r2
(8.99 x 109
(1.0 >< 10 3
C)
(18.1)
Theforce calculated in Example 2 corresponds to about 2000 pounds and is so large
bec
ausecharges of ± 1.0
only in the
C are enormous. Such large charges are encountered
mostsevere
transferred
C
be
25
can
as
conditions, as in a lightning bolt, where as much
e magnitudeof a
by a vertical bar to the
variable is sometimes called the absolute value and is symbolized
leftandto the
of
the variable q, which is
value
thevalue right of the variable. Thus, Iq I denotes the magnitude or absolute
of q without
•
then
Iql = 2.0 c.
c,
=
-2.0
q
If
its algebraic plus or minus sign. For example,
(magnitude and sign) when the
object is in equilibrium.
Related Homework: Problem 61
For an interactive solution, go to
www.wiley.com/college/cutnell
544 | CHAPTER 18 I ELECTRIC FORCES AND ELECTRIC FIELDS
between the cloud and the ground. The typical charges produced in the
much smaller and are measured convenientlyin microcoulombs(l microcoulombs
1
= 10-6 C).
Coulomb's law has a form that is remarkably similar to Newton's lawof gravitas
tion (F = GmlR12/r2).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 betweenthem. In
addition, the force is proportional to the product of an intrinsic propertyof each
in Coulomb's law and the
objects, the magnitudes of the charges 1and
difference between the twolaws
major
a
is
there
But
ml and
in the gravitation law.
attractive,
depending on
The electrostatic force can be either repulsive or
not the charges have the same sign; in contrast, the gravitational force is alwaysanat.
tractive force.
Section 5.5 discusses how the gravitational attraction between the earth anda
lite provides the centripetal force that keeps a satellite in orbit. Example 3 illustrates
that
famous
in
a
role
similar
model
a
plays
of
the atomcr.
the electrostatic force of attraction
ated by the Danish physicist Niels Bohr (1885—1962).
ANALYZING MULTIPLE-CONCEPT PROBLEMS
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 5.29 X 10-11m, as Figure 18.11
shows. The mass of the electron is 9.11 ><10-31kg. Determine the speed of the electron.
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
Fc = mv2/r, where m and v are, respectively, the mass and speed of the electron and r is
Figure 18.11 In the Bohr modelof the
the radius of the orbit. This equation can be solved for the speed of the electron. Since
hydrogen atom, the electron (—e)orbits
the mass and orbital radius are known, we can calculate the electron's speed provided
the proton (+e) at a distanceof
that a value for the centripetal force can be found. For the electron in the hydrogen
r = 5.29 X 10-11 m. The velocityofthe
atom, the centripetal force is provided almost exclusively by the electrostatic force that
electron is v.
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 gravitationalforce•
However, the gravitational force is negligible in comparison to the electrostatic force.
Knowns and Unknowns
Description
Electron charge
Electron mass
Proton charge
Radius of orbit
Unknown Variable
Orbital speed of electron
The data for this problem are:
Symbol
Value
-1.60 x 10-19c
m
9.11 X 10-31 kg
+1.60 x 10-19c
5.29 x 10-11 m
9
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 FC = mv2/r (Equation 5.3). By solving this equation for the speed, we obtain Equation 1 at the right. The mass and radius in this expres-
sion are known. However, the magnitude of the centripetal force is not known, so we will
evaluate it in Step 2.
Coulomb's Law As the electron orbits the proton in the hydrogen atom, it is
proton by the electrostatic force. The magnitude F of the electrostatic
attractedto the
Coulomb's law as F = klqlllq21/r2
(Equation18.1),where and Iq21
forceis givenby
the
charges,
r
is
the
of
orbital
radius,
and k = 8.99 X 109N • m2/C2.
arethemagnitudes force is provided almost
entirely
by
centripetal
the electrostaticforce, it follows
Sincethe
=
I—el
and
Iq21
=
l+el.
Furthermore,
With
these substitutions,Equation
F.
thatFC=
18.1becomes
Fc=k
—ell + el
(1)
Fc=k
2
Allthevariables on the right side of this expression are known, so we substitute it into
1, as indicated in the right column.
Equation
solution Algebraically combining the results of the modeling steps, we have
STEP 1 STEP 2
I—ell + el
r2
m
m
mr
Thespeedof the orbiting electron is
kl —ell + el
mr
(8.99 x 109 N. m2/c 2)l -1.60 x 10-19Cll +1.60 x 10-19cl
(9.11 X
Related Homework:
10 -31
X 10 -11 m)
- 2.19x
Problems 16, 20
Sincethe electrostatic force depends on the inverse square of the distance between The physics of
thecharges,it becomes larger for smaller distances, such as those involved when a strip adhesion.
Ofadhesivetape is stuck to a smooth surface. Electrons shift over the small distances betweenthe tape and the surface. As a result, the materials become oppositely charged.
Sincethe distance between the charges is relatively small, the electrostatic force of attractionis large enough to contribute to the adhesive bond. Figure 18.12 shows an image of
thestickysurface of a piece of tape after it has been pulled off a metal surface. The image
wasObtainedusing an atomic-force microscope and reveals the tiny pits left behind when
microscopicportions of the adhesive remain stuck to the metal because of the strong adhesivebonding
forces.
gm
1.5
THE FORCE ON A POINT CHARGE
DUETo TWO OR MORE OTHER POINT CHARGES
Up to now, we have been discussing the electrostatic force on a point charge
(magnitudeIqi l) due
Suppose that a third point
to another point charge (magnitude11/21).
charge(magnitude
Iq31) is also present. What would be the net force on due to both
andq3?It is
convenient to deal with such a problem in parts. First, find the magnitude and
directionOfthe force
exerted on by (ignoring (13).Then, determine the force exerted
onqi by (ignoring
The net force on 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.
