Chapter 14: Electric Charge, Force and Energy
! How does a photocopier work?
! Why is your hair attracted to a plastic comb after brushing?
! How is a pop can different from a plastic bottle on a microscopic level?
Make sure that you know how to:
1. How to identify a system and construct an energy bar-chart for it.
2. How to identify a system and construct a force diagram for it.
3. How to convert a force diagram and an energy bar-chart into a mathematical statement.
Chapter opening
You have probably used a photocopy machine many times. You place a document on a
glass surface, close the lid, push the start button, and a second later a copy appears on a new sheet
of paper. The toner inside the copier produced the same image on the new sheet as on the
original. How does the toner stick to the paper and how does the toner “know” where to stick? By
the end of the chapter you’ll be able to answer these questions.
Lead In Chapters 9-13 we learned that all objects are made of tiny particles. What holds
those particles together? Could they be hold together by the gravitational forces? Actually, we
can basically rule that out right away. Gravitational interactions lead to the attraction of objects. If
all particles inside an object were attracted to each other, what would prevent the object from
shrinking indefinitely? In this chapter we will learn about a new type of interaction (other than
gravitational) that is responsible for many phenomena that we observe in our everyday world.
14.1 Observations of electrostatic interactions
If you rub a hanging balloon with wool and then bring the wool near the balloon, the wool
attracts the balloon. (Fig. 14.1a) If you bring a second balloon rubbed with wool near the first
balloon, they repel. (Fig. 14.1b) Many other objects behave in a similar way. If instead of
balloons, we use tubes made of foam and rub them with different materials, we observe similar
effects: sometimes attraction, sometimes repulsion. Let’s do some more observational
experiments; then try to devise an explanation for our observations. We will use balloons and rub
them with different materials and record the observations of their interactions with each other.
See Observational Experiment Table 14.1.
Figure 14.1Rubbing different materials causes a new interaction
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Observational Experiment Table 14.1 Results of experiments with rubbed objects.
Observational experiment
(a) Hang each of two small balloons from two thin
threads. Rub each balloon with felt. Bringing the
balloons closer to each other increases the angle between
the threads.
Analysis
See a force diagram for each rubbed balloon. An
unknown force points away from each balloon.
This force is stronger the closer the balloons are.
(b) Repeat experiment (a) only this time rub two small
balloons with plastic wrap. Bringing the balloons closer
to each other increases the angle between the threads.
The force diagram is the same as in
the previous experiment.
(c) Hang from a string the balloon rubbed with felt. Bring See a force diagram for the balloon. The force is
the felt close to the balloon. The balloon is attracted to
stronger the closer the balloon is to the felt.
felt.
(d) Hang from a string the balloon rubbed with plastic
wrap. Bring the plastic wrap close to the balloon. The
balloon is attracted to the plastic.
(e) Bring the balloon rubbed with felt near the balloon
rubbed with plastic wrap. The balloons are attracted to
each other.
!
!
!
!
Same as the previous experiment.
See the force diagram for each of rubbed balloon.
An unknown force pulls each balloon towards the
other. This force is stronger the closer the
balloons are.
Patterns
Balloons rubbed with the same material repel each other.
Balloons rubbed with a material are attracted to that material.
A balloon rubbed with felt attracts a balloon rubbed with plastic wrap.
All of these effects are stronger the closer the balloons are to each other.
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Additionally, sometimes we hear a cracking sound when the balloons are close. If in a dark room,
we sometimes see sparks. We also find that more vigorous rubbing intensifies the interaction.
We can do similar experiments using different objects such as foam tubes, combs, plastic
and glass rods, rubbing them with silk, felt, etc. Usually whenever two different materials are
rubbed together, they either repel or attract other objects, behaving similarly to the balloons in
Observational Experiment Table 14.1.
People have observed similar effects for many years. The Greeks first saw that spindles
made of amber used for weaving attracted the hair of women working with the spindles. The
word for amber in Greek is “electron”; thus this attraction was called electrical. Over time, the
new property acquired by the materials due to rubbing came to be called electric charge. The
objects were said to acquire positive electric charge or negative electric charge.
By agreement, the charge that appears on a balloon rubbed with plastic wrap or a glass
rod rubbed with silk is called positive; the charge that appears on a balloon rubbed with wool or
felt or on a plastic rod rubbed with felt is called negative. With these ideas we can reinterpret the
patterns that we found in the observational experiments (Observational Experiment Table 14.1):
1. Materials rubbed against each other acquire an electric charge.
2. Two objects with the same charge repel each other.
3. Objects that have different types of charge attract each other.
4. Two objects made of different materials rubbed against each other acquire opposite
charges.
5. The magnitude of the force that the charged objects exert on each other is greater if the
objects are closer to each other.
6. Sometimes more vigorous rubbing leads to a greater force exerted by the rubbed objects
on each other.
Remember, we have NOT observed electric charge in any of these experiments. We
observed attractive and repulsive forces and used the idea of electric charge to explain what is
happening. This idea does not explain why electric phenomena happen or what electric charge is.
We also learned that the model of two electric charges explains all of the experiments that we
observed so far. Does it mean that there are only two types of electric charge or might there be
three types? Let’s test that hypothesis.
Is there a third type of charge?
Imagine that rubbing a balloon with some new type of material (Lycra for example) gives
the balloon a third type of charge. Then according to our pattern – two balloons rubbed with
Lycra should repel – and they do. But they should also attract both the balloon rubbed with felt
and the balloon rubbed with plastic (if we assume that all differently charged objects attract each
other). However, this never happens. If the balloon rubbed with Lycra attracts the balloon rubbed
with felt, it always repels the balloon rubbed with plastic, or vice versa. No matter how many
such experiments we perform, if an object acquires charge and repels one type of charged object,
it always attracts the other type. No experiment every performed has required the existence of a
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third type of electric charge in order to explain its outcome. As far as we know there are only two
types of electric charge, and we call them positive and negative.
Conceptual Exercise 14.1 Mysterious Scotch Tape Take two 9-inch long pieces of Scotch
Magic Tape and place them sticky side down on a plastic or wooden tabletop. Now pull on one
end of each tape to remove them from the table. Bring the pieces of tape near each other. Figure
14.2a depicts what you will observe. They repel. Decide whether or not the repulsion of the tapes
is due to the electric interaction.
Figure 14.2(a) Experiments with Scotch Tape pulled from tabletop
Sketch and Translate The tapes do not resemble rods or balloons in any way and we did not rub
them with any material. However, the tapes repel each other, which is the way like-charged
objects behave (2 in the list above); thus electric charge on the pieces of tape is a possibility. To
test this hypothesis, we need to design an experiment whose outcome we can predict using the
idea that the tapes are electrically charged. If they are, and we bring them one at a time next to a
glass rod rubbed with silk (positively charged) and a plastic rod rubbed with felt (negatively
charged), then they should attract one and repel the other.
Simplify and Diagram When we perform the experiment (Fig. 14.2b and c), you see that each
tape is repelled by the plastic rod rubbed with felt and attracted to the glass rod rubbed with silk.
Does it mean that the tapes are electrically charged? Actually, we cannot say for sure. Perhaps
there is another interaction at work here that has some similarities to the electric interaction.
That’s possible, but our experiment does not rule out our hypothesis that the pieces of tape are
electrically charged.
Figure 14.2(b)(c)
Try It Yourself: Place one piece of tape on top of the table and then another piece on top of the
first. Pull them together off the table and then separate them. You observe that the tapes attract
each other. Design an experiment to find out which one is positively and which one is negatively
charged.
Answer: A single piece of tape pulled off a table and a plastic rod rubbed with felt repel—thus
they are both negatively charged. Hold such a plastic rod near the two pieces of tape described
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above to decide the signs of their charges. You find that the bottom tape is negative and the top
tape is positive.
Tip! A tape pulled off a table repels a balloon rubbed with plastic; thus we will consider both
objects negatively charged for future experiments.
Is the electrical interaction just a magnetic interaction?
Electrically charged objects interact with each other: they repel or attract. We already
know about one type of interaction that makes objects attract each other—gravitation. Clearly, the
electric interaction is different than the gravitational interaction because we observe both electric
repulsion and attraction. Do you recall observing any other interaction that involved repulsion? If
you ever played with magnets you probably know that magnets have poles (North and South) and
same poles repel while different poles attract. Electrically charged objects also attract and repel
each other. Maybe the electric interaction is actually the magnetic interaction, just described
using different terminology? Let’s test this hypothesis.
Testing Experiment Table 14.2 Testing the electric = magnetic interaction hypothesis
Testing experiment
Bring a negatively
charged plastic rod near
the North pole of a
magnet.
Prediction
If the electric interaction is the same as
the magnetic interaction, then the
negatively charged rod should either
attract or repel the North pole of the
magnet.
Outcome
The negatively
charged rod
attracts the
North pole.
Bring the negatively
charged plastic rod near
the South pole of a
magnet.
If electrically charged objects behave
like magnets, the negatively charged
rod should now repel the South pole of
the magnet.
The
negatively
charged rod
also attracts
the South
pole.
Conclusion
Since the outcome of the second experiment was inconsistent with the prediction, we disproved the
hypothesis that the electric interaction is the same as the magnetic interaction.
The same charged rod attracts both the North and the South poles. This disproves our hypothesis
that electric interactions are the same as the interactions of magnets. We’ll continue our
investigation into a mechanism for the electric interaction in the next section.
Review Question 14.1
How do people know that a plastic rod rubbed with felt is negatively charged and the glass rod
rubbed with silk is positively charged?
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14.2 Explanations of electrostatic interactions
The goal of this section is to develop an explanation for electric interactions. However,
before we proceed to devise the explanations, let’s investigate another phenomenon associated
with the electric interaction.
Do uncharged objects interact with charged objects?
So far we found that we need to rub objects for them to become electrically charged and
participate in electric interactions. However, our everyday observations show that lightweight
objects, like shreds or paper, without any prior rubbing, are attracted to charged objects. A simple
experiment with a plastic comb and small pieces of paper or pieces of aluminum foil allows us to
observe this phenomenon. When we bring a plastic comb rubbed against felt or against a
synthetic sweater (making the comb negatively charged) close to small pieces of paper, the pieces
of paper stand up and some of them jump to the comb. (Fig. 14.3) If instead of the paper we have
small pieces of aluminum foil, they jump on the comb even more readily than the paper. Some of
the aluminum foil pieces the fall off, then jump to the comb again. It is fascinating to observe this
dance that the aluminum foil pieces do. We will first focus on the features of this experiment that
are common to both the paper and foil pieces – an attraction to the comb. Let’s do some
observational experiments to investigate this further.
Figure 14.3 Paper pieces attracted to charged comb
Observational Experiment Table 14.3 Interactions of charged and uncharged objects
Observational experiment
(a) Hang a negatively charged balloon near a wall (the wall
is uncharged). The balloon is attracted
to the wall.
Analysis
A force diagram for the balloon.
(b) Hang a positively charged balloon near the uncharged
wall. The balloon is attracted to the wall.
A force diagram for the balloon.
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(c) Repeat experiment (a) only this time, bring the
negatively charged balloon near an uncharged metal plate.
The balloon has a stronger attraction to the plate than it did
with the wall.
A force diagram for the balloon.
(d) Repeat experiment (b) only this time, bring the
positively charged balloon near an uncharged metal plate.
The balloon has a stronger attraction to the plate than it did
with the wall.
A force diagram for the balloon.
Pattern
The positively and negatively charged balloons were attracted to uncharged materials (the wall and the
metal). The intensity of interaction was much stronger with the metal plate than the non-metallic wall.
From the experiments in Observational Experiment Table 14.3, we can infer that both types of
charged objects are attracted to uncharged objects. Thus, when we observe that some object 1 is
attracted to another charged object 2, it does not necessarily mean that object 1 is oppositely
charged compared to object 2? Object 1 could either have the opposite charge as object 2, or
object 1 could be uncharged. To decide whether or not object 1 is charged, we need to observe its
repulsion from some charged object. The only way two objects will repel is if they have the same
electric charge.
Let’s summarize all the patterns we observed so far about the electric interaction:
1. When two objects are rubbed against each other, one of them might become positively
charged or negatively charged. The other object acquires the opposite charge.
2. Like charged objects repel; oppositely charged objects attract.
3. Uncharged objects are attracted to objects with both charges; uncharged metal objects are
attracted much stronger than nonmetal objects.
We also found through the testing experiments that the electric attraction or repulsion of two
objects is NOT the same as the attraction or repulsion of the magnets. Any new explanation for
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the electric interaction must incorporate all three of the patterns. We will first examine the
historical explanations.
One fluid model of electric charge
Benjamin Franklin (1706-1790) was one of the first persons to propose an explanation for
the observations of electrically charged objects. Franklin proposed that a special weightless fluid
(electric fluid) is contained in all objects and consists of tiny weightless particles that repel each
other. According to Franklin, too much electric fluid in an object results in a positive electric
charge on the object and a deficiency of this fluid results in a negative charge on the object. If you
rub two different types of material against each other, the electric fluid may move from one
material to the other. One material will lose some positive fluid and the other one will gain that
fluid. The material that lost the positive fluid becomes negatively charged and the material that
gained positive fluid becomes positively charged.
