chapter 20 electrostatics and coulomb’s · pdf fileelectric charge has again been...

20.2 Separation of Electric Charge by RubbingTo help understand some of the characteristics of electric charge, let us

perform the following experiment. We attach a pith ball, a spongy material likecork, by a piece of string to the ceiling, and it hangs freely, as shown in figure20.1(a). We move an amber rod toward the pith ball, eventually touching it [figure20.1(b)]. We observe that nothing happens to the pith ball during the touchingprocess. Now we rub the amber rod with a piece of fur and then cause the rod totouch the pith ball, figure 20.1(c). This time the pith ball flies away, as seen infigure 20.1(d).

How can this behavior be explained? When the rod was not rubbed and wastouched to the pith ball, nothing happened to the pith ball. But when the rod wasrubbed with fur and touched the pith ball, the pith ball was repelled from the rod.We must conclude that something happened to the rod during the rubbing process.Let us assume that rubbing the rod caused an amount of something, which we willcall electric charge, to be deposited on the rod.

The type of charge deposited on the rod could be arbitrarily called an “A”charge, or a “B” charge, or a “blue” charge, or a “red” charge, but we will, againarbitrarily, call it negative charge. When the negatively charged rod touches thepith ball, some of this negative charge on the rod flows onto the pith ball, leavingthe pith ball negatively charged. The pith ball then flies away from the rod,indicating that like charges repel each other. It is because like charges repel eachother that the original charges on the rod flowed to the pith ball.

Chapter 20 Electrostatics and Coulomb’s Law“I wish to give an account of some investigationswhich have led to the conclusion that the carriers ofnegative electricity are bodies, which I have calledcorpuscles, having a mass very much smaller thanthat of the atom of any known element, and are of thesame character from whatever source the negativeelectricity may be derived.” J.J.Thomson

20.1 Introduction In chapter 1 we attempted to explain the world in terms of the fewest number

of fundamental quantities. Up to now only length, mass, and time have concernedus. Now we will consider the fourth fundamental quantity, electric charge. Ourknowledge of electric charge is not new. The earliest known experiments onelectrostatics were performed by Thales of Miletus (ca. 624-547 B.C.) around 600B.C., when he found that amber, when rubbed with fur, attracted light objects.Today, we say that the amber possesses an electrical charge. (The word electric isderived from the Greek word elektron, meaning amber.) The study of electric chargesat rest under the action of electric forces is called electrostatics.

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The thought that should now occur to us is, “Does rubbing any materialgenerate electric charge?” Let us repeat the experiment, only this time with a glassrod. First, we touch the unrubbed glass rod to a new pith ball. As before nothinghappens to the pith ball. Then we rub the glass rod with silk and touch it to the pithball, figure 20.1(e). We see that the pith ball flies away, as in figure 20.1(f). Electriccharge has again been generated by rubbing the rod. But is this the same chargethat was generated when the amber rod was rubbed? At this point we cannotanswer the question. To be completely general, let us assume that it is a differentelectric charge, one that we will arbitrarily call positive charge. The positive chargeappears to flow from the glass rod to the pith ball during contact, leaving the pithball also positively charged. The pith ball then flies away from the glass rod, againindicating that like charges repel each other. Is this really a different charge, thatis, a positive one? The experimental results are the same for both rods. Both pithballs are repelled from the rod.

String

Pith ball

(a)

String

Pith ball

(b)

Amber rod

String

Pith ball

(c)

Amber rodRubbed

(d)Repulsion

(e)

Glass rodrubbed withsilk

(f) Repulsion

(g)Attraction

Figure 20.1 Separation of charge by rubbing.

Chapter 20 Electrostatics and Coulomb’s Law

20-2

Let us now hang two separate pith balls from the ceiling, and charge the oneon the left negatively by touching it with the rubbed amber rod. The pith ball to theright we charge positively by touching it with the glass rod rubbed with silk. If thecharges on both balls are really the same, then the balls should repel each other,just as the balls were repelled from the rods in figures 20.1(d) and 20.1(f). Weobserve, however, that the two pith balls are attracted to each other, as shown infigure 20.1(g). Therefore rubbing the glass rod with silk does indeed produce adifferent charge, a positive one, than that produced by rubbing an amber rod withfur, the negative charge. Also note that the unlike charges attract each other. Thisexperiment can be continued by rubbing different rods with various materials, butonly these two types of charges, negative and positive, are ever found. We concludethat there are two and only two types of electric charges in nature -- negativecharge and positive charge. As a result of these experiments, the fundamentalprinciple of electrostatics can be stated: Like electric charges repel eachother, whereas unlike electric charges attract each other. To give a moremodern description of electrostatics we need to discuss atomic structure.