1.0
0.5
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. (CourtesyLouis
Scudiero and J. Thomas Dickinson,
Washington State University.)
546 | CHAPTER 18 I ELECTRIC FORCES AND ELECTRIC FIELDS
nA,
0.20 m
-4.0 PC
0.15m
+3.0
q3
-7.0 pc
(a)
(b) Free-body diagram for
Figure 18.13 (a) Three charges lying
along the x axis. (b) The force exerted
on qi by is F12,while the force
exertedon
by
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 Cll.
Reasoning Part b of the drawing shows a free-body diagram of the forces that act onq.
Since and have opposite signs, they attract one another. Thus, the force exertedonq,
by is F12, and it points to the left. Similarly, the force exerted on by is F13andisalso
an attractive force. It points to the right in Figure 18.13b.The magnitudes of these forcescan
be obtained from Coulomb's law. The net force is the vector sum of F12and Fn.
Solution The magnitudes of the forces are
is Fn.
llq21
109
r 12
Problem solving insight
When using Coulomb's law (F =
k Iql llq21/r2), rememberthat the symbols
qtl and Iq21stand for the charge
magnitudes. Do not substitute negative
numbers for these symbols.
(8.99 x
IL/lIIq31
x
X 10-6 C)
X 10-6
(8.99 X 109
r 13
10 -6
(0.20 m) 2
X 10-6 C)
(0.15
= 2.7 N
= 8.4N
Since F12points in the negative x direction, and F13points in the positive x direction,thenet
force F is
(-2.7 N) + (8.4 N)
The plus sign in the answer indicates that the net force points to the right in the drawing.
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 Cll.
Need more practice?
Interactive LearningWare 18.2
A positive charge is placed on the x
axis (q = +7.00 gc, x = 0.600m),
and another positive charge is
placed on the y axis (q = +9.00
y = 0.400 m). A third charge (q —
-6.00 gc, m = 5.00 x 10-8 kg) 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
For an interactive solution, go to
www.wiley.com/college/cutnell
Reasoning The force exerted on by is F12and is an attractive force becausethetwo
charges have opposite signs. It points along the line between the charges. The force exertedon
qi by is F13 and is also an attractive force. It points along the line between qi and(13.
Coulomb's law specifies the magnitudes of these forces. Since the forces point in differentdirections (see Figure 18.14b), we will use vector components to find the net force.
Solution The magnitudes of the forces are
Iql Il q21
(8.99 x 109
r 12
F 13= k Iqlllq31
X 10-6
X 10-6 C)
x 10-6
X 10-6 C)
(0.15
(8.99 X 109
r 13
(0.10
F12 sin 730
730
COS 73 0
is F =
+ F13.The angle that F
makes with the +x axis is 0.
= 18N
The net force F is the vector sum of F12and F13,as part b of the drawing shows. The comop
nents of F that lie in the x and y directions are Fr and Fy, respectively. Our approachto findin
g
F is the same as that used in Chapters I and 4. The forces F12and F13are resolved intox andy
components. Then, the x components are combined to give Fx, and the y componentsare
-6.0pc
Figure 18.14 (a) Three charges lying in
a plane. (b) The net force acting on
= 9.6 N
+4.0{IC
= 0.10 m
(a)
-5.0 pc
(b) Free-body diagram for qi
18.6 | THE ELECTRIC FIELD | 547
Fy. Once Fx and Fy are known, the magnitude and
direction of F can be decombinedto give
using trigonometry.
termined
x component
y component
Force
F12
+(9.6 N) cos 730 = +2.8 N
+(9.6 N) sin 730 = +9.2 N
ON
F13
+21 N
= +9.2 N
O
(21
tan -1
tan -l
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.
F and the angle 0 of the net force are
Themagnitude
F = Fx2+ F 2
Problem solving insight
+ (9.2
=
9.2 N
21 N
CHECK YOUR UNDERSTANDING 2
Thedrawing shows three point charges arranged in differentways. The charges are +q,
-q, and —q;each has the same magnitude, one being positive and the other two negative.
Ineach part of the drawing the distance dis the same. Rank the arrangements in descendingorder (largest first) according to the magnitude of the net electrostaticforce that acts on
thepositive charge. (The answer is given at the end of the book.)
d
-q
d
(a)
—q
—q
d
900
d
(b)
—q
(c)
Background:The electrostatic force that one point charge exerts on another is described by
Coulomb'slaw. Like any force, this force is a vector quantity, and the rules of vector
additionapply.
Forsimilar questions (including calculational counterparts), consult Self-Assessment Test
18.1,which is described next.
ELF-ASSESSMENT TEST 18.1
www.wiley.com/college/cutnell
Testyour understanding of the
material in Sections 18.1-18.5:
• The Origin of
Electricity
• Charged Objects and the Electric Force
• Conductors
and Insulators
• Charging by Contact and by Induction
• Coulomb's Law
THE ELECTRIC FIELD
DEFINITION
Ebonite rod
As We know, a charge can experience an electrostatic force due to the presence of
Othercharges.
For instance, the positive charge qo in Figure 18.15 experiences a force F,
Which
is the vector
forces exerted by the charges on the rod and the two
spheres.It is useful sum of the
to think of CIO
as a test charge for determining the extent to which the
sum)undingcharges
to select
generate a force. However, in using a test charge, we must be careful
one with a very
small magnitude, so that it does not alter the locations of the
othercharges.
The next example illustrates how the concept of a test charge is applied.
e
Example 6 | A
Test Charge
positive test
forceF
charge shown in Figure 18.15 is = +3.0 X 10-8 C and experiences a
X 10-8 N in the direction shown in the drawing.(a) Find theforce per coulomb
Figure 18.15 A positive charge qo
experiences an electrostatic force F due
to the surrounding charges on the
ebonite rod and the two spheres.