Franklin proposed a model common for his time—an invisible weightless fluid that
explained the observed phenomena (caloric was another example of this type of explanation.)
This model could also explain the attraction of an uncharged object to charged objects. Imagine
that you bring a negatively charged object B close to an uncharged object A. If the positive
electric fluid already present in the uncharged object A can move, then it will move towards
object B, making the positive side of A closer to negatively charged B; A and B should attract.
The same will happen if B is positively charged. The fluid in A will move away from it, making
the negatively charged end of A close to the positive charge on B; they should attract.
Two fluid model of electric charge
At about the same time, a different competing fluid model existed. In this second model
there were two electric fluids—positive and negative. In an object that was not rubbed, there was
a balance of the fluids. After rubbing, one object loses positive fluid and becomes negative while
the other one gains the positive fluid and becomes positive. Or one object loses negative fluid and
becomes positive while the other one gains negative fluid and becomes negative. So the net fluid
remained the same. Objects with an excess of the same fluid repelled each other. Objects with an
excess of different fluids attracted. Both models could account for all of the observations that had
been at the time.
Particle model of electric charge
Experiments conducted by J.J. Thomson and Robert Millikan in the early 20th century
1909 suggested that electric charge cannot be associated with a weightless fluid but is carried by
the particles that comprise matter. Robert Millikan and his graduate student Harvey Fletcher
placed tiny droplets of oil between two oppositely charged horizontal metal plates. Oil droplets
fell downward between the plates due to the gravitational force exerted by Earth on them. While
falling through the air, they acquired some electric charge due to their friction with the air, similar
to balloons becoming charged through rubbing. It was possible to adjust the electric charge on the
plates so that the droplets drifted down at constant speed (the upward electric force due to the
charge on the plates exactly balanced the downward gravitational force). Then, if the charge of
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the droplets changed, the charge of the plates had to be adjusted again to maintain the steady
speed of the droplets (Fig. 14.4).
Figure 14.4 Oil drop experiment
Thousands of measurements demonstrated that the adjustments of the charge on the
plates were always equal to a multiple of the same number (number x 5 or number x 7). This
meant that the electric charge of each oil droplet changed only by multiples of some discrete unit
of electric charge. The explanation for this pattern is that electric charge is carried by microscopic
particles – removing or adding the particles to an object changes the charge on the object, and that
each charged particle had a charge equal to some smallest indivisible amount of electric charge.
Contemporary model
Today we understand that the explanation of charging by rubbing is actually very
complicated. A simple version of it is as follows. (The experimental evidence for this is provided
in Chapters 26, 27 and 28.) Atoms, the basic units of matter, consist of three types of particles:
electrons, protons, and neutrons. Two of these particles—electrons and protons—have electric
charge, and the neutron has zero charge. The protons and neutrons are located in the nucleus of
the atom, which is very small (about 10-15 m) compared with the size of the atom (how positively
charged protons can be contained inside such a small nucleus is a the question that will be
answered in chapter 28). Electrons surround the nucleus giving the atoms their full size (about 1010
m)—see Fig. 14.5. The magnitude of the electric charge of an electron equals that of a proton
(even though the electron is about 2000 times lighter than the proton). The proton is said to have
a positive electric charge; the magnitude of the charge is given the symbol e. An electron has a
negative charge whose value is represented by the symbol –e. Atoms have zero net charge since
they contain equal numbers of protons and electrons. The mass of a material is primarily due to
the protons and neutrons located in the atomic nuclei. The electrons, though they are responsible
for the size of atoms, contribute very little to the mass of the material.
It is possible for an electron to leave a material and move to a different material. When
such a transfer occurs, it happens because the atoms in the material the electron transfers to, hold
on to their electrons more tightly than the atoms in the material the electron transferred from. An
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atom that has lost one electron has a net charge of +e (it has one more proton than electron) and is
called a positive ion (a positively charged sodium ion is shown in Fig.14.6). An atom that gains
an extra electron has a net charge of –e and is called a negative ion. This explains why rubbed
materials acquire opposite charges. One material gains electrons and becomes negatively charged.
The other loses an equal number of electrons and with this deficiency of electrons becomes
positively charged.
We can represent this process using integers. A neutral object can be represented with a
zero. A zero can be made up of a sum of positive and negative numbers – for example +10 + (–
10) = 0 or +17 + (–17) = 0. When we rub object 1 with a second object 2 that pulls electrons off
object 1, the negative number of object 1 becomes smaller, and the positive number stays the
same.
For example, suppose we remove three electrons from an object originally with charge
+10 + (–10) = 0; now the object has charge +10 + (–10) – (–3) = +10 + (–7) = +3. The object is
positively charged. If on the other hand we rub this object with a different type of object that
deposits electrons on it (for example it deposits 3 electrons), the negative number becomes
bigger, but the positive number stays the same: +10 + (–10) + (–3) = +10 + (–13) = –3.
Every time you rub or contact in other ways materials that hold on to their electrons with
different strengths, some electrons are likely to transfer from one material to the other. Sometimes
you do not even need to rub the materials against each other – contact alone might be enough.
The small electrical shock you receive when you touch another person after shuffling across a
synthetic rug is the result of removal of electrons from the rug through your shoes and onto your
body. Some of these electrons leave your body quickly when you touch the other person. The
spark that is sometimes seen is caused by the changes in the properties of air as the result of the
electron transfer (we will learn the details of how the spark is produced in the next chapter.)
Sometimes when you rub two objects against each other, no transfer of electrons occurs.
This happens if the electrons in both materials are bound equally strongly by the nuclei in each
material.
Figure 14.5 Simplified model of carbon atom
Figure 14.6 Simplified model of a sodium ion Na+
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Conceptual Exercise 14.2 Shirt and sweater attract You pull your sweater and shirt off
together and then pull them apart. You notice that they attract each other. (Fig. 14.7a) Explain the
mechanism of such attraction and suggest an experiment to test your explanation.
Figure 14.7(a) A sweater and shirt are pulled apart
Sketch and Translate As the shirt and the sweater are being separated, they rub against each other.
Due to this rubbing they might get charged. The charges can explain their interaction.
Simplify and Diagram According to our simplified model of charging through rubbing, some of
the electrons from one garment might transfer to the other garment; for example, the electrons
from the sweater transfer to the shirt, or vice versa. (Fig. 14.7b) If this is indeed what happens,
then a Scotch tape strip pulled off a table should be repelled by one of the garments and attracted
to the other. We can test this explanation by actually performing the experiment. The outcome of
this experiment matches the prediction.
Try It Yourself: You rub a balloon with the sweater. Now how does the balloon interact with the
sweater and with the shirt?
Answer: The balloon is attracted toward the sweater and repelled from the shirt.
Conductors
From the experiments in the Observational Experiment Table 14.3 we found that
electrically charged objects are attracted to uncharged objects and the charged objects are
attracted more strongly to metal objects than they are to nonmetal objects. How can we explain
this using the above model of atomic structure and electron transfer?
Suppose (this is just an idea) that in metals, some electrons are not bound to their
respective atoms. Then they can move freely throughout the metal. We will call them “free
electrons”. If you bring a positively charged object next to a metal rod, the free electrons in the
metal should move closer to the positively charged object leaving the other side of the metal rod
with a deficiency of electrons, and therefore positively charged. (Fig. 14.8a) The negatively
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charged side of the rod is closer to the positively charged object and thus should be attracted
stronger to it than the more distant positively charged side of the rod is repelled by it.
Figure 14.8 Metal rod attracted to charged objects
The same phenomenon could occur if a negatively charged object is brought near an
uncharged metal rod. This time free electrons in the metal rod move away from the negatively
charged object, leaving that end of the rod positively charged. The positive side of the metal rod
is attracted more strongly to the negative object than the more distant negative side of the rod is
repelled. The net effect is the rod is attracted to the negatively charged object (Fig. 14.8b.)
When a positively charged object touches a metal rod, some of the metal’s free electrons
could move onto this object, making the metal slightly positively charged and causing it to be
repelled from the less plus charged object. (Fig. 14.9) A similar scenario works if the object is
negatively charged, only this time the free electrons move from that object onto the metal rod.
This scenario is a possible mechanism that explains the behavior of the metal rod in the
Observational Experiment Table 14.3 and the behavior of the aluminum foil bits near an
electrically charged plastic comb. Let’s test this hypothetical mechanism with a new experiment.
Figure 14.9 A neutral metal ball and a plus-charged object
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Testing Experiment Table 14.4 Are there free electrons in a pop can?
Testing experiment
Place a metal pop can lengthwise on a cardboard
box. Hang a 10-cm long piece of negatively
charged Scotch tape (charged by pulling it from
a tabletop) from a wooden dowel so its bottom
end is near one end of the can. It is attracted to
the can.
Prediction
If negatively charged particles move
freely in the can, some of them will
move away from the negatively
charged plastic rod toward the end
of the can near the negatively
charged tape. The tape should be
repelled from this end of the can.
Outcome
The outcome is
consistent with
the prediction.
Then bring a strongly negatively charged plastic
rod (rubbed with felt) near the other end of the
can without touching.
Conclusion
This experiment supports our hypothesis that there are free electrons inside metals. We can repeat the
experiment bringing a positively charge glass rod (rubbed with silk) near the can. Using the same logic we
predict that the negatively charged tape on the other end should be attracted even more to the can. This is
exactly what happens. (U14.23) These two experiments give us more confidence in our free electron model
of metals.
If we repeat these experiments with other metals—the results are similar. Thus, the
explanation that negative charge can move freely in metals accounts for all of our experiments.
We call metals electric conductors, materials in which some its charged particles can move. For
metals the moving charges are negatively charged free electrons.
Now we can explain why a charged plastic bar is attracted to both sides of a magnet. A
magnet is a piece of metal. Thus, it behaves like the can in Table 14.4 (see Fig. 14.10). If the bar
is charged negatively, free electrons in the magnet move away from the end that is closer to the
bar. This end becomes positively charged and the magnet and the bar are attracted toward each
other. If you move the bar to the opposite end, the effect is the same. This phenomenon actually
has nothing to do with the magnetic properties of the magnet. The same thing happens with an
un-magnetized piece of metal.
Figure 14.10 Magnet attracted to negative bar
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Conceptual Exercise 14.3 Aluminum foil bit and a charged bar You have a plastic bar rubbed
with felt and an aluminum foil bit attached to a string that hangs from a dowel. Predict what will
happen if you slowly bring the bit toward the bar.
Sketch and Translate The situation is sketched in Fig. 14.11a.
Figure 14.11(a) Aluminum foil bit attracted to charged bar, touches it, and is then repelled
Simplify and Diagram The bar is negatively charged; thus according to our simplified model, the
free electrons in the bit should move to the right away from the bar, leaving the left side of the bit
that is closer to the bar positively charged (Fig. 14.11b). As the negatively charged part of the bit
is farther away than the positively charged part, the net force that the bar exerts on the bit is
toward the bar on the left. When the bit touches the bar (Fig. 14.11c), it should pick up some
electrons from the bar, become negatively charged, and is then repelled by the bar. As a result,
the bit should accelerate away from the bar (Fig. 14.11d). When we perform the experiment we
observe that this predicted behavior indeed occurs.
Figure 14.11(b)(c)(d)
Try It Yourself: If you repeat the experiment using a small piece of Styrofoam instead of
aluminum foil bit, you observe that it is also attracted to the bar but then sticks to it without
coming off. Explain.
Answer: We can explain the attraction by the same mechanism that we had for the metal objects;
however, we then encounter a problem with the second part of the experiment. We address this
difficulty shortly.
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We explained the “jumping off” part of the experiments above by using the idea that the
aluminum foil bit can acquire some of the electric charge present on the object to which they
were attracted. Apparently, this is not the case with Styrofoam.
We have a mechanism that explains why a neutral metal object is attracted to a charged
object. The free electrons accelerate either towards the charged object (if it is positive) or away
from it (if it is negative). This leaves a region of opposite charge close to the charged object,
which results in attraction. But, we do not yet have a mechanism that explains why a neutral
nonmetal object is attracted to a charged object. Let’s investigate this.
Electrical attraction of uncharged nonmetal materials to charged objects
In the above Try It Yourself example, we observed that the Styrofoam bit behaved
differently from the aluminum bit. A charged bar attracts the Styrofoam bit but the bit does not
swing back after touching the charged bar. It remains touching the charged bar. We can explain
this by assuming that electrons from the foam bit do not transfer to the bar. The reason could be
that plastic, glass and other materials (including Styrofoam) do not have free electrons, and the
positive and negative charges within the material are uniformly distributed throughout (Fig.