20.3 Atomic StructureIn an attempt to find simplicity in nature, the Greek philosophers Leucippus

and Democritus suggested in the fifth century B.C. that matter is composed of verysmall particles called atoms. The word atom comes from a Greek word that meansthat which is indivisible. However, it was not until the early nineteenth centurythat John Dalton (1766-1844), an English chemist, proposed that to every knownchemical element there corresponds an atom of matter. Every material in the worldis just some combination of these indivisible atoms. However, in 1897 J. J.Thompson (1856-1940), an English physicist, discovered the electron, a negativelycharged particle having a mass me = 9.1095 × 10−31 kg. Although it had been knownthat there was such a thing as negative electrical charge, it was not known whatthe carrier of that negative charge was. This newly discovered electron, however,was the basic or elementary particle carrying the smallest amount of negativecharge. All other negative charge that occur in electrostatic experiments aremultiples of the electronic charge.

The finding of the electron, however, presented a rather difficult problem.Where did it come from? The only place it could come from is the interior of theindivisible atom, which could not then be indivisible. The indivisible atom musthave some structure. Because the atom is generally neutral, there must be somepositive charge within the atom to neutralize the negative electron. In the early1900’s Ernest Rutherford (1871-1937), a British physicist, bombarded atoms withalpha particles (positively charged particles) and by observing the effects of thecollision, developed the nuclear model of the atom. His model of the atom consistedof a small, dense, positively charged nucleus with negative electrons orbiting aboutit, somewhat in the manner of the planets orbiting about the sun in the solar


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system. Rutherford found this positive particle of the nucleus and named it theproton in 1919. The proton has a positive charge equal in magnitude to the chargeon the electron. The mass of the proton, mp = 1.6726 × 10−27 kg, is about 1836 timesgreater than the electron mass.

In 1920 Rutherford suggested that there is probably another particle withinthe nucleus, a neutral one, to which he gave the name the neutron. The neutronwas discovered some twelve years later in 1932 by the English physicist, JamesChadwick. In terms of these particles, or building blocks, the different atoms areformed. The difference between one chemical element and another is in the numberof protons, and electrons within it. As seen in figure 20.2(a), the chemical elementhydrogen contains a nucleus which consists of one positive proton about whichorbits the lighter, negative electron. Sometimes the electron is to the right and

e−

e−

4p+4n

e−e−

(d) Beryllium

e−

3p+3n e−

(c) Lithium

e−

p+ e−

(a) Hydrogen

2p+2n

e−e−

(b) Helium

Figure 20.2 Atomic structure.

sometimes to the left of the nucleus. Sometimes it is above and sometimes below thenucleus. By symmetry, the electron’s mean position coincides with the position ofthe positive nucleus. Therefore, the atom as a whole acts as though it wereelectrically neutral. The next chemical element helium is formed by the addition ofanother proton to the nucleus, and another electron to the orbit, figure 20.2(b). Twoneutrons are also found in the nucleus of helium. The next chemical element,lithium is formed by the addition of another proton, another electron, and another


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neutron, figure 20.2(c). In this way all of the chemical elements are formed,although in the higher elements there are usually more neutrons than protons.Because each element contains the same number of electrons and protons, eachelement is electrically neutral.

Although this model of the atom is quite useful, it is not completely correct.An electron moving in a circle is an accelerated charge, and it has been found thatwhenever a charge is accelerated, it radiates energy. Therefore the radiatingelectron should lose energy and spiral into the nucleus and the atom should cease toexist. The world, which is made up of atoms, should also cease to exist. Since theworld continues to exist, the above model of the atom can not be completely correct.