548 CHAPTER 18
ELECTRIC FORCES AND ELECTRIC FIELDS
Using the result of part (a), predict the forcethatac
that the test charge experiences. (b)
hogt
if it replaced qo.
of +12 ><10-8 C would experience
apply a force F to the test charge
Reasoning The charges in the environment is F/qo. If is replaced by a newc
test charge
force per coulomb experienced by the
force per coulomb times q.
then the force on this new charge is the
charge is
Solution (a) The force per coulomb of
6.0 x 10-8 N = 2.0 N/C
F
3.0 x 10-8 c
that the surrounding charges can exert 2.0
(b) The result from part (a) indicates
a fore
charge of +12 ><10-8 C would experience
force per coulomb of charge. Thus, a
whose magnitude is
as that experienced by the test charge,since
The direction of this force would be the same
have the same positive sign.
in Example 6(a) is one illustrationof
The electric force per coulomb, F/qo, calculated
The idea is calledtheelectric
an idea that is very important in the study of electricity.
field.
electric
the
of
field. Equation 18.2 presents the definition
DEFINITION OF THE ELECTRIC FIELD
bya
The electric field E that exists at a point is the electrostatic force F experienced
itself:
charge
the
by
small test charge* qo placed at that point divided
(182)
The electric field is a vector, and its direction is the same as the directionof theforce
F on a positive test charge.
SI Unit of Electric Field: newton per coulomb (NIC)
Equation 18.2 indicates that the unit for the electric field is that of forcedivided
by
charge, which is a newton/coulomb (NIC) in SI units.
It is the surrounding charges that create an electric field at a givenpoint.Any
itive or negative charge placed at the point interacts with the field and, as a result,expen•
ences a force, as the next example indicates.
Example 7 | An Electric Field Leads to a Force
In Figgre 18.16a the charges on the two metal spheres and the ebonite rod createanelectnc
field E at the spot indicated. This field has a magnitude of 2.0 N/C and is directedasinthe
Of(a)
drawing. Determine the force on a charge placed at that spot, if the charge has a value
= +18 X 10-8 C and (b) = -24 X 10-8 C.
011 the
Reasoning The electric field at a given spot can exert a variety of forces, depending
magnitude and sign of the charge placed there. The charge is assumed to be smallenough
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 magnitudesof qoandE'
F = IqolE = (18 x 10-8
N/C) =
Since qo is positive, the force points in the same direction as the electric field,as part
drawing indicates.
*As long as the test charge is small enough that it does not disturb the surrounding charges,it maybe either
positive or negative. Compared to a positive test charge, a negative test charge Of the same magnitude
experiences a force of the same magnitude that points in the opposite direction. However, the sameelectnc
is given by Equation 18.2, in which F is replaced by —F and qo is replaced by —CIO.
18.6 | THE ELECTRIC FIELD | 549
(a)
Figure 18.16 The electric field E 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 E, while (b) the force on a
negative charge points opposite to E.
(b)
(b)Inthiscase, the magnitude of the force is
F = IqolE
(24 X 10-8
N/C)
(18.2)
Theforceon the negative charge points in the direction opposite to the force on the positive
charge—thatis, opposite to the electric field (see part b of the drawing).
At a particular point in space, each of the surroundingcharges contributes to the net
electricfield that exists there. To determine the net field, it is necessary to obtain the variouscontributionsseparately and then find the vector sum of them all. Such an approach is
anillustrationof the principle of linear superposition, as applied to electric fields. (This
principleis introduced in Section 17.1, in connection with waves.) Example 8 emphasizes
thevectornature of the electric field, and Example 9 illustrates that a charged particle
accelerates
in an electric field.
V
Example 8 | Electric Fields Add as Vectors Do
Figure18.17shows two charged objects, A and B. Each contributes as follows to the net electricfieldat point P: EA = 3.00 N/C directed to the right, and EB = 2.00 N/C directed downward.Thus, EA and EB are perpendicular. What is the net electric field at P?
ReasoningThe net electric field E is the vector sum of EA and EB•.E = EA + EB. As
illustrated
in Figure 18.17, EA and EB are perpendicular, so E is the diagonal of the rectangle
shownin the drawing. Thus, we can use the Pythagorean theorem to find the magnitude of E
andtrigonometryto find the directional angle 0.
Solution The magnitude of the net electric field is
= (3.00 N/C)2 + (2.00 N/C)2 =
Thedirectionof E
is given by the angle 0 in the drawing:
0 = tan -1
tan -1
2.00 N/C
3.00 N/C
Figure 18.17 The electric field
contributions EA and EB, which come
from the two charge distributions. are
added vectorially to obtain the net field
E at point P.
ANALYZING MULTIPLE-CONCEPT PROBLEMS
Example9 | A Proton Accelerating in an Electric Field
In a
a
vacuum, a proton
(charge = +e, mass = 1.67 X 10-27kg) is moving parallel to a
uniformelectric field
that is directed along the +x axis (see Figure 18.18). The proton
startsWitha velocity
Of +2.5 X 104m/s and accelerates in the same direction as the
electricfield, which has a value of +2.3 X 103NIC. Find the velocity of the proton
Whenits
displacement is +2.0 mm from the starting point.
+2.00 mm
Figure 18.18 A proton, moving to the right,
accelerates in the presence of the electric
field Ex.