14.12a.) All the electrons are tightly bound to their atoms/molecules. If a charged object (let’s
make it positive) is brought nearby, it exerts forces on the negatively charged electrons and on the
positively charged nuclei inside the atoms. Negatively charged electrons are pulled slightly closer
to the charged bar, and the nuclei are pushed slightly away. This causes a slight atomic-scale
separation of the negative and positive charges in the material, a process called polarization.
When the atoms are in this polarized state, they are called electric dipoles. An electric dipole is an
object that is overall electrically neutral but has its negative and positive charges separated.
Polarization leads to a small accumulation of charge on the surface of the object (depicted in Figs.
14.12b and c). Due to this process the side of the Styrofoam foam bit closer to a positive bar
becomes slightly negative and is attracted to the bar; but as it cannot transfer its electrons to the
bar, it does not bounce back.
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Figure 14.12 A charged bar causes polarization of atoms in non-conducting material-simplified model
Our simple model of the polarization of non-conducting materials assumed that before
the material is near a charged object, its atoms or molecules are not dipoles but have the positive
and negative electric charge inside distributed uniformly. This is true for some materials.
However, some molecules (for example, the water molecule) have its electric charges distributed
inside in a way that makes it a dipole even without the polarizing influence of an external charged
object. Objects comprised of molecules with natural dipoles interact with charged objects more
strongly than objects comprised of molecules without natural dipoles.
Let’s summarize what we have learned about the electric properties of materials.
Conductors and insulators Materials can be divided into two groups: conductors and
insulators (also known as dielectrics). Both materials are comprised of oppositely
charged particles with a total charge of zero. In conductors some of the electrically
charged particles can move freely. In dielectrics, the particles cannot move freely from
one place to another; but the positive and negative charges of individual atoms can be
redistributed slightly so that there is slightly more negative charge on one side of the
atom than on the other side, a process called polarization.
This difference between conductors and dielectrics can be tested in the following
experiment. Imagine that you have two touching aluminum cups (conductors) on a table and two
touching plastic cups (dielectrics). You bring a negatively charged rod next to each pair of cups
(Fig. 14.13a for metal cups and b for plastic cups). While holding the rod nearby, you separate the
cups wearing a rubber glove (Fig. 14.13c for metal and d for plastic). What will be the difference
between the metal cups and the plastic cups? If our understanding of the microscopic structure is
correct, then the metal cups will be oppositely charged; the cup closer to the rod will be positively
charged and the cup farther from the rod will be negatively charged. The plastic cups should
remain uncharged. We can use a piece of the scotch tape pulled off a table to check this
prediction. This tape is attracted to one of the metal cups and repelled from the other. The
charged tape is attracted slightly to either plastic cup (either plastic cup becomes slightly
polarized due to the nearby negatively charged tape). Thus we gain confidence in our model.
Etkina/Gentile/Van Heuvelen Process Physics Chapter 14 1/e
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However, a new question arises. It was important that you wore a plastic glove for the
experiment. Why? We will address this question in the next section.
Figure 14.13 Effect of charged rod on metal and plastic cups
Tip! It is important to note that electric charge is defined by the effect it produces. You cannot
see electric charge; it has no known weight, color, length, or width. Yet electric charge and the
interactions between electrically charged objects are among the most important physical
processes of nature. Almost every chemical and biological process in your body and in the
environment around you involves a rearrangement of particles with electric charge.
Review Question 14.2
Summarize the fluid and the particle models of charging by rubbing.
14.3 Electroscope
So far we have learned about electric charges, charging by rubbing, and the differences
between conductors and dielectrics. However, we have not learned any quantitative relationships.
One reason is because we haven’t encountered a device that can detect or measure electric charge.
In this section we will learn about one such instrument, called an electroscope. A sketch of an
electroscope is shown in Fig. 14.14. It consists of a metal rod that protrudes from the outside
through a non-conducting support into a glass-fronted enclosure below. A very lightweight
needle-like metal rod is connected on a pivot near the bottom of the larger rod. The needle rod is
free to rotate around its pivot. Let us explore how this device can detect and measure electric
charge.
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Figure 14.14 An electroscope
Imagine that we bring a negatively charged object near the top of the electroscope rod
without touching it. If our understanding of conductors is correct, then some of the free electrons
in the rod move away from the negatively charged object on top leaving the top with a positive
charge. Excess electrons move to the lower part of the rod and to the needle, which as a result
both become negatively charged. They repel each other, and the needle deflects away from the
rod (Fig. 14.15a). If you then take the negatively charged object away from the top of the
electroscope, the needle returns to its original vertical position.
Now, repeat the experiment only this time, touch the top of the electroscope with the
charged rod. We observe a deflection again; only this time when the rod is removed, the needle
stays deflected. (Fig. 14.15b) We can explain this phenomenon if we assume that touching the
negatively charged object to the top of the electroscope rod transfers electrons onto the top
positively charged end of the electroscope rod. If we then move the negatively charged object
away from the electroscope, the electroscope rod and needle are now negatively charged
throughout and again repel each other. If we repeat the experiments with a positively charged
object, we observe the same outcomes.
Figure 14.15 Charging an electroscope
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If instead of just touching the tip of the electroscope with the charged rod, we carefully
rub the charged rod against the top of the electroscope as if scrubbing the charges from the rod,
we see that the angle of the deflection increases – the angle of deflection is related to the amount
of charge on the electroscope.
We can also use the electroscope to determine the sign of the charge on an object.
Suppose that you rub a comb against the sleeve of your sweater. Is it positively or negatively
charged? To answer this question, you first touch (charge) the electroscope with a known
charge—for example, a plastic bar rubbed with wool. With this bar, the electroscope becomes
negatively charged (Fig. 14.16a). Now, if you touch the electroscope with the comb and the
deflection of the leaves decreases, the comb has a positive charge (Fig. 14.16b). If the deflection
increases, the comb has a negative charge (Fig. 14.16c).
Figure 14.16 Use electroscope to determine sign of charge on comb (b and c)
We can also use the electroscope to decide if a particular material is a conductor or a
dielectric. For this experiment we need two electroscopes. We charge one electroscope by
rubbing it with a plastic bar rubbed with wool, making the electroscope negatively charged (Fig.
14.17a). Now connect the top of the first electroscope by a metal rod to another uncharged
Etkina/Gentile/Van Heuvelen Process Physics Chapter 14 1/e
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electroscope; the needle of the second electroscope deflects (Fig. 14.17b). Some of the electrons
transferred from the charged electroscope through the metal rod to the uncharged electroscope.
The needle of the second electroscope deflects, indicating that the second electroscope is now
negatively charged as well.
Now, repeat the experiment only using a plastic rod, a piece of paper, or a plastic bottle to
connect the electroscopes. The needle of the second electroscope does not deflect. No electrons
were able to transfer through the nonmetal object (Fig. 14.17c). We understand why: a plastic
dielectric material does not allow electrons to move freely through it—it is a non-conductor.
Figure 14.17 Using two electroscopes to decide if a material is a conductor (b) or non-conductor (c).
Is the human body a conductor or a dielectric?
Suppose you have an electroscope that is positively charged; its needle is deflected as in
Fig. 14.18a. If you touch the charged electroscope with your hand, the needle goes back to
vertical, meaning that the electroscope is now uncharged. Why did this happen? One explanation
is that the human body is a conductor similar to a metal and has movable electrically charged
particles inside it. Some of those particles are positively and negatively charged ions, but some
are free electrons that can leave your skin. As your hand approaches the positively charged
electroscope, the free electrons on the surface of your hand and in your arm are attracted toward
the electroscope (Fig. 14.18b). When you touch the electroscope, electrons from the surface of
your hand travel to the electroscope causing it to discharge—to have a net charge of zero. A
similar process happens when you touch a negatively charged electroscope, only this time the
electrons transfer from the electroscope to your skin. In these explanations we assume that the
human body has a large number of free electrons that it can transfer to the electroscope or that it
can receive a large number of electrons from the electroscope. Now you understand why you had
Etkina/Gentile/Van Heuvelen Process Physics Chapter 14 1/e
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to wear a plastic glove in the experiment with the metal and plastic cups in the previous section –
you did not want electrons to transfer to or from your hand.
Figure 14.18 Your body, a conductor, discharges an electroscope
Grounding
Instead of touching a charged electroscope with your hand, you can also discharge it by
touching a wire between the electroscope and a metal pipe that goes into the ground (Fig. 14.19).
This process is called grounding. To explain why a grounded electroscope discharges, we assume
that the earth is a large conductor. If the electroscope is charged positively, negative charges in
the ground are attracted toward the electroscope and travel from the ground to the electroscope
causing it to discharge—its net charge becomes zero. A similar explanation works for a
negatively charged electroscope, except that electrons in the electroscope now travel through the
grounding wire into the ground.
Figure 14.19 Grounding an electroscope
If you charge an electroscope and observe it for a time interval, you observe that its
needle slowly returns to vertical even if nothing touches it. You observe that the electroscope
remains charged longer in dry air than it does in humid air. In fact if you put a steamer next to a
charged electroscope, it discharges almost instantly. To explain this phenomenon we need to
remember that the atmosphere is comprised of many different atoms and molecules. Let’s focus
on the water molecule. It consists of two hydrogen atoms and one oxygen atom. The molecule has
Etkina/Gentile/Van Heuvelen Process Physics Chapter 14 1/e
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a bent shape and behaves as an electric dipole – a molecule where its centers of positive and
negative charges are spatially separated (Fig. 14.20). The hydrogen side of the molecule is
slightly positive while the oxygen side is slightly negative. This natural polarization is why water
is an example of a polar molecule. Polar molecules are great solvents as they easily attach to other
molecules. When these molecules are near a charged electroscope (negative, for example), they
will be attracted to it. The positive side of the molecule contacts the electroscope and removes an
electron from it. The water molecules in the air are responsible for the discharging of the
electroscope. Since the electroscope discharges much more slowly in dry air, it means dry air is a
much better dielectric.
Figure 14.20 Schematic sketch of water molecule
Review Question 14.3
You have a charged electroscope on your desk. How can you decide whether it is positively or
negatively charged?
14.4 Physical quantities and Coulomb’s Law
In the last sections we used the concept of electric charge, a property of the smallest
atomic particles of matter, to explain the electric forces that rubbed objects exert on each other.
We define electric charge as the property of objects that allows them participate in electrostatic
interactions. The greater the electric charge of an object, the greater the intensity of its
interactions with other charged objects. We also know that there are two kinds of charge, which
we call positive and negative. Millikan’s experiments led to the understanding that electric charge
is quantized – you can only change the charge of an object in finite increments, not continuously.
A microscopic model of electric charge indicates that atoms are comprised of positively and
negatively charged particles, each of which has a single unit of electric charge. The total charge
of the atom is zero.
We also know how to measure electric charge qualitatively using an electroscope.
Methods of measuring electric charge quantitatively are based on ideas we have not yet
investigated. We will learn about them in Chapter 16.
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Electric charge conservation
Electric charge in addition to the features mentioned above exhibits a conservation
property similar to that of mass, momentum, and energy. Imagine that we have two identically
constructed electroscopes, one charged and one not. We note the reading of the charged
electroscope (Fig. 14.21a). When we connect it to the uncharged electroscope, the reading on the
first electroscope decreases and the reading on the second increases. The reading on each is half
the original reading (Fig. 14.21b). For example, if the original reading was 12 units, the new
readings will be 6 and 6. If the electroscopes are different, the sum of the readings will still equal
the original reading (Fig. 14.21c). For example, if the original reading is 12, then the final
readings can be 7 and 5 or 4 and 8.
Figure 14.21 Electric charge is conserved
Think back to charging by rubbing. We know that when a plastic rod is rubbed by felt,
the rod becomes negatively charged. The felt becomes positively charged. If we could measure
charges, we would find that the magnitudes of these acquired charges are identical: for example if
the rod acquires a charge of –5 units, the felt acquires a charge of +5 units. If the charge of both
was zero before the rubbing, then after the rubbing it was still zero as –5 + (+5) = 0. These and
many other experiments led physicists to believe that electric charge is a conserved quantity – it
changes in a predictable way in non-isolated system and is constant in isolated systems. In
summary:
Electric charge Electric charge (symbol q or Q ) is a property of objects that participate in
electrostatic interactions. Since electric charge is quantized—you can only change an object’s
charge by increments, not continuously. Electric charge is conserved. The unit for electric
charge is the coulomb (C). The charge of one proton is e # $1.6 % 10"19 C . The charge of one
electron is "e # "1.6 % 10"19 C .
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With this quantitative definition of electric charge, we can now try to develop a quantitative
relationship that describes the electric force that one charged object exerts on another.
Electric force
Recall that when we hold rubbed objects nearer each other, they interact stronger than
when they are farther apart. Thus, the forces that these objects exert on each other must depend
on the distance r separating the charged objects. Also, we learned that if we rub the objects
together more vigorously (leading to greater transfer of electric charge between them), their
interaction is stronger. So the force that charged objects exert on each other must also depend on
the electric charges q1 and q2 on the two interacting objects (we use q as a symbol for the
electric charge of an object.)