550 | CHAPTER 18 I ELECTRIC FORCES AND ELECTRIC FIELDS
Reasoning Since we know the initial velocity and displacement of the proton, we can determine its final velocityfrom
tion of kinematics, provided the proton's acceleration can be found. The acceleration is given by Newton's secondlawasan
the
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
Proton charge
Proton mass
Initial velocity of proton
Electric field
Displacement of proton
Symbol
Value
Comment
+1.60 x 10-19c
m
vox
1.67 X 10-27 kg
+2.5 X 104 m/s
x
+2.3 X 103 N/C
+2.0 mm
2.0 mm = 2.0 X 10-3 m
Unknown Variable
Final velocity of proton
Modeling the Problem
STEP 1 Kinematics To obtain the final velocity vx of the proton we employ Equation
3.6a from the equations of kinematics: vx2 = v0x2 + 2axx. We have chosen this equation
because two of the variables, the initial velocity vot and the displacement x, are known.
Taking the square root of each side of 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 ax is not known, we will obtain an expression for
it in Step 2.
STEP 2 Newton's Second Law Newton's second law, as given in Equation 4.2a,
states that the acceleration ax of the proton is equal to the net force EFx acting on it
divided by the proton's mass m: ax = EFx/m. Only the electrostatic force Fx acts on the
proton, so it is the net force. Setting SFx = Fx in Newton's second law gives
ax —
m
m
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 Fx given by Fx = qoEx(Equation 18.2), where qo is
the charge. Setting = e for the proton, we have
Fx = eEx
+ 2ax X
ax =
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
Vx= +
Algebraically combining the results of the three steps, we have
m
eEx
m
x
(l)
(2)
m
18.6 THE ELECTRIC FIELD
551
velocity of the proton is
Thefinal
+ Vox+ 2 m x
vs
-+
(2.5 X 104m/s)2 +2
(1.60 X 10-19
X 103N/C)
(2.0 X 10-3 m)
1.67 X 10-27kg
+3.9 X 104m/s
denotes that the final velocity points along the +x axis.
wherethe + sign
Problems 38, 40, 66
Related Homework:
POINT CHARGES
A more complete understanding of the electric field concept can be gained by
consideringthe field created by a point charge, as in the following example.
Example 10 | The Electric Field of a Point Charge
(a)
Thereis an isolated point charge of q +15 'LC in a vacuum at the left in Figure 18.19a. Using
a testcharge of qo = +0.80 /-LC,determine the electric field at point P, which is 0.20 m away.
Reasoning Following the definition of the electric field, we place the test charge qo 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:
Iqollql
r2
(8.99 x 109
x 10-6
(0.20
x 10-6 C)
q
(b)
q
= 2.7 N
Equation18.2 gives the magnitude of the electric field:
2.7 N
0.80 x 10-6 c
3.4 x 106 NIC
Iqol
Theelectric field E points in the same direction as the force F on the positive test charge.
Sincethe test charge experiences a force of repulsion directed to the right, the electric field
vectoralso points to the right, as Figure 18.19b shows.
(c)
Figure 18.19 (a) At location P, a
positive test charge qo experiences a
repulsive force F due to the positive
point charge q. (b) At P the electric
field E 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.
The electric field produced by a point charge q can be obtained in general terms from
Coulomb'slaw. First, note that the magnitude of the force exerted by the charge q on a
testcharge is F
klqllqol/r2. Then, divide this value by Iqol to obtain the magnitude
Ofthe field. Since Iqol is eliminated algebraically from the result, the electric field does
notdependon
the test charge:
Pointcharge
q
klql
2
(18.3)
As in Coulomb's
law, the symbol Iql denotes the magnitude of q in Equation 18.3, withOutregard to
whether q is positive or negative. If q is positive,then E is directed away from
as in Figure 18.19b.
On the other hand, if q is negative,then E is directed toward q, since a
negativecharge
attracts
a positive test charge. For instance, Figure 18.19c shows the electric
fieldthat Would
exist at P if there were a charge of —qinstead of +q at the left of the drawIng•Example I I
reemphasizes the fact that all the surroundingcharges make a contribution
tothe
electric field that
exists at a given place.
d
V
Example 11 | The Electric Fields from SeparateCharges May Cancel
TwoPOSitivepoint charges,
+4.0 pc, are separated in a vacuum by a
+16 'LC and
distanceof 3.0
m, as Figure 18.20 illustrates. Find the spot on the line between the charges
Wherethe
net electric
field is zero.
3.0 m
Figure 18.20 The two point charges q,
and create electric fields El and E2
that cancel at a location P on the line
between the charges.
552 CHAPTER 18
ELECTRIC FORCES AND ELECTRIC FIELDS
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 El equals that of E2. However,
since is smaller than (Il, this location must be closer to (12,in order that the field of the
smaller charge can balance the field of the larger charge. In the drawing, the cancellationspot
is labeled P, and its distance from is d.
Solution At P, El = E2, and using the expression E = klql/r 2, we have
Problem solving insight
Equation 18.3gives only the magnitude of
the electric field produced by a point
charge. Therefore, do not use negative
numbers for the symbol Iql in this
equation.
—q
1
k(4.o X 10-6 C)
(3.0 m —
Rearranging this expression shows that 4.0(3.0 m — d) 2 = d 2. Taking the square root of each
side of this equation reveals that
2.0(3.Om - d)
The plus and minus signs on the right occur because either the positive or negative root can be
taken. Therefore, there are two possiblevalues for d: +2.0 m and +6.0 m. The value +6.0 m
corresponds to a location off to the right of both charges, where the magnitudes of El and E2
are equal, but where the directions are the same. Thus, El and E2 do not cancel at this spot.
The other value for d corresponds to the location shown in the drawing and is the zero-fieldIocation: d = +2.0 m
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.
c
4
k(16 x 10-6 c)
3
(a)
V
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?
Reasoning and Solution In Figure 18.21a, the charges at corners 1 and 3 are both +q. The
positive charge at corner I 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 l.