Charles Coulomb in 1785 discovered the relationship between these quantities. He did
this experimentally. His apparatus used a torsion balance similar to the apparatus used by Henri
Cavendish who determined the value of the constant in the Law of Universal Gravitation (Chapter
4). Coulomb hung a light glass rod from a thin wire. At the ends of the rod he attached small
metal balls. He also had a third identical metal ball that he could hold near one of the balls on the
ends of the glass rod (Fig 14.22). He then charged the third ball and touched one of the balls
attached to the rod. Since the two balls now had the same electric charge, they repelled. The wire
twisted until the torque exerted by the wire on the glass rod balanced the torque exerted by the
third ball. Coulomb would measure the angle of this twist and used this angle to determine the
electric force exerted by the third ball on the ball attached to the glass rod. He measured the
distance between the repelling balls and varied it to see how the force of repulsion depended on
the distance between them.
Figure 14.22 Coulomb’s apparatus
At Coulomb’s time scientists did not know how to directly measure electric charge. So
Coulomb used identical metal balls and achieved fractions of charge by touching a charged ball to
an identical uncharged ball. So he could have charges of q,
1
1
q, q, and so forth (see Table
2
4
14.5). Below in Table 14.5 you see the data such as Coulomb might have collected indicating
how the repulsive force that one metal ball exerted on another depended on their separation and
Etkina/Gentile/Van Heuvelen Process Physics Chapter 14 1/e
14- 24
on the charges on the balls. Realistically this data would not have followed such clear patterns
and would have included experimental uncertainty.
Table 14.5 Coulomb’s data
Experiment
Charge
1
2
3
4
5
6
7
8
9
10
1 (unit)
½
¼
1
1
½
¼
1
1
1
q1
Charge
1(unit)
1
1
½
¼
½
¼
1
1
1
q2
Distance
r
1(unit)
1
1
1
1
1
1
2
3
4
Force
Fq1 on q2
1 (unit)
½
¼
½
¼
¼
1/16
¼
1/9
1/16
In Coulomb’s experiments the distance between the metal balls was much larger than the
diameter of the balls so it was reasonable for him to model the balls as point-like objects. To find
patterns in the data presented in Table 14.5, let’s focus on one variable at a time. In experiments 1
and 2 the only quantity that changes is the magnitude of the charge q1 . We see that when the
charge is halved, the force exerted by q1 on q2 is also halved. Looking at experiment 3, dividing
the charge by four reduced the force to one-fourth. The force seemed to be directly proportional
to the magnitude of the charge q1 . Experiments 4 and 5 show the same pattern for the magnitude
of the charge q2 .
If a quantity is directly proportional to two independent quantities, it has to be
proportional to their product. Thus, we infer from the table that Fq1 on q2 & q1q2 . We can now
check this expression using experiments 6 and 7 – in fact reducing both charges by 2 leads to a
decrease of the force by 1/4; reducing each charge by 4 makes the force 1/16 the value.
To find a relation between the distance between the objects and the force that they exert
on each other, we repeat the same procedure for experiments 1 and 8-10. We see that doubling
the distance decreased the force by ¼; and tripling the distance reduced the force by 1/9. It
appears that the force is inversely proportional to the separation between q1 and q2 squared
( Fq1 on q2 &
1
). If so, then if we quadruple the distance, the force should decrease by 1/16. The
r2
results of experiment 10 are consistent with that.
We can now combine these patterns into one mathematical expression that shows how
the force that one electrically charged point-like object exerts on another depends on the
magnitudes of their charges and on the distance between them. The relation was named after
Coulomb and is written as follows:
Etkina/Gentile/Van Heuvelen Process Physics Chapter 14 1/e
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Coulomb’s law The magnitude of the electric force that point-like object 1 with electric charge
q1 exerts on point-like object 2 with electric charge q2 when they are separated by a distance r
is given by the expression:
Fq1 on q2 # k
q1q2
.
r2
(14.1)
k is a proportionality constant that depends on the system of units. In the SI system,
k = 9.0 % 109
N • m2
. When using the above to determine the magnitude of the electric
C2
force, do not include the signs of the electric charges q1 and q2 .
Notice that Coulomb worked with very small charged objects in rarified air; thus the law
is only valid for these conditions. We do not know yet how to determine the electric force that
extended objects exert on each other, or how to handle situations where the objects are immersed
in any substance other than air.
Mathematically, the expression for electric force that two charged objects exert on each
q1q2
is analogous to the expression for the gravitational force that any two
r2
mm
objects with mass exert on each other Fm1 on m2 # G 1 2 2 . The difference is that the gravitational
r
other Fq1 on q2 # k
force is always an attractive force, whereas the electric force can be an attractive or a repulsive
force. The proportionality constants also have very different values.
Magnitude of electric force compared to gravitational force
N • m2
; in other words, if we could take
C2
two objects that have 1.0-C charges and separate them by a distance of 1.0 meter , they would
The symbol k in Coulomb’s law has units of
repel each other with a force of magnitude 9.0 % 109 N . However, 1 C is a huge charge and it is
impossible to put this much charge on each of two object separated by 1 m . Realistically we can
take two charged objects with charges of 10-8 C (characteristic of the charge transferred by
rubbing) and separated them by 1 m ; the electric force that each of them exerts on the other
equals to 9.0 %10-7 N . That’s a very weak force since the amount of charge that objects acquire
through rubbing is extremely small.
Let’s compare the electric force to the gravitational force exerted by the proton on the
electron in a hydrogen atom. A proton has charge +1.6 % 10-19 C and mass 1.7 % 10 –27 kg . An
electron has a charge of "1.6 % 10-19 C and mass 9.1%10 –31 kg . They are separated by about
10-10 m .
Etkina/Gentile/Van Heuvelen Process Physics Chapter 14 1/e
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Electrical force:
Fq1 on q2
"19
q1q2
C) 2
9
2
2 (1.6 % 10
# k 2 # (9.0 % 10 N!m / C )
# 2.3 % 10"8 N .
2
"10
r
(10 m)
Gravitational force:
Fm1 on m2 # G
"27
m1m2
kg)(9.1% 10"31 kg)
"11
2
2 (1.7 x 10
#
(6.67
%
10
N
!
m
/kg
)
# 1.0 % 10"47 N .
2
"10
r2
(10 m)
If we divide the electric force by the gravitational force, the result is about 2 % 1039 . The electric
force that the proton exerts on the electron is about 2 % 1039 times greater than the gravitational
force that the proton exerts on the electron! What about the gravitational force exerted by Earth
on the electron?
FE on m2 # G
24
mE m2
kg)(9.1% 10"31 kg)
"11
2
2 (6.0 x 10
#
(6.67
%
10
N
!
m
/kg
)
# 8.9 % 10"30 N .
2
2
6
r
(6.4 x 10 m)
This gravitational force is 17 orders of magnitude larger, but still 22 orders of magnitude weaker
than the electric force exerted by the proton on the electron. That is why physicists can
confidently ignore gravitational forces when dealing with atomic size particles.
We just used Coulomb’s law and the law of universal gravitation for a system of two
objects that are about 1/109 the separation of the systems used to construct these ideas in the first
place. It would be very special if either of those ideas were still relevant. Amazingly, Coulomb’s
law still does apply, though we will find in later chapters that Newton’s 2nd law does not.
Physicists don’t know if the law of universal gravitation still works the same way at the size scale
of atoms. Because the gravitational interaction between atomic scale particles is so weak, it is not
yet possible to make measurements of it. A deviation from the 1 r 2 distance dependence in the
law of universal gravitation at small length scales would be a profound discovery. It turns out that
the exponent in the denominator (2 in this case) is related to the number of spatial dimensions of
our universe. So, for example, if at short distance scales the exponent changed from 2 to 3, it
could mean there exists a hidden extra dimension for which only the gravitational interaction is
sensitive.
Tip! Since Coulomb’s law contains the product of the two charges involved, the electrical force
that an electron exerts on the proton in a hydrogen atom is exactly the same in magnitude as the
electrical force that the proton exerts on the electron. In other words, Coulomb’s law is consistent
with Newton’s third law.
Now, let’s return to the application of Coulomb’s law to some everyday size objects.
Conceptual Exercise 14.4 Interactions of charged balloons Two equal mass small balloons A
and B have electric charges $ q (on A) and $3q (on B). (a) Compare the magnitude of the
Etkina/Gentile/Van Heuvelen Process Physics Chapter 14 1/e
14- 27
electrical force that the balloon with a smaller electric charge exerts on the balloon with the larger
charge ( FA on B ) to the force that the larger charged balloon exerts on the smaller charged balloon
( FB on A ). Justify your answer. (b) If the balloons hang from equal length strings with their tops
attached together, will one string hang at a greater angle from the vertical than the other? Justify
your answer.
Sketch and Translate A labeled sketch of the situation is shown in Figure 14.23a. We choose each
balloon as a separate system of interest. Balloon A interacts with Earth, the string, and balloon B.
Balloon B interacts with Earth, the string, and balloon A.
Figure 14.23(a) Equal mass balloons charged differently
Simplify and Diagram Model the balloons as point-like objects. Force diagrams are shown for
each balloon in Fig. 14.23b. (a) The electric forces that the balloons exert on each other have the
same magnitude according to Newton’s third law and also according to Coulomb’s law
( FA on B # kqA qB / rAB 2 # kqB qA / rAB 2 # FB on A ). (b) Since the balloons have the same mass,
Earth exerts the same force on each balloon. Thus the forces exerted by the strings should also be
the same. Therefore the angles of the strings relative to the vertical should also be the same.
Figure 14.23(b)
Try It Yourself This time balloon A has twice the mass of the balloon B. The charges on the
balloons are the same as in the exercise. Compare the angles that the strings holding the balloons
make with the vertical.
Etkina/Gentile/Van Heuvelen Process Physics Chapter 14 1/e
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Answer The string for the balloon B makes an angle with the vertical whose cosine is twice the
cosine for the angle that the string for the balloon A makes with the vertical.
Quantitative Exercise 14.5 Coulomb’s law and proportional reasoning You have three
identical metal spheres A, B and C each on an insulating stand (Fig. 14.24a). You charge sphere
A with charge q . Then you touch sphere B to A and separate them by distance d (Fig. 14.24b).
(a) Write an expression for the electric force that the spheres exert on each other. What
assumptions did you make? (b) Suppose you now take sphere C and touch it to the sphere A (Fig.
14.24c) and then separate spheres A and B at a distance
d
(Fig. 14.24d). Now write an
2
expression for the magnitude of the force that spheres A and B exert on each other. Assume that
when a sphere with a charge q touches an uncharged sphere of the same size, each of them ends
up with a charge of
q
. The size of the spheres is small compared with distance between them, so
2
you can model them as point-like objects.
Figure 14.24 Reasoning with charge and Coulomb’s Law
Represent Mathematically (a) After sphere A touches B, the charge on each sphere is
After C touches A, the charge on A halves again so that A now has charge
between A and B is
q
. (b)
2
q
. The distance
4
d
. We use Coulomb’s law to write an expression for the force that each
2
charge exerts on the other for each part of the problem.
Solve and Evaluate (a) The magnitude of the electric force that A and B exert on each other is:
FA on B # FB on A # k
(b)
(q / 2)(q / 2)
q2 / 4
q2
#
k
#
k
.
d2
d2
4d 2
q d
Now, the force that A and B exert on each other is:
4 2
Etkina/Gentile/Van Heuvelen Process Physics Chapter 14 1/e
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FA on B* # FB* on A
(q / 4)(q / 2)
q2 / 8
q2
#k
#k 2
#k 2 .
(d / 2) 2
2d
d /4
Although one of the charges got smaller, the decrease in the distance between A and B more than
compensated for this, and the force became twice as great as in the first case.
Try It Yourself: Repeat the previous problem only start with charge q on sphere A and no charge
on spheres B and C. Touch B to A. Then touch C to B. Separate spheres B and C by distance
d
.
2
Write an expression for the force that B exerts on C.
Answer: FA on B # FB on A # k
(q / 4) 2
q2
.
#
k
(d / 2) 2
4d 2
Example 14.6 Net electric force The metal spheres on insulating stands from the previous
example have the following electric charges: qA = +2.0 % 10-9 C , qB = +2.0 % 10-9 C , and
qC = – 4.0 % 10-9 C . The spheres are placed at the corners of an equilateral triangle whose sides
have length d # 1.0 m 1.0 m with C at the top of the triangle. What is the magnitude of the total
electric force that spheres A and B exert on C.
Sketch and Translate A labeled sketch of the situation is shown in Fig. 14.25a. All three angles
inside the triangle are 600 .
Figure 14.25(a) Finding net electric force
Simplify and Diagram Assume that the distance between the spheres is much bigger than their
radii so that they can be reasonably modeled as point-like objects. Choose sphere C as the system
of interest and draw a force diagram for it. We are only interested in the magnitude of the net
electric force exerted on C. Thus, we include in the diagram only the electric forces exerted on C
!