Thus, the two fields have opposite directions. The magnitudes of the fields are identicalbecause 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
18.2la. 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
(b)
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.
of these fields are equal because each charge has the same magnitude and is located at the
same distance from the center of the rectangle. As a result, the two fields combine to give the
net electric field Eu shown in the drawing.
In Figure 18.21b, the charges on corners 2 and 4 are identical to those in part a Ofthe
drawing, so this pair gives rise to the electric field labeled E24.The charges on corners I and 3
are identical to those on corners 2 and 4, so they give rise to an electric field labeled E13,
which has the same magnitudeas E24.The net electric field E is the vector sum of E24and
E13 and is also shown in the drawing. Clearly, this sum is greater than Eu alone. Therefore'
the netfield in part b is larger than that in part a.
Related Homework: Conceptual Question 12, Problem 34
Need more practice?
nteractive LearningWare 18.3 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 +4.0 x 10-9 C 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
—7.00'LC charge were placed at this spot, what would be the electrostatic force (magnitude
and direction) exerted on it?
Related Homework: Problem 59
For
4. an interactive solution, go to www.wiley.com/college/cutnell
18.7
THE PARALLEL
ELECTRIC FIELD LINES
553
PLATE CAPACITOR
which gives the electric field of a point charge, is a very useful reEquation 18.3,
calculus, this equation can be applied in a variety of situaintegral
aid of
stilt.Withthe
distributed over one or more surfaces. One such example
are
point charges
tionswhere
importance
is the parallel plate capacitor. As Figure 18.22
practical
thathasconsiderable
device consists of two parallel metal plates, each with area A. A charge +q is
shows,this
over one plate, while a charge —qis spread uniformly over the other
spreaduniformly
between the plates and away from the edges, the electric field points
plate.In the region
plate toward the negative plate and is perpendicular to both. It can be
fromthe positive
16 in Section 18.9) that the electric field has a magnitude of
shown(see Example
(18.4)
parallelplate
capacitor
Area = A
—q
Figure 18.22 A parallel plate capacitor.
of free space. In this expression the Greek symbol sigma (c)
where€0is the permittivity
the charge density.
denotesthe charge per unit area (u = q/A) and is sometimes called
has the same value at all places between the
Exceptin the region near the edges, the field
distinct contrast to
plates.The field does not depend on the distance from the charges, in
point charge.
thefieldcreated by an isolated
CHECK YOUR UNDERSTANDING 3
A positivepoint 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 electricfieldat corner A is zero. (a) At which corner is the second charge located? (b) Is the secondcharge positive or negative? (c) Does the second charge have a greater, a smaller, or
thesamemagnitude 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 magnitudeand direction depend on
themagnitudes and algebraic signs of the charges that create it.
Forsimilarquestions (including calculational counterparts), consult Self-Assessment Test
18.2,whichis described at the end of Section 18.9.
c
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
strengthof 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.
Sincethe electric field is the electric force per unit charge, the electric field lines are
sometimes
called lines offorce.
TOintroduce 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
1
2
8
5
(a)
(b)
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.
554 CHAPTER 18
ELECTRIC FORCES AND ELECTRIC FIELDS
—q
Figure 18.24 The electric field lines are
directed radially inward toward a
negative point charge —q.
Edge view
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 magnitudeand
direction at all points.
repulsive force, as the arrows in the drawing indicate. Therefore, the electricfieldc
by the charge +q is directed radially outward. The electric field lines are lines
show this direction, as part b of the drawing illustrates. They begin on the charge+qand
point radially outward. Figure 18.24 shows the field lines in the vicinityof 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 towardnegaåve
charges.
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
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 chargeas froma+q
charge.
The pattern of electric field lines also provides information about the magnitude
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 thecharges
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 factno
matter how many charges are present, the number of lines per unit area passingperpen.
dicularly 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 samenumber
of lines per unit area everywhere, and the electric field has the same strengthat allpoints
For example, Figure 18.25 shows that the field lines between the plates of a parallelplate
capacitor are parallel and equally spaced, except near the edges where theybulgeout.
ward. The equally spaced, parallel lines indicate that the electric field has the samemagnitude and direction at all points in the central region of the capacitor.
Often, electric field lines are curved, as in the case of an electric dipole. Anelectric
dipole consists of two separated point charges that have the same magnitude butopposite
signs. The electric field of a dipole is proportional to the product of the magnitudeofone
of the charges and the distance between the charges. This product is called the dipolemo•
ment. Many molecules, such as 1--120
and HCI, have dipole moments. Figure 18.26depicts
the field lines in the vicinity of a dipole. For a curved field line, the electric fieldvectorat
a point is tangent to the line at that point (see points 1, 2, and 3 in the drawing).Thepat•
be
tern of the lines for the dipole indicates that the electric field is greatest in the region
tween and immediately surrounding the two charges, since the lines are closesttogether
there.
Notice in Figure 18.26 that any given field line starts on the positive chargeand
on the negative charge. In general, electric field lines always begin on a positivecharge
1
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 l, 2, or 3,
the field created by the dipole is tangent
to the line through the point.
2
3
18.7 ELECTRIC FIELD LINES
555
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.
andend on a negative charge and do not start or stop in midspace.Furthermore, the
numberof lines leaving a positive charge or entering a negativecharge is proportional
to the magnitude of the charge. This means, for example, that if 100 lines are drawn
leavinga +4 /JC charge, then 75 lines would have to end on a —3kLCcharge and 25 lines
ona —1/.LCcharge. Thus, 100 lines leave the charge of +4 'LC and end on a total charge
of —4'LC, so the lines begin and end on equal amounts of total charge.