!
by A and by B (Fig. 14.25b). We see that the vector sum of FA on C and FB on C points straight
down.
Etkina/Gentile/Van Heuvelen Process Physics Chapter 14 1/e
14- 30
Figure 14.25(b)
Represent Mathematically Each of the spheres A and B exerts a force on sphere C of magnitude:
FA on C # FB on C # k
q A qC
.
d2
where d is the length of a side of the triangle. Each force vector points along the line connecting
the respective spheres. We see in Fig. 14.25b that the x-components of the two force vectors
cancel each other. Thus, the resultant force has only a y-component. Using trigonometry the net ycomponent force that A and B exerts on C is:
FNet on C y # – FA on C sin(600 ) – FB on C sin(600 )
' FNet on C y # –2 FA on C sin(600 ).
Use the expression for the magnitude of the force FA on C to write the expression for the total
force:
FNet on C y # –2 FA on C sin(600 ) # –2k
q A qC
sin(600 ) .
2
d
Solve and Evaluate Inserting the appropriate values gives
FNet on C y # –2k
q A qC
sin(600 )
2
d
(2.0 % 10"9 C)(4.0 %10"9 C)
# –2(9.0 x 10 N ( m / C )
sin 600 # –6.2 % 10"8 N.
2
1.0 m
9
2
2
The force has magnitude 6.2 %10"8 N and points straight down. This is a very small force. If the
spheres have relatively ordinary masses (a few hundred grams) and are mounted on stands on
which the surface exerts a regular friction force, we would probably not observe any effects of
these electrical interactions.
Try It Yourself: Draw an equilateral triangle with two corners on the bottom and the third corner
above the middle of the base. Determine a set of charges at the corners of the triangle so that the
net force on the top charge is horizontally to the left.
Answer: The bottom left could be " q , the bottom right $ q , and the top center charge could be
any positive charge.
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Review Question 14.4
You have two identical small metal foil bits hanging on threads. You charge one of them, and
then touch them together. Then you place them at a distance d from each other and measure the
force with which they repel each other. Qualitatively, how will the force change if you keep the
distance the same but after charging touch one of the bits with your hand? What assumptions did
you make in coming up with your answer?
14.5 Electric Potential Energy
In the previous section we learned to describe the interactions of electric charges using
forces. We learned in Chapter 6 that in addition to forces we could use the ideas of work and
energy to describe interactions. In this section we will investigate how to extend work-energy
ideas so that they incorporate the electric interaction.
When you have a system of two objects exerting gravitational forces on each other (an
apple-Earth system, for example), the system possesses gravitational potential energy. Does a
system of electrically interacting objects posses some sort of electric energy? We know that
Coulomb’s law for electric force is mathematically very similar to the law of universal
gravitation. Thus, it is reasonable to suggest that a system of electrically charged objects also
possess a new type of energy – electric energy. Let’s investigate this idea using the following
hypothetical situation.
Electric potential energy: qualitative
Consider the ‘electric cannon’ in Fig. 14.26a. A positively charged cannonball is held
near another fixed positive charge object in the barrel of the cannon. The cannonball, the other
charged object, and Earth are the system. When you release the trigger, the positively charged
cannonball is released. Since the cannonball is repelled from the fixed like charged object, the
cannonball accelerates out the end of the barrel (Fig. 14.26b). As this happens both the system’s
kinetic ( K ) and gravitational potential ( U g ) energies increase. Where did this extra energy come
from? If we assume that the system is isolated, then some other type of energy has to decrease to
compensate for the increase in these two energies. This must be the electric energy whose
existence we suggested above. We will use the symbol U q to represent it. Since this energy
seems to be analogous to the gravitational potential energy of two objects with mass, we’ll call it
electric potential energy.
Figure 14.26(a)(b) Electric potential energy of two like-charged objects
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The situation shown in part (a) is similar to that of an object pressed against a compressed
spring. In this case though, it’s two like charged objects held in close proximity to each other. We
can say that the system comprised of the two charged objects has electrical potential energy. It
seems that the electrical potential energy of like charged objects increases as they are forced
nearer each other, and that it decreases as they are allowed to move apart. In the case of the two
like-charged objects plus Earth system, as the cannonball moves farther from the fixed charged
object, the electric potential energy of the system decreases. Some of the electric potential energy
of the system has been transformed into kinetic and gravitational potential energy.
We can represent this process using a bar chart (Fig. 14.26c). The initial state is when the
cannonball is near the other fixed charged object; the final state is when the cannonball is moving
at the end of the barrel. Remember that for gravitational potential energy we needed to specify a
zero-level so that we knew how high to draw the energy bars. For this process we will assign the
zero level for gravitational potential energy to be at the initial position of the cannonball. We
must also make a zero-level assignment for electric potential energy. We will declare that when
two charged objects are so far apart that they essentially do not interact, they have zero electric
potential energy. (We could choose the zero of the gravitational potential energy to be at infinity
too, but traditionally for the processes that involve small changes in elevation the zero is chosen
at the lowest position of the system above ground to make the simplified expression for the
gravitational potential energy U g # mgh easy to use.)
Figure 14.26(c)
Let us consider a different hypothetical device – an electric nutcracker (Fig. 14.27a). The
system consists of the two oppositely charged blocks, one of which can slide without friction.
When the negatively charged block is released and moves nearer the nut, the kinetic energy of the
system increases (Fig. 14.27b); thus some other type of energy must decrease. Electric potential
energy is clearly involved in this process. This process is analogous to a stretched spring. Instead
of the coils of a spring being stretched, the two oppositely charged objects have been pulled apart.
This means that the electrical potential energy of unlike charges increases as they move farther
apart (like stretching a spring). Thus, the electrical potential energy of the system of oppositely
Etkina/Gentile/Van Heuvelen Process Physics Chapter 14 1/e
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charged objects decreases as they move closer together. For the nutcracker, the decreasing
electrical potential energy of the system is transformed into kinetic energy.
Figure 14.27(a)(b) Electric potential energy of unlike charged objects
Remember our zero-level convention for electric potential energy: when two charged
objects are very far apart their electric potential energy is zero. But, we just found that for two
unlike charges, their electric potential energy decreases as they get closer. This means that the
electric potential of energy of unlike charges is always less than zero, which means it is negative.
The bar chart in Fig. 14.27c represent this process. The initial electric potential energy of the
system is negative. As the objects come closer together, the kinetic energy of the system increases
and the electric potential energy becomes even more negative.
Figure 14.27(c)
Changing the magnitude of the electric changes and their separation should affect the
magnitude of the electric potential energy of the system. If the cannonballs started closer together,
or if the blocks started farther apart, the system should have more electric potential energy. If the
cannonballs or blocks had greater electric charge, again the system should have more electric
potential energy.
Electric potential energy: quantitative
To derive an expression for the change in electrical potential energy of two electric
charges whose separation changes, we use the generalized work-energy principle:
W # )U system
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where W is the work done on the system by objects in the environment, and )U system is the
resulting change in the system’s energy. The system’s energy can change in many ways: the
objects can change motion, elevation above Earth (if Earth is a part of the system), the stretch of
the springs if any are involved, the temperature or structure change, or (if electrically charged
objects are involved) the values of their electric charges or the separation between them might
change. We can represent each of these possible changes using the generalized work-energy
principle:
W # )U system # )K $ )U g $ )U s $ )U int $ )U q
where all energy changes are familiar except )U q , which represents the change in the electrical
potential energy of the system.
To derive an expression for )U q we choose a hypothetical system that has the following
feature: when work is done on this system, only its electrical potential energy changes. The
system shown in Fig. 14.28a consists only of the point-like charged objects 1 and 2 with like
charges q1 and q2 , initially separated by a distance ri . The like charges repel each other exerting
an electrical force on each other whose magnitude is given by Eq. (14.1):
Fq1 on q2 # k
q1q2
.
r2
Imagine that a person wishes to prevent object 2 from flying away from 1 by exerting a
force on 2 toward 1 of the same magnitude that the charged object 1 exerts on 2. (Fig.14.28a). If
the person pushes object 2 just a tiny bit harder, object 2 can be displaced a small distance
)r1 toward object 1. (Fig. 14.28b)
Figure 14.28(a)(b) Work done in pushing two like-charged objects closer together
We use Eq. (6.1) to calculate the work )W1 done by the pushing force during this small
displacement (the force and displacement are in the same direction; thus the cosine of the angle
between them is 1):
)W1 # FP on 2 )r1 #
kq1q2
)r1
ri 2
This calculation is only approximate though because the force needed to push object 2 changes
(increases) as its separation from object 1 changes (decreases). If )r1 is small, the equation is a
good approximation for the work done during that small displacement of object 2. If we wish to
Etkina/Gentile/Van Heuvelen Process Physics Chapter 14 1/e
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push object 2 farther however, the person must exert a larger force since it is now closer to object
1. The work done during the next small displacement )r2 , shown in Fig. 14.28c, is
approximately:
)W2 # FP on 2 )r2 #
kq1q2
)r2
(ri " )r1 ) 2
What happens to the amount of work if we continue to push the charged objects closer together?
For each step closer, the amount of work needed to move 2 closer to 1 increases.
Figure 14.28(c)(d)
The total work done in moving the charge from an initial separation ri to a final
separation rf (Fig. 14.28d) equals the sum of the work for each small step. This procedure is
similar to what is needed to arrive at Eq. (6.12) in Chapter 6 for the gravitational potential energy
of two objects. This type of infinitesimal addition is done easily using calculus. The result of
such a calculation is:
W # )W1 $ )W2 $ )W3 $ ... #
kq1q2 kq1q2
1 1
"
# k q1 q2 ( " ).
rf
ri
rf ri
At each step of the process represented in Fig.14.28, there is no acceleration and therefore no
change in kinetic energy; nor are there changes in gravitational, elastic, or internal energies. Thus
we can reason that the only energy change of the system due to this work is the electric potential
energy change. By substituting the above expression for W into Eq. (6.8), and zeros for all energy
changes except the electrical potential energy change, we find
kq1q2 (
1 1
" ) # 0 $ 0 $ 0 $ )U q $ 0 $ ...
rf ri
Therefore,
)U q # kq1q2 (
1 1
" )
rf ri
' Uq f "Uq i
kq q kq q
# 1 2" 1 2
rf
ri
.
This equation applies to any situation in which two charges in a system move closer to
each other or farther apart. This is the expression for the electric potential energy change.
Evidently the electric potential energy of a system with two charged point-like objects separated
by distance r is:
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Uq #
k q1 q2
.
r
This equation is also consistent with the convention that we established earlier: the electric
potential energy of interaction of two charged objects is zero when the distance between them is
infinite.
The change in electrical potential energy )U q when two point-like objects with
charges q1 and q2 are moved from an initial separation ri to a final separation rf is:
)U q # U q f " U q i # kq1q2 (
1 1
" ).
rf ri
(14.2)
Electrical potential energy is measured in units of joules. Equation (14.2) is valid for both
positively and negatively charged objects provided the signs of the charges are included.
Tip! Notice two things: (a) that electric potential energy is proportional to
1
1
and not to 2 (as it
r
r
is in Coulomb’s force law.) and (b) mathematically the expression for the electrical potential
energy is similar to the expression for the gravitational potential energy of a system with two
objects with mass ( U g # "G
m1m2
).
r
Example 14.7 The sign of electric potential energy What is the sign of the electric potential
energy of two oppositely charged objects (with positive charge $ q and negative charge –Q )
separated by distance ri ? Will the energy decrease or increase if you pull the objects farther
apart? Explain.
Sketch and Translate First, what is the sign of the electrical potential energy when the oppositely
charged objects are a distance ri apart—we’ll call it the initial state? A sketch of this initial
situation is shown in Fig. 14.29a. The system is the two charged objects. Then you pull the –Q
charged object so it is farther from the positively charged object—the final situation (Fig.
14.29b). As the direction of the force exerted on the –Q -charged object is the same as the
direction of its displacement, positive work is being done on the system. This means the electric
potential energy of the system must increase as the distance between the charges increases. If we
pulled –Q infinitely far from $ q , the final electric potential energy of the system would be
zero. Thus, the initial electric potential energy must be negative.
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Figure 14.29(a)(c) Electric potential energy of oppositely charged system
Simplify and Diagram The two objects were modeled as point-like particles, and we assumed that
only the electric potential energy energy changed. A bar chart representing the process is shown
in Fig. 14.29c.
Figure 14.29(c)
Represent Mathematically We now use the bar chart and the generalized work-energy principle to
represent the process mathematically:
Ui $ W # U f
'
k ($ q )(– Q)
k ($ q )(–Q)
+W =
.
ri
rf
Solve and Evaluate Notice that if ri is less than rf , the electric potential energy term on the left
side is more negative than that on the right side—consistent with the bar chart. Note that for
unlike charges, the product of q1q2 is negative. This means the electric potential energy is
negative and changes from a larger negative value to a smaller negative value as the charges get
farther apart. In other words, the positive work done on the system results in an increase in the
system’s energy. This outcome is consistent with our conceptual understanding of the process.
Try It Yourself: Determine the change in electric potential energy that occurs if the electron in a
hydrogen atom moves from 0.53 %10-10 m from the proton nucleus to infinitely far from the
nucleus. Both particles have charge of magnitude 1.6 % 10-19 C , the electron negative and the
proton positive.
Answer: 44 % 10-18 J .
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Tip! If it is difficult for you to work with negative numbers, help yourself by using integer
examples. For example, suppose that +5 units of positive work is done on a system of two
oppositely charged objects causing their energy to become zero units. This can be represented by
an equation x $ 5 # 0 , where x is the initial electric potential energy of the system. Solving for
x , we find it to be x # –5 , a negative number.
Graphing the electric potential energy versus distance
So far we represented electric potential energy with bar charts and mathematically. A
graphical representation is another useful representation that can help conceptualize this abstract
physical quantity. Consider first how the electric potential energy changes as the separation r of
two like-charged objects varies. Because of the
1
dependence in the electric potential energy, the
r
energy approaches positive infinity when the separation approaches zero (see Fig. 14.30a). The
energy becomes less positive and approaches zero as the like charges move far
apart. (-10) (+3) (-10) + (+3) = +7 (-10)
For a system with two oppositely-signed charged object, the energy approaches negative
infinity when their separation approaches zero. The energy becomes less negative (increases) and
approaches zero as they move far apart (Fig. 14.30b). If the system is comprised of more than two
interacting charged objects, their electric potential energy becomes more complicated. We’ll
handle that case now.
Figure 14.30 Electric potential energy versus distance separating charges
Tip! Let’s use integers again: if you start with a value of (–10) and add a positive number to it
(+3) , the result will be (–10) + (+3) = –7 , a smaller magnitude negative number than the initial
(-10) .
Electric potential energy of multiple charge systems
Suppose a system has several interacting electrically charged objects. How do we
determine the electric potential energy of such a system? Each pair of charged objects has an
associated electric potential energy, and the total electric potential energy of the system is the sum
Etkina/Gentile/Van Heuvelen Process Physics Chapter 14 1/e
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of the energies of all pairs. Suppose that the system has three charged objects labeled 1, 2 and 3.
Then, the total electric potential energy is:
U q # U q 12 $ U q 13 $ U q 23
' Uq #
(14.3)
kq1q2 kq1q3 kq2 q3
$
$
r12
r13
r23
where r12 is the distance between object 1 and 2, and similarly for the other terms. The signs of all
charges must be included.
Review Question 14.5
How can we reduce the electric potential energy of a system of two electrically charged objects?
14.6 Generators produce charge separation
Most of the time, macroscopic objects are electrically neutral, meaning they have
approximately the same number of protons as electrons. We learned in the previous section that if
we managed to move some of the electrons from one object to another, we would increase the
electric potential energy of those two objects (unlike charges are separated). How can this be
done in practice? Actually, it’s pretty easy. Just rub two objects that are made of different
materials together. We learned earlier in the chapter that this makes one of the objects positive
and the other negative. However, the magnitude of the resulting charge of the objects is quite low.
To charge objects more significantly we need to use specialized equipment. In this section you
learn about two such devices, a Van de Graff generator and a Wimshurst generator.
Van de Graaff Generator
A Van de Graff generator is a machine that looks like a plastic cylinder with a moving
belt inside and a hollow metal dome at the top. You push an “on” button and the belt starts
moving. When the belt is moving you can hear a cracking sound and see little sparks around the
dome. Some generators have an attachment - a metal ball on an insulating stand. The sparks
between this ball and the dome can be large (see the sparks jumping all over from the dome in
Fig. 14.31). This means that the dome of the generator is electrically charged. How does it get
charged?
Figure 14.31 A charged Van de Graff dome discharging to surrounding pieces of metal
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A Van de Graff generator belt moves on two rollers (one on the top inside of the metal
dome and the other on bottom - see Fig. 14.32). Brush assemblies rub against the belt as it moves
over the bottom roller, removing electrons from the belt. The positively charged region of the belt
then moves upward into the dome where electrons transfer from the dome to the belt (leaving the
dome positively charged.) These electrons are then carried by the belt back down to the bottom
roller where they are removed and transferred to the ground using what is known as a grounding
wire. This charge separation (positive on the dome and negative on the ground) greatly increases
the electric potential energy of the dome/ground system. The charge accumulating on the dome is
significant—on the order of 10 –5 C . We discuss other important properties of Van de Graff
generators in the next chapter.
Figure 14.32 Operation of a Van de Graff generator
Conceptual Exercise 14.8 Why does hair stand on end? The woman shown in Fig. 14.33a has
her hands on the dome of a Van de Graff generator and her feet on an electrically insulated
footstool. Why is her hair standing on end?
Figure 14.33(a) Hair standing on end when woman touches charged Van de Graff dome
Sketch and Translate Standing hair can be due to the repulsion of individual strands from each
other and from the body. For this to happen, the hair and body need to have the same electric
charge. A strand of hair is then repelled by her body, and by the other strands of hair. How does
this happen?
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Simplify and Diagram If we assume that the human body is a fairly good conductor of electric
charge, the woman’s body is acting like an extension of the dome. Electrons flow from her to the
dome, leaving her and each strand of hair positively charged. This makes them repel from her
body and from the other strands--they stand on end. A microscopic model of the process is
sketched in Fig. 14.33b .
Figure 14.33(b)
Try It Yourself: You place a stack of aluminum pie plates on top of an uncharged Van de Graff
generator and then turn on the generator. Predict what happens to the plates and explain your
prediction.
Answer: If we assume that electrons transfer from the plates to the positively charged dome, then
the now positively charged plates should repel one another and the dome. We should observe
them them fly up off the dome one at a time. They do.
Wimshurst Generator
The Wimshurst generator (Fig. 14.34) is another device that can produce large charge
separations. The device, invented in the 1880s, has two plastic discs that are rotated in opposite
directions by a hand-crank and drive belt. Charge separation in the Wimshurst machine occurs by
a process called induction. If one of the disks has any electric charge (a tiny amount due to some
random processes during handling), then when the disks rotate with respect to each other, the
charged disk polarizes the uncharged disc producing opposite sign charge on its two sides. Both
discs have metal plates that are connected by collecting combs to large containers, where excess
charge is stored. The containers are called capacitors, studied the in the next chapter.
Figure 14.34 A Wimshurst generator
These storage chambers are connected to metal rods with small metal spheres on their
ends—called spark gap electrodes. As the discs revolve, electric charge is accumulated on the
Etkina/Gentile/Van Heuvelen Process Physics Chapter 14 1/e
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storage chambers, which results in one of the electrodes becoming positively charged and the
other negatively charged. When the two electrodes are brought near one another, sparks as long
as two or three inches can be generated.
Conceptual Exercise 14.9 Detecting signs of charge on Wimshurst electrodes How do we
know that the Wimshurst electrodes are oppositely charged?
Sketch and Translate To determine the signs of charges on the electrodes, we need a third object
that we know is electrically charged. One easy way to achieve this is to stick a piece of Scotch
tape on a table and quickly pull it off (we used this phenomenon to help construct our qualitative
understanding of the electric interaction early in this chapter.) Attach this piece of charged tape
to a wooden dowel, charge the Wimshurst with its metal electrodes far apart so there are no
sparks, and bring the tape first near one Wimshurst electrode and then near the other. Keep your
body far from the Wimshurst generator—the reason for attaching the tape to the dowel.
Simplify and Diagram The tape should be attracted toward one Wimshurst electrode (the
positively charged electrode as the tape is negatively charged) and repelled from the other (the
negatively charged electrode). (Fig. 14.35) That is in fact what happens.
Try It Yourself: What happens to the tape if it is placed exactly in the middle between the charged
electrodes?
Figure 14.35 Detecting type of charge on Wimshurst electrodes
Answer: If the electrodes have equal magnitude electric charge, the magnitudes of the electric
forces that they exert on the tape are equal. But, the positive electrode attracts the tape and the
negative electrode repels it—the forces point in the same direction. The tape bends strongly
toward the positive electrode.
Review Question 14.6 How could you determine the sign of the electric charge on the dome of a
Van de Graff generator?
14.7 Skills: analyzing processes involving electric force and electric
energy
While solving problems with electrically charged objects, you can use the same problem
solving strategy as we used in the dynamics and energy chapters. The key here is to incorporate
Etkina/Gentile/Van Heuvelen Process Physics Chapter 14 1/e
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the new electric interactions into Newton’s 2nd law and the generalized work-energy principle,
including the expressions for calculating the electric force and the electric potential energy. The
strategy we use is described and illustrated in the following example.
Example 14.10 Where to put the charge? Two electrically charged objects q1 # $1.0 %10"9 C
and q2 # $2.0 % 10"9 C are separated by 1.0 m. Where should you place a third electrically
charged object so that the net electric force exerted on it by the first two objects is zero?
Sketch and Translate
! Sketch the process described
in the problem statement. Label
the physical quantities.
! Choose an appropriate system
of interest.
Simplify and Diagram
! Decide whether you can
consider the charged objects to be
point-like. Decide what other
interactions you will consider and
what interactions you can ignore.
! Construct a force diagram for
the system by drawing force
arrows for each object in the
environment that exerts a force
on the system. If you are using
the work-energy principle,
construct an energy bar chart.
The situation is pictured below.
We do not yet know where to place the third object, but it could b
in any of the three regions shown in the figure. If we choose the
object of unknown electric charge q as the system, then it interac
with objects 1 and 2.
Consider all objects to be smaller than the distances between them
Force diagrams for the charged object q in the three possible
regions are shown at the right. The net electric force exerted on q
the vector sum of the two forces due to q1 and q2 .
(I) If q is positioned to the left of both charged objects, the forces
due to q1 and q2 both point left and cannot add to zero.
(III) If q is positioned to the right of the two charges, both forces
point right and cannot add to zero
(II) If q is between the charges, the forces point in opposite
directions and can add to zero. If q is closer to the smaller charg
the reduced distance will compensate for the smaller q2 .
Represent Mathematically
! Use the force diagram to help
apply the component form of
Both forces are horizontal (their y components are zero.) Thus, w
use only the horizontal x component of Newton’s 2nd law. With th
positive x direction toward the right and taking the signs of the
components into account, we get:
Etkina/Gentile/Van Heuvelen Process Physics Chapter 14 1/e
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Newton’s second law to the
process (or use the energy bar
chart to apply the generalized
work-energy equation).
! If necessary, use kinematics
equations to describe the motion
of the object.
Fq1 on q $ ( " F q2 on q ) # 0
Using the force law for electric force and the fact that the total
distance between the charges is 1.0 m , the fact that q2 # 2q1 , a
labeling the distance between q1 and q as r , the above equation
becomes:
kq1 q
r
#
2
k 2q1 q
[(1.0 m) " r ]
2
.
Divide both sides of the equation by kqq1 to get:
1
2
#
.
2
[(1.0 m) " r ]
r
Take the square root of both sides of the equation:
2
1
r
Solve and Evaluate
! Rearrange the equation and
solve for the unknown quantity.
Check if your answer is
reasonable with respect to sign,
unit, and magnitude. Also make
sure the equation applies for
limiting cases, such as objects
having very small or very large
charge.
#
2
(1.0 m) " r
,
Rearranging the last equation, we get:
2 r # (1.0 m) " r or r # 0.41 m.
The location where a net electric force of zero will be exerted on
is a distance 0.41 m from charged object 1. The result looks
reasonable, as the unknown charge q should be closer to the
smaller magnitude charge q1 than to the larger charge q2 .
The distance that we found does not depend on the magnitu
or sign of the charge q . The net force exerted on q by two other
objects will be zero independent of the sign of q .
One limiting case is if one of the charges q1 or q2 is zero. I
q1 # 0 , then only q2 could exert a force on q and the net force
could NOT be zero. The force equation becomes
0 $ ( " F q2 on q ) # 0 which has no solutions as the right and left
sides are never equal to each other. So this limiting case is
consistent with our mathematics.
Try It Yourself: Suppose we placed q slightly closer to q1 and slightly farther from q2 than the
position calculated in this example (just a tiny bit from the desired position). What happens to q ?
Answer: The magnitude of the force that q1 exerts on q increases slightly and the magnitude of
the force that q2 exerts on q decreases slightly. The net force exerted on q is no longer zero—it
will accelerate away from the desired position. The desired position is said to be an unstable
equilibrium point. If q is just to either side of that point, the forces exerted on it by the other
objects tend to cause it to accelerate away from that point.
The next example involves an energy approach.