The electric field lines are also curved in the vicinity of two identical charges. Figure
18.27shows 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
thatthe electric field is relatively weak between the charges.
Some of the important properties of electric field lines are considered in Conceptual
Example 13.
V
Conceptual
Example 13 | Drawing Electric Field Lines
Figure18.28a 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
thingswrong with this drawing. What are they?
Reasoningand Solution One aspect of Figure 18.28athat is incorrect is that electric field
linescross at point P. Field lines can never cross, and here's why.An electric charge placed at
P experiencesa 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
thatpoint. 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
anypoint, 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 negativecharges. Remember that the number of field lines leaving a positive charge or entering a
(a)
negativecharge is proportional to the magnitudeof the charge.The —2qcharge has half the
magnitudeof 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 —qcharges, 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
evenlySpaced.This would
indicate that the electric field everywhere in this region has a constant
magnitudeand direction,
as is the case in the central region of a parallel plate capacitor. But the
electricfield between
the
+4q and —qcharges is not constant everywhere. It certainly is stronger
InPlacesclose to
the +4q or —qcharge than it is midway between them. The field lines, therefore,
shouldbe drawn
with a curved nature, similar (but not identical) to those that surround a dipole.
Igure18.28b shows
more nearly correct representations of the field lines for the four charges.
Related
Homework: Problems 25, 26
-q
—W
(b)
Figure 18.28 (a) Incorrectly and
(b) correctly drawn electric field lines.
566 | CHAPTER 18 | ELECTRIC FORCES AND ELECTRIC
FIELDS
Learning Aids
Discussion
Topic
Electric field of a point charge
PROBLEMS
directionof the force on a positive test charge.
The direction of the electric field is the same as the
(NIC). The source of the electric field
coulomb
per
newton
the
is
field
electric
The SI unit for the
spot.
that
surrounding
at any spot is the charged objects
charge q is
The magnitude of the electric field created by a point
Ex.10-u18
klql
(18.3)
Excess charge carried by a
conductor at equilibrium
IS 1841
10.There is an electric field at point P. A very small charge is
placedat this point and experiences a force. Another very small
chargeis then placed at this point and experiencesa force that differsin both magnitude and direction from that experienced by the
firstcharge.How can these two differentforces result from the singleelectricfield that exists at point P?
Il. Ihree point charges are fixed to the comers of a square,one to a
comer.in such a way that the net electric field at the empty comer is
zero.Do these charges all have (a) the same sign and (b) the same
magnitude(but, possibly, different signs)? Justify your answers.
(18.4)
Electric field of a parallel plate
Direction of electric field
Strength of electric field
foryour answer.
nonconducting rod, positive charges are spread evenly,
9. On a thin,
the same amount of charge per unit length at every
is
there
that
so
pint. On another identical rod, positive charges are spread evenly
overonly the left half, and the same amount of negative charges are
spreadevenly over the right half. For each rod, deduce the direction
of theelectric field at a point that is located directly above the midpointof the rod. Give your reasoning.
ILW 18.3
and r is the distance from the
where Iql is the magnitude of the charge and has no algebraic sign
a negative charge.
toward
and
charge
positive
a
from
charge. The electric field points away
plate,
the magnitude of
each
on
For a parallel plate capacitorthat has a charge per unit area of
is
plates
the
between
the electric field
Electric field lines
simple harmonic motion on being released, as will the
notexhibit
spring. Explain why not.
the
on
panicle
point charges are fixed to opposite corners of a
Identical
8. ssm
does a third point charge experience the greater force,
square.Where
empty corners or at the center of the square?Account
atone of the
FIELDLINES Electricfield lines are lines that can be thought of as a "map," inso18.7 ELECTRIC
far as the lines provide informationabout the directionand 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 C 18.2
passing perpendicularlythrough a surface is the greatest—that is, where the lines are packed together most tightly.
12. Review Conceptual Example 12 as an aid in answering this ques-
18.8 THE ELECTRIC FIELD INSIDE A CONDUCTOR: SHIELDING Excess negativeor positive EL 14
tion.Supposein figure 18.21 that charges +q are placed on corners
I and3 of the rectangle, and charges —qare placed on comers 2 and
4. Whatis 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
wherethe 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
geater than the magnitude of the positive charge. Is there any place
onthe line through the charges where the electric field is zero? Jus-
charge resides on the surface of a conductorat equilibrium under electrostatic conditions. In such
a situation, the electric field at any point within the conductingmaterial is zero, and the electric
field just outside the surface of the conductoris perpendicular to the surface.
through a surface is related to the magnitude E of the
18.9 GAUSS' LAW The electric flux
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)
cos
Gauss' law states that the electric flux througha closed surface (a Gaussian surface) is equal to the
net charge Q enclosed by the surface divided by €0, the permittivity of free space:
=
Gauss' law
tify your answer.
ELL 16
cos 4)AA
14. ssm Drawings I and
Il showtwo examples of
(18.7)
IS 18.51
electric
fieldlines.Decide
whichof the following
statementsare true and
use Self-AssessmentTest 18.2to evaluate your understanding of Sections 18.6—18.9.
whichare false, defending
CONCEPTUAL QUESTIONS
SSm Solutionis in the StudentSolutionsManual.
5. Blow up a balloon and rub it against your shirt a
left with a charge? Account for your answer.
remain stuck to the ceiling. Why?
6. A proton and an electron are held in place on the x axis. fie
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 naturaloils
and drying it thoroughly.run a plastic comb through it. Small bits of
paper will be attracted to the comb. Explain why
the electric field becomesstronger. (c) The electric field in I is the
same everywherebut becomesstronger in Il as you move from left
to right. (d) The electric fields in both I and Il could be created by
negativecharges located somewhere on the left and positive charges
somewhereon the right. (e) Both I and Il 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 betweenthe 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 throughthe surface is also zero: that is.