Etkina/Gentile/Van Heuvelen Process Physics Chapter 14 1/e
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Example 14.11 Radon decay in lungs Suppose that a radon atom in the air in a house is inhaled
into the lungs. While in the lungs the nucleus of the radon atom undergoes radioactive decay
emitting a so-called * (alpha) particle (comprised of two protons and two neutrons). During this
process, the radon nucleus turns into a polonium nucleus (we will study this process in detail in
Chapter 28). The polonium nucleus has charge +84e and mass 3.5 %10 –25 kg . The * particle
has charge +2e and mass 6.7 % 10 –27 kg . Suppose the two particles are initially separated by
1.0 % 10 –15 m and are at rest. How fast is the * particle moving when it is very far from the
polonium nucleus (very far for such a process is 1 mm but we can assume infinity). * *
Sketch and Translate Draw a sketch showing the initial and final states of the system (Fig.
14.36a). Choose the radon nucleus as the system (which after its radioactive decay results in the
positive polonium nucleus and positive * particle as the system.) To solve the problem we need
to determine how fast the * particle is moving when far from the polonium nucleus. Note that
the like-charged objects repel.
Figure 14.36(a) Radioactive decay of radon nucleus
Simplify and Diagram Model the nuclei (including the alpha particle) as point-like objects and
neglect the gravitational attraction between them. Assume that they are at rest at the start of the
process and that we can ignore the final kinetic energy of the massive polonium nucleus. A workenergy bar chart for the process is shown in Fig. 14.36b. In the initial state, the system has electric
potential energy; in the final state, the * particle has kinetic energy and since the particles are far
apart they have zero electric potential energy.
Figure 14.36(b)
Represent Mathematically Each non-zero bar in the bar chart turns into a non-zero term in the
generalized work-energy equation.
Etkina/Gentile/Van Heuvelen Process Physics Chapter 14 1/e
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k ($84e)($2e) 1
# m* v f 2
ri
2
Rearranging the above, we get an expression for the final speed of the * particle:
vf #
2k (84e)(2e)
m* ri
Solve and Evaluate We can now substitute known information into the above equation to make an
estimate of the final speed of the alpha particle.
vf #
=
2k (84e)(2e)
m* ri
2(9.0 %109 N•m 2 /C2 )(84.0 % 1.6 % 10 –19 C)(2.0 % 1.6 x 10 –19 C)
# 1.0 % 108
–27
–15
(6.7 x 10 kg)(1.0 %10 m)
N•m/kg
That is an unusual unit for velocity—or is it. Let us check if it is a correct unit. Note that
1 N = 1 kg • m/s 2 . Thus,
N • m/kg = (kg • m/s 2 ) • m/kg = m 2 /s 2 .
When we take the square root we do get the standard unit for speed, m/s. Thus, we estimate that
the final alpha particle speed will be about 108 m/s . When radon nuclei undergo radioactive
decay in our lungs, these high-energy alpha particles can do damage to lung tissue—a reason for
keeping radon concentration low in homes.
Try It Yourself: Remember that we assumed in the above example that the massive polonium
nucleus remained at rest. Use the given information and the idea of momentum conservation to
determine how fast the polonium nucleus was actually moving following the decay process in
terms of the speed of the alpha particle. Assume that neither particle interacts with the
environment. How would their kinetic energies compare?
Answer: vpolonium = (2/84) valpha . The kinetic energies, which depend on the speed squared, would
be: K polonium = 0.00057 K alpha . We were justified in ignoring the polonium final kinetic energy in
this example.
Example 14.12 Electric cannonball launcher The barrel of the cannonball launcher shown in
Fig. 14.37 is tilted at a 37 0 angle. The 1.0-kg ball travels 2.0 m up the barrel before leaving it.
We want the ball 1.0 kg to leave the barrel traveling at a speed of 50 m/s . The ball starts
0.50 m from the fixed charged block at the bottom end of the cannon. What product of electric
charge q1q2 (on the block and on the cannonball) is needed to make the ball leave the barrel at
this speed? Assume that g # 10 m/s 2 .
Etkina/Gentile/Van Heuvelen Process Physics Chapter 14 1/e
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Figure 14.37(a) Electric cannon
Sketch and Translate Choose the bock, the ball, the gun, and Earth as the system. The initial state
is just before firing; the final is just as the ball leaves the barrel.
Simplify and Diagram Model the charged block and the ball as point-like objects. From Fig.
14.37 this looks like a questionable way to model these parts of the system. However, we do not
have detailed information about their sizes and even if we did we have not developed a way to
incorporate those details into our analysis. Next draw an energy bar chart for the process (Fig.
14.37b.) Note that the system initially has only electrical potential energy due to the ball and
block’s electric charges. By the choice of the origin of the vertical axis, the system has zero
gravitational potential energy in its initial state. In the final state, the ball is moving and has
kinetic energy. It is also at a higher elevation and thus the system now has gravitational potential
energy. Finally, the two charged objects are farther apart but still have some electric potential
energy since they are not very far apart. The relative magnitudes of the final energies are
unknown, so we just make choices that are reasonable and consistent with the process.
Figure 14.37(b)
Represent Mathematically Now turn each bar into a term in the generalized work-energy
equation:
U qi # K f $ U gf $ U qf
or
+ kq1q2 ,
+ kq1q2 ,
1
2
0
.. .
. # m2 v f $ m2 glbarrel s in 37 $ -2
/ ri 0
/ rf 0
The subscript ‘1’ refers to the charged block, and ‘2’ refers to the charged ball,
where ri and rf are the distances between the cannonball and the fixed charge before and after the
shot, respectively.
Etkina/Gentile/Van Heuvelen Process Physics Chapter 14 1/e
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Solve and Evaluate Next move all terms containing q1q2 to the left side of the equation and solve
for it:
1
m2 v2 2 $ m2 glbarrel in 370
q1q2 # 2
.
+1 1 ,
k- " .
- ri rf .
/
0
Now plug in the values for the known quantities paying special attention to the units:
1
(1.0 kg)(2500 m 2 /s 2 ) $ (1.0 kg)(10 m/s2 )(2.0 m)0.60
2
q1q2 #
# 8.8 %10"8 C2
1
1
(9.0 % 109 Nm 2 / C2 )(
"
)
0.50 m 2.5 m
In the most simple case, the charge divides evenly ( ! 3.0 %10"4 C ) between the charged objects.
Is this a reasonable value for the electric charge of an object of these sizes? We will learn how to
decide this in the next chapter.
Try It Yourself: Suppose the 3.0 %10-4 C cannonball and block were initially separated by
0.30 m instead of 0.50 m and that the ball travels 2.2 m before leaving the barrel. Now, how
fast would it leave the barrel? Assume again the g = 10 N/kg .
Answer: 68 m/s .
Review Question 14.7
Why is it important to choose a system of interest when analyzing a process using Newton’s
second law or the work-energy principle?
14.8 Putting it all together
We apply the concepts introduced in this chapter and some of those from earlier chapters
to analyze three processes – the creation of sparks, energy transformations in stars, and a
photocopy machine.
Charge Separation Produces Sparks in Air
Sparks are produced between the dome of a charged Van de Graff generator and a
grounded metal ball held 10 cm from the side of the dome. Similarly, sparks are produced when
the charged metal electrodes of a Wimshurst machine are brought within several centimeters of
each other. We will learn later in the book that a spark or a flash of light happens when electrons,
removed from atoms by some process, return to the positive ions thus releasing the extra energy
they had while free to produce a flash of light. What causes the electron to be freed? Obviously,
the opposite charges on the electrodes or on the dome-grounded ball of the Van de Graff must
cause the sparks. But what is actually happening to the air that results in a flash and a sharp
cracking sound?
Etkina/Gentile/Van Heuvelen Process Physics Chapter 14 1/e
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In some books you find the following explanations:
“air molecules are torn apart” or “electrons are ripped loose from their molecules” or
“air breaks down, ionizing and supporting sparks”.
We can summarize these statements as the following explanation:
Explanation 1: A Van de Graff or Wimshurst generator causes a spark because
the charge on these devices causes electrons to be pulled off of atoms and
molecules in the air.
Let’s evaluate this explanation.
Example 14.13 Rough estimate of the force that Van de Graff’s dome exerts on electron in
atom We learn in the next chapter that the charge on a Van de Graff dome used in physics
lectures is Q 1 +10 –5 C or less. We choose values for other quantities needed in the estimate
so that we can compare the force that the charge on the Van de Graff dome exerts on an electron
in an atom and the force that the nucleus of that atom exerts on the electron. If the explanation 1
is correct, then the magnitude of these two electric forces should be comparable.
Sketch and Translate Consider a hydrogen atom as one electron circling around a single proton
nucleus. The atom is near a charged Van de Graff dome (Fig. 14.38a). The radius of the dome is
about 10 cm = 0.1 m . The radius of a hydrogen atom is about 10 –10 m . The magnitude of the
electron charge and the proton charge is e # 1.6 %10 –19 C and the dome charge is about
Q 1 10 –5 C .
Figure 14.38(a) H atom near Van de Graff
Simplify and Diagram Choose the electron as the object of interest. The proton nucleus exerts on
"!
the electron a force F Proton on Electron and the charge on the Van de Graff dome exerts on the
"!
electron a force F VandeGraff on Electron . A force diagram for the electron is shown in Fig. 14.38b.
Figure 14.38(b)
Etkina/Gentile/Van Heuvelen Process Physics Chapter 14 1/e
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Represent Mathematically The magnitude of each force is calculated using Coulomb’s law:
Fq1 on q2 #
kq1q2
.
r2
Solve and Evaluate Using our approximate numbers, we can now estimate the magnitudes of
these two forces:
ke 2
r2
C)2 (1010 )(3 x 10–38 )
1
N 1 10 –8 N.
–20
(10 )
FProton on Electron =
(9 x 109 N•m 2 /C2 )(1.6 x 10 –19
=
(10 –10 m) 2
kQe
R2
C) (1010 )(10 –24 )
1
N 1 10 –12 N.
–2
(10 )
FVandeGraff on Electron =
#
(9 x 109 N•m 2 /C2 )(10 –5 C)(1.6 x 10 –19
(0.1 m)2
The Van de Graff charge exerts far too small a force to pull the electron off the hydrogen atom.
We have to reject this explanation.
Try It Yourself: The average distance of an electron from the proton nucleus in a hydrogen atom
is actually 0.53 %10 –10 m . The radius of a Van de Graff dome used in lectures is closer to
0.15 m . How do these two adjustments affect the calculations.
Answer: The FProton on Electron force will increase and the FVandeGraff on Electron force will decrease,
disproving the explanation even more convincingly.
Is there an alternative explanation for the sparking in the air that occurs due to charge on
the Van de Graff dome? Cosmic rays, high-energy particles from space, continually rain down on
Earth’s atmosphere. These cosmic rays have sufficient energy when they hit atoms and molecules
in the air to ionize some of the atoms and molecules so that free electrons are produced leaving
behind positively charged ions. Free electrons and ions are also produced in the air by highenergy particles produced during the radioactive decay of nuclei in our environment (such as the
decay of radon that we learned about earlier in the chapter.) The positive charge on the dome of a
Van de Graff generator attracts the free electrons causing them to accelerate toward the dome.
While moving toward the dome, these newly freed electrons collide with other atoms and
molecules in the air causing them to ionize as well and then produce even more free electrons.
When electrons kicked out from the atoms return to the atoms, we see a spark .
To characterize how far the electrons or ions travel between individual collisions,
physicists use a physical quantity called the mean free path—the average distance between
collisions. (Fig. 14.39) If electrons accelerate too little during their mean free path trips, their
energy when colliding with an atom or molecule in the air is too small to knock electrons out of
the atoms and molecules.
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Figure 14.39 Electron mean free path
If the acceleration is large, the electron gains enough energy during their mean free path trips
between collisions to ionize air molecules. Let’s summarize this explanation.
Explanation 2: Free electrons in the air liberated from their atoms/molecules by cosmic rays and
radioactive particles gain enough energy when accelerated by their attraction to the positive
charge on the dome to cause an electron to be knocked out of an air molecule during the next
collision.
Example 14.14 Can explanation 2 account for sparking caused by a Van de Graff? We will
learn in Chapter 27 that the energy needed to remove an electron from a hydrogen atom is about
2 %10"18 J . The mean free path of an electron in air is about 10 –6 m . The radius of the dome is
0.15 m . Could a free electron in the air gain enough kinetic energy to ionize an air molecule as
the electron travels that short distance toward the charged dome of a Van de Graff generator with
about +10 –5 C of charge on it?
Picture and Translate The situation is sketched in Fig. 14.40a.
Figure 14.40(a) Free Electron gains kinetic energy during mean free path
Simplify and Diagram We assume the electron is near the surface of the dome and travels one
mean free path distance. The system will be the electron and the dome. We model both as pointlike objects. This is reasonable for the electron, and reasonable for the dome as well, but only
because it’s spherical in shape. We will discuss that in more detail in the next chapter. An energy
Etkina/Gentile/Van Heuvelen Process Physics Chapter 14 1/e
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bar chart representing the process is shown in Fig. 14.40b. The electron-dome system starts with
negative electric potential energy and zero kinetic energy. The electron accelerates toward the
dome and its kinetic energy increases. Being closer to the dome, its negative electric potential
energy has decreased.
Figure 14.40(b)
Represent Mathematically Use the bar chart to help represent the process mathematically with the
generalized work energy equation:
k (Q)(– e)
1
k (Q)(– e)
= mvf 2 +
ri
2
rf
Solve and Evaluate Our question concerns the kinetic energy
1
mv f 2 of the electron—is it as
2
large as or does it exceed the energy needed to ionize an air molecule ! U 1 2 % 10 –18 J?
+1
+r –r ,
1
k (Q)(– e) k (Q)(– e)
1,
mvf 2 =
–
= –kQe - – . = –kQe - i f .
rf
ri
ri 0
2
/ rf
/ ri rf 0
+ 10 –6 m ,
= –(9 x 109 N•m 2 /C2 )(10–5 C)(1.6 x 10 –19 C) 2 .
/ (0.15 m) 0
+ 10 –6 ,
1 (1010 )(10–5 )(10–19 ) - –2 . 1 10–18 N•m = 10–18 J.
/ 10 0
This is comparable to the energy needed to ionize a hydrogen atom. This result supports
explanation 2. This does not prove that this explanation is correct but does not disprove it either.
Note that some electrons will travel farther than the mean free path distance (since it is an
average) and will have even more kinetic energy than we estimated, making the explanation even
more plausible.
We can test the explanation further by repeating the experiment in a special chamber with
a lower air density. When the air density is reduced by half, the mean free path distance (rf " ri )
doubles. Thus, the electron’s kinetic energy should be twice as great when it collides. If
explanation 2 is correct, the charge of the dome should be one half of the charge in the regular
density air for the dome to start discharging. Physicists indeed find that the Van de Graff charge
Q needed to cause a spark can be reduced by half when the air density is reduced by half. This
adds support for the second explanation.
Etkina/Gentile/Van Heuvelen Process Physics Chapter 14 1/e
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Try It Yourself: Why are we ignoring the positive ions in the air and considering only the
acceleration of the free electrons due to the positive charge on the dome?
Answer: The positive ions are much more massive than the electrons and have mean free path
lengths that are much smaller than those of the electrons. Hence, they do not acquire enough
kinetic energy between collisions to knock electrons off atoms during their next collision.
Stellar Energy
In Section 9.10 we solved a problem concerning the energy output of the Sun. Recall that
each second the Sun radiates 4 %1026 J of energy. According to geological estimates, the age of
Earth is 4.5 billion years. The Sun must be at least this old. Thus the energy needed for it to shine
for 4.5 billion years (about 5 billion years or 5 x 109 years using one significant figure) must be
about:
E 1 (4 % 1026 J/s)(5 % 109 years)
# (4 % 1026 J/s)(5 %109 years)(3 % 107 s/year) = 6 % 1043 J .
Our Chapter 12 estimate of the internal thermal energy of the particles in the Sun indicated that
the Sun’s thermal energy could provide the energy needed for the Sun’s rate of emission for only
4 % 107 years—about 0.01 times the energy already emitted. Thus we know that thermal energy
could not be the basic source of the Sun’s radiative energy.
Scientists are now convinced that a process called nuclear fusion is the mechanism
behind what keeps the Sun and other stars shining for billions of years. During nuclear fusion,
small nuclei, like the nuclei of hydrogen atoms, combine together to form larger nuclei, such as
helium (when elementary particles such as protons are at the distances that are smaller than
10 –15 m they start interacting via an additional mechanism which produces attraction as opposed
to repulsion; this attraction is much stronger than Coulomb’s repulsion, therefore the particles
join together). A product nucleus actually has slightly less mass than the sum of the masses of the
smaller nuclei that combine to form it. We will learn in Chapters 25 and 28 that mass can be
transformed into energy (and vice versa). This reduction of mass and consequently the release of
the energy is a process that can fuel the Sun.
For two light nuclei to fuse they must come within about 10 –15 m m of each other. As
the nuclei have positive electric charge and repel each other, they do not come that close under
normal conditions. However, if the particles are moving really fast, they can get this close. This
can occur at very high temperature, millions of degrees K. Now, the question is: “Is the
temperature inside the Sun high enough to make fusion possible?”
Example 14.15 How hot should the Sun be? Estimate the temperature of a gas consisting of
hydrogen nuclei so that during the nuclei collisions they come within 10 –15 m of each other,
close enough to fuse together to form a new nucleus.
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Sketch and Translate A sketch of the situation is shown in Fig. 14.41a. Assume that the two
hydrogen nuclei are moving very fast when very far from each other and momentarily stop when
they are 10-15 m apart. (On a microscopic scale, “very far” can be a centimeter or so.) We choose
both hydrogen nuclei as the system.
Figure 14.41(a) Protons getting closes enough to fuse
Simplify and Diagram Model the hydrogen nuclei as point-like objects and assume that the nuclei
only interact with each other and not with the environment. Neglect their gravitational interaction
since it is so much weaker than their electric interaction. An energy bar chart for the process is
shown in Fig. 14.41b. In the initial state, the system only has kinetic energy; in the final state, it
only has electric potential energy.
Figure 14.41(b)
Represent Mathematically Use the bar chart and the generalized work-energy principle to
construct a mathematical description of the process:
1
1
q2
mvi 2 $ mvi 2 # k
2
2
rf
Recall from Chapter 9 that the average kinetic energy of a particle at a temperature T is
1
3
mv 2 # k * T where k * is the Boltzmann constant (we put a star next to it so you do not
2
2
confuse it with the Coulomb’s law constant that has the same letter). Assume that vi # v ,
meaning that the initial speed of each hydrogen nuclei is the average speed of all the hydrogen
nuclei in the Sun’s core. Combining these ideas with the previous equation, we get:
3
q2
2( k * T ) # 3k * T # k
2
rf
Solve and Evaluate Dividing both sides of the equation by 3k * , we have:
Etkina/Gentile/Van Heuvelen Process Physics Chapter 14 1/e
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T #k
q2
3k * rf
Before inserting the values for the known quantities and constants, let us check that the units are
correct. Recall that the units for the Boltzmann constant are J/K and J = N•m.
+ N!m 2 , C2
N!m 2
N!m 2
#
K
=
K=K
.
2
J•m
(N!m)m
/ C 0 (J/K)m
The units are correct. Now, let’s finish the estimation. We will work without units, since we
already checked them. Let’s do this without using a calculator (for example, 1.62 1 3 ):
T #k
"38
q2
(1.6 x 10"19 )2
10 3 x 10
# (9 % 109 )
1
1 1010 K .
10
"23
"15
"38
3k * rf
3(1.4 x 10 )(10 )
4 x 10
The order of magnitude is 1010 or 10 billion degrees K.
Try It Yourself: How is the above temperature calculation changed if the mass of the initial
particles to be fused are twice that in the previous example.
Answer: The temperature depends only on the desired final separation and the charge of the
particles. Doubling the mass has no effect on the temperature required for fusion to occur.
The actual temperature in the core of the Sun is about 107 K . Does it mean that no
nuclei can come close enough for fusion? Remember though that 107 K is the temperature at the
Sun’s center corresponding to the average speed of the nuclei. Many nuclei are moving faster
than this (and many slower.) Even at this relatively low temperature, there are a tiny fraction of
“fast” nuclei that do have the kinetic energy required to fuse with other nuclei. If the temperature
of the Sun were 1010 K, the fusion would occur quickly like an explosion, and our Sun would
have had a very short life. If the temperature were even lower than 107 K, fusion would occur too
slowly to provide the needed energy for life on Earth. It’s a remarkable set of conditions needed
for life to evolve as it has—a Sun hot enough for fusion but cool enough so fusion is sustained for
billions of years.
A photocopy machine
At the beginning of the chapter we considered a photocopy machine. What is the
connection between that machine and the topic of this chapter? It turns out, the mechanism that
gets the toner in the copy machine to stick to the paper is very similar to how a balloon rubbed on
a wool sweater allows it to attract tiny bits of paper off a tabletop. A copy machine has a drum
that becomes charged in a pattern that is a copy of the document being duplicated. The drum is
then coated in toner, which sticks only to places where the drum is charged. The drum then turns
and presses against the copy paper transferring the toner particles to the paper, producing the
desired image. The temperature of the toner is then increased, baking it into the page making it
permanent.
Let’s analyze this process in more detail. The operation of a photocopier is illustrated in
Etkina/Gentile/Van Heuvelen Process Physics Chapter 14 1/e
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Fig. 14.42. The photocopier has a drum that is covered with a special photoconductive material.
This material is a semiconductor that starts in a non-conducting state but is converted to an
electrical conducting material in places where it is radiated by light (you will learn more about
semiconductors in chapter 27.) When we turn on the photocopier, the drum becomes positively
charged on the surface beneath the photoconductive material and negatively charged on the
outside of the photoconductive material. This is accomplished by a component of the copier
called a corona wire or charge roller (Fig. 14.42a). Thus, the photoconductive material has
positive charge on the inside and negative on the outside. The negative charge does not leave the
drum as the photoconductor is in its non-conducting state.
When you want to photocopy a page of text, you place it face down on a glass surface.
The page has white spaces with no text and pictures, and black spaces with the text and pictures.
You press the start button and a strong source of light moves across the page from below the page
but above the drum (Fig. 14.42b). This light reflects off bright regions of the page but does not
reflect off dark regions. The reflected light reaches the photoconductive material and converts it
from a non-conductor to a conductor. The electrons on the surface of the photoconductive
material that has absorbed the light now move inside to the drum leaving the surface neutral in
places where light was reflected from the original page being copied. Drum areas below the text
or figures on the page remain negatively charged. Thus the drum has an electrical image of the
text and figures (Fig. 14.42b).
The next step is covering the drum with the toner, which is positively charged. The toner
sticks to the “electrical image” on the drum (Fig. 14.42c). Then a blank negatively charged paper
wraps around the drum and pulls off the toner (Fig. 14.42d). The drum and the paper are heated
and pressed together to make this transfer more effective and to bake the toner on the surface of
the paper. Finally, a rubber material wipes the drum clean of remaining toner and illuminates it
with light to remove all remaining charge. In summary, much of the copy machine operation
depends on electric interactions between charged objects, plus some other phenomena (for
example, light affecting the photoconductive material) that we will study in later chapters.
Figure 14.42(a) Photocopy process
Etkina/Gentile/Van Heuvelen Process Physics Chapter 14 1/e
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Summary: Role of electricity in everyday life
In this chapter we learned about the interactions of electrically charged objects. These
interactions are easily observed in everyday life – your hair sticks out when combed with a plastic
comb; you see sparks when taking a sweater out of a drier; you make a photocopy or use a laser
printer. But the most important role of electric interactions is at the microscopic scale. Atoms
interact electrically to form molecules, which through further electric interactions form liquids
and solids. Electric interactions are also the mechanism behind the friction force, the buoyant
force, the elastic force, the normal force, because they all involve the microscopic contact
between atoms and molecules. However, as objects tend to have an equal number of positive and
negative charges, objects tend to be electrically neutral. Earth, the Sun, other stars, are all
electrically neutral and so do not interact via electric forces. Their primary interactions are
gravitational. Thus, the electric interaction dominates at microscopic scales and the scale of our
everyday lives on earth (though gravitational interactions start to become important at that scale
as well). Gravitational interactions dominate on the scale of solar systems, galaxies, and the
universe as a whole.
Review Question 14.8
How do we know that a Van de Graff or Wimshurst generator does not pull electrons off of atoms
and molecules in the air to produce sparks?
Summary
Word Description
Pictorial or Physical
Representation
Mathematical
Representation
Electric charge is a physical quantity that
characterizes a property of objects that causes
them to participate in electrostatic interactions.
a) Charge comes in two types – positive and
negative.
b) Like charged objects repel and unlike
charged objects attract.
c) The smallest charge is the charge of an
electron – e # –1.6 % 10 C .
d) Electric charge is a conserved quantity
-19
Conductors are materials in which electrically
charged particles can move freely.
Dielectrics are materials in which electrically
charged particles cannot move freely.
Etkina/Gentile/Van Heuvelen Process Physics Chapter 14 1/e
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However, the electric charges in the atoms and
molecules in the material can rearrange
slightly (called polarization) allowing them to
participate in electric interactions.
Coulomb’s force law is used to determine the
magnitude of the electric force that point-like
objects with charges q1 and q2 exert on each
other when separated by a distance r.
Fq1 on q 2 = Fq 2 on q
#
k q1 q2
r2
Force diagram Choose a system object and
draw arrows representing the forces that
external objects exert on the system object.
The directions of the arrows determine the
signs of the force components when using
Newton’s second law.
The electric potential energy Uq of point-like
objects with charges q1 and q2 separated by a
distance r is positive for like charges and
negative for unlike charges (include the signs
of the charges when using this equation.)
Etkina/Gentile/Van Heuvelen Process Physics Chapter 14 1/e
Uq =
14- 59
kq1q2
r