= O. Does the fact that
= O imply that the electric field É at
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 q, q:, and '13.
A Gaussian surface is drawn
around q, and '12. (a) Which
charges determine the electric
flux throughthe Gaussiansurface? (b) Which charges produce the electric field at the
point P? Justify your answers.
Gaussian surface
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?
PROBLEMS
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
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 happensto 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
11
yourchoicein each case. (a) In both I and Il the electric field is the
sameeverywhere.(b) As you move from left to right in each case,
| 567
Of
times. In so doing you give the balloon a net electric charge.
touch the balloon to the ceiling. On being released, the balloonwill
ton is at x =
d, while the electron is at x = +d. Theyarereleased
simultaneously, and the only force that affects their motionsis the
electrostaticforce of attractionthat each applies to the other.
particle reaches the origin first? Give your reasoning.
7. A particle is attached to a spring and is pushed so that the spnng
is compressed more and more. As a result, the spring exerts a greater
and greater force on the particle. Similarly, a charged partw
experiences a greater and greater force when pushed closer
of
closer to another particle that is fixed in position and has a charge
the same polarity In spite of the similarity. the charged particle
Problemsthat are not marked with a star are consideredthe easiest to solve. Problems that are marked witha single star C) are more difficult,while
thosemarkedwith a
double star ( •e) are the most diffcult.
Noteto Instructors: Most of the homeworkproblemsin this chapterare availablefor assignment via an online homeworkmanagementprogram such
as WileyPLUS
or WebAssign,and those marked withthe icon GO are presented in a guided tutorialformat that provides enhanced interactivity. See Pmface
for addittonal
details.
ssm Solutionis
in the Student Solutions Manual.
This icon represents a biomedical application.
www Solution is
available on the World Wide Web at www.wiIey.com/college/cutnelJ
Section18.1The
Origin of Electricity,
18.2Charged Objects and the Electric Force,
Section183
Conductors and Insulators,
Section18.4
Charging by Contact and by Induction
1.
ssm How many
electrons must be removed from an electrically
neutralsilver
dollar to give it a charge of +2.4 gC?
2. A plate carries a charge of —3.0/LC,while a rod carries a charge
of +2.0 gC. How many electrons must be transferredfrom the plate
to the rod, so that both objects have the same charge?
3. A metal sphere has a charge of +8.0 /.cc. What is the net charge
after 6.0 X 1013electrons have been placed on it?
4. Four identicalmetallic objects carry the following charges: + 1.6.
+6.2, —4.8.and —9.4pc. ne objects are brought simultaneously
570 | CHAPTER
18 I ELECTRIC
FORCES
ELECTRIC
AND
FIELDS
CONCEPTS
electric field E in the drawing is uniform and has a magnitude
42. The drawing shows
an electron entering the
lower left side of a paral-
of 250 NIC. Find the electric flux through (a) surface I and
(b) surface 2.
lel plate capacitor and ex-
0.150 cm
iting at the upper right
side. The initial speed of
the electron is 7.00 x
106 m/s. The capacitor is
48. A rectangular surface (0.16 m ><0.38 m) is oriented in a uni.
form electric field of 580 NIC. What is the maximum possibleelec.
tric flux through the surface?
49. A vertical wall (5.9 m X 2.5 m) in a house faces due east.A
2.00 cm long, and its plates are separated by O.150 cm. Assume that
the electric field between the plates is uniform everywhere and find
its magnitude.
43. A small plastic ball with a mass
of 6.50 X 10-3 kg and with a charge
of +0.150
is suspended from an
insulating thread and hangs between
1 30.00
uniform electric field has a magnitude of 150 NIC. This field is par-
allel to the ground and points 350 north of east. What is theelectric
flux through the wall?
50. A charge Q is located inside a rectangular box. The electricflux
= +1500 N•m2/C,
through each of the six surfaces of the box is:
= +4600 N.m2/C,
= -1800
= +2200 N.m2/C,
= -3500 N. m2/C,
= -5400 N.m2/C. Whatis Q?
the plates of a capacitor (see the drawing). The ball is in equilibrium, with
the thread making an angle of 30.00
with respect to the vertical. The area of
each plate is 0.0150 m2. What is the
magnitude of the charge on each plate?
51. Interactive Solution 18.51 at www.wiley.com/college/cutnell
offers help with this problem in an interactive environment.A solid
nonconducting sphere has a positive charge q spread uniformly
44. Two particles are in a uniform electric field whose value is
= 1.4 x 10-5 kg
+2500 NIC. The mass and charge of particle I are
and q, = —7.0 gC, while the correspondingvalues for panicle 2 are
= 2.6 ><10-5 kg and = +18 '.LC.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 acceler-
a point within the sphere at a radius r has a magnitude of
throughout its volume. The charge density or charge per unit volume
3
ate 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
4 reoR 3
(Hint: For a Gaussian surface, use a sphere of radius r 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 www.wiley.com/college/
cutnell for a perspective that is useful in solving this problem.Two
spherical shells have a common center. A —1.6 X 10-6 C chargeis
spread uniformly over the inner shell, which has a radius of 0.050m.
at the midpoint M between the
charges has a magnitude EM.The
field directly above the midpoint +q
at point P has a magnitude Ep. The ratio of these two field magnitudes is EM/Ep = 9.0. Find the angle a in the drawing.
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 +3.5 X 10-6 C charge, (b) a single
—2.3 X 10-6 C charge, and (c) both of the charges in (a) and (b).
drawing
47. ssm The
shows an edge-on view
of two planar surfaces
that intersect and are
mutually perpendicular.
Surface I has an area of
1.7 m2, while surface 2 has
an area of 3.2 m2. The
Use Gauss' law to show that the electric fieldat
therefore, is
Surface 1
350
Surface 2
A +5.1 x 10-6 C charge is spread uniformly over the outershell,
of
which has a radius of 0.15 m. Find the magnitude and direction
the electric field at a distance (measured from the commoncenter)
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 anx,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.
magnitude of the field is 1500 NIC. (a) Find the electricflux
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 thecubinet
cal surface is zero, as Gauss' law predicts, since there is no
charge within the cube.
Q
54. A long, thin, straight wire of length L has a positive charge
elee
distributed uniformly along it. Use Gauss' law to show thatthe
tric field created by this wire at a radial distance r has a magnitude
sudace•
of E = A/(27T€or),
where A = Q/L. (Hint: For a Gaussian
use a cylinder aligned with its axis along the wire and notethat
cylinder has a flat surface at either end, as well as a curved
face.)
ADDITIONAL PROBLEMS
55. An electric field of 260 000 NIC points due west at a certain
spot. What are the magnitude and direction of the force that acts on
a charge of —7.0 'LC at this spot?
eachOther
56. The force of repulsion that two like charges exert on
is 3.5 N. What will the force be if the distance between the
is increased to five times its original value?
57.
Conceptual Example 14 deals with the hollow spherical conduc-
18.31. The conductor is initially electricallyneutral,
torin Figure
charge +q is placed at the center of the hollowspace.
a
then
and
has a net charge of +2q instead of
supposethe conductor initially
What is the total charge on the interior and on the exbeingneutral.
when the +q charge is placed at the center?
teriorsurface
from a point charge, the magnitude of the elec58. At a distance r,
charge is 248 NIC. At a distance r2 from the
tric field created by the
a magnitude of 132 NIC. Find the ratio r2/rl.
charge,the field has
59. ssm Background pertinent to this problem is available in
Interactive LearningWare 18.3 at www.wiley.com/college/cutnell.
A 3.0-/JCpoint charge is placed in an external uniform electric field
of 1.6 X 104NIC. At what distance from the charge is the net electric field zero?
60.In a vacuum, two particles have charges of q, and 012,where
= +3.5 gC. They are separated by a distance of 0.26 m, and particle I experiences an attractive force of 3.4 N. What is (magnitude and sign)?
61. ssm Interactive LearningWare 18.1 at www.wiley.com/college/
cutnelloffers some perspective on this problem. Two tiny spheres have
thesame mass and carry charges of the same magnitude. The mass of
eachsphereis 2.0 X 10-6 kg. The gravitational force that each sphere
exertson the other is balanced by the electric force. (a) What algebraic
signscan the charges have? (b) Determine the charge magnitude.
62. A charge of q = +7.50 1-CCis located in an electric field. The x
andy components of the electric field are Ex = 6.00 X 103 NIC and
Ey= 8.00 X 103 NIC, respectively. (a) What is the magnitude of the
forceon the charge? (b) Determine the angle that the force makes
withthe +x axis.
63.Two charges are located along the x axis: qi = +6.0 /LC at
= +4.0 cm, and q2 = +6.0 gC at x2 = —4.0 em. Two other
at = +5.0 cm,
chargesare located on the y axis:
= +3.0
and qg= —8.0
at
= +7.0 cm. Find the net electric field
(magnitudeand direction) at the origin.
& CALCULATIONS
| 571
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.
d
65. A proton is moving parallel to a uniform electric field. The electric field accelerates the proton and thereby increases its linear mo-
mentumto 5.0 X 10-23kg•m/s from 1.5 X 10-23kg•m/s in a time
of 6.3 X 10-6 s. 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 q = 7.5 'LC and
mass m = 9.0 X 10-5 kg, is placed in a constant electric field. Starting from rest, the object accelerates to a speed of 2.0 X 103 m/s in a
time of 0.96 s. Determine the magnitude of the electric field.
67. ssm In the rectangle
the drawing, a charge
+3.0PC
4d
is to be placedat the
empty comer to make the
net force on the charge at
+3.0 pc
+3.0 pc
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.
CONCEPTS & CALCULATIONS
Note: Each of these problems consists of Concept Questions
followedby a related quantitative Problem The Concept Questions
involvelittle or no mathematics. Theyfocus on the concepts with which
theproblemsdeal. Recognizing the concepts is the essential initial step in
anyproblem-solving
technique.
69.
metal spheres have
GO Concept Questions Two identical
charges of q, and (12.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
spherethe same?
Why?
Problem Four
identical metal spheres have charges of
-2.0 pc, qc = +5.0 gc, and
= +12.0
Go
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 is Iql = 8.6 LLC.and the distance between them is 3.8 mm. Determine the magnitude of the net
your anforce on charge 2 for each of the three drawings. Verify that
swers are consistent with your answers to the Concept Questions.
2
= -8.0 ,zc,
(a) Two ofthe
separated.
spheresare brought
together so they touch and then they are
is
spheres are they, if the final charge on each of the two
+5.0 1..tC?(b) In
are brought
a similar manner, which three spheres
togetherand then
of the three is
separated, if the final charge on each
one of
+3.0 1...tC?
(c) How many electrons would have to be added to
thespheres
neutral?
in pan (b) to make it electrically
70.
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
charges
Concept Questions The drawings show three
signs. In all
different
but
that have the same magnitude,
and 3 is
casesthe distance
between 2
d between charges I and 2 and
3
Concept Questions Suppose you want to neutralize the
gravitational attraction between the earth and the moon
on each. (a) Should the charges
by placing equal amounts of charge
positive and the other negaone
or
negative.
both
positive,
be both
71. GO
of the charge? Why or why not?
and the moon to find the magnitude