Also since like charges repel each other, the protons in the nucleus shouldalso repel each other and the nucleus should blow itself apart. Hence, the wholeworld should blow itself apart. But it does not. Therefore, there must be some otherforce within the nucleus holding the protons together. This new force is called thestrong nuclear force. Because the electron is the smallest unit of charge ever found,the fundamental unit of charge, the Coulomb, named after the French physicistCharles A de Coulomb (1736-1806), is defined in terms of a certain number of theseelectronic charges. That is, 1 Coulomb of charge is equal to 6.242 × 1018 electroniccharges, and the charge on one electron is 1.60219 × 10−19 Coulomb. The unit forelectrical charge, the Coulomb, will be abbreviated as a capital C, in keeping withthe SI convention that units named after a person are symbolized by capital letters.Because electric charge only comes in multiples of the electronic charge, it is said thatelectric charge is quantized. Also, the total net charge of any system is constant, aresult known as the law of conservation of electric charge.

Although no electric charges have ever been found carrying fractionalportions of the electronic charge, the latest hypothesis in elementary particlephysics is that protons and neutrons are made up of more elementary particlescalled quarks. (Murray Gell-Mann, 1964) It is presently proposed that there are sixquarks: the (1) up (u), (2) down (d), (3) strange (s), (4) charm (c), (5) bottom (b), and(6) top (t) quark. The charges on these quarks are fractional as shown in table 20.1.

2/3 t

−1/3 b 2/3 c

−1/3 s

−1/3d

2/3 u

CHARGE (Fraction ofElectron Charge)

QUARK

Table 20.1 Quarks and their charges.


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A proton is assumed to be made up of two “up” quarks and one “down” quark,as shown in figure 20.3. The neutron is assumed to be made up of one “up” quark

u

u

d

charge2/3 +2/3 - 1/3 = 1

Figure 20.3 The quark configuration of a proton.

and two “down” quarks as shown in figure 20.4. Of course, individual quarks havenot yet been found, and indeed most theories of particle physics predict that anisolated single quark cannot exist. There is, however, strong indirect evidence fromscattering experiments that quarks do indeed exist. The quark model has alsopredicted the existence of other elementary particles which have been found,indicating that the quark hypothesis is on very good experimental ground. Ofcourse, if the existence of quarks are definitely confirmed, the next question thatwould then have to be asked is, “What are quarks made of?”

u

dd

charge2/3 - 1/3 - 1/3 = 0

Figure 20.4 The quark configuration of a neutron. In terms of the atomic concept of electric charge, the generation of charge by

rubbing is explained as follows. When an amber rod is rubbed with fur, the rubbingcauses electrons to be stripped from the fur atoms and deposited on the amber rod,making the amber rod negative. The fur, which now has a deficiency of electrons,becomes positive. Hence, charge has not been created, it has merely been separated.When the pith ball is touched by the negative amber rod, electrons flow from the rodto the pith ball, leaving the pith ball negatively charged.

When the glass rod is rubbed with silk, electrons are stripped from the atomsin the glass rod and are deposited on the silk cloth making the silk cloth negative.The glass rod, having a deficiency of electrons, becomes positive. When the glass rod


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touches the neutral pith ball, electrons from the originally neutral pith ball flowonto the glass rod, neutralizing some of its positive charge. The deficiency ofelectrons on the pith ball causes the pith ball to become positive. This correctlyexplains the previous experiment where, at that time, we incorrectly assumed thatpositive charges flowed from the glass rod to the pith ball. In all these cases ofcharging bodies by rubbing, the charge that moves from one body to another isalways the negative electronic charge. Recall that the electron is 1/1836 lighter thanthe proton and hence easier to move. Also the heavier protons are tightly bound intothe nucleus and are not as easy to detach from the atom as the weakly boundelectrons.

We use the letter q to represent the electric charge on a body and, in general,the net charge q on a body consisting of both protons and electrons is given by

q = (Np − Ne)e

where Np is equal to the number of protons, Ne is equal to the number of electrons,and e is equal to the basic unit of charge, namely 1.60 × 10−19 C.

20.4 Measurement of Electric Charge The ElectroscopeOne of the traditional means for measuring electric charge is the electroscope, asseen in figure 20.5(a). An electroscope is a device made of two strips of thin gold

(a) (b)

Figure 20.5 An electroscope.

leaf or aluminum foil fastened to the end of a metallic rod. The metallic rod ishoused in an insulated enclosure. The top of the rod is connected to a metallic ballat the top of the electroscope. If a negatively charged rod is touched to the metal ballon the top of the electroscope, electrons move from the negative rod to the ball of theelectroscope, down through the metallic connecting rod to the aluminum leaves,some moving to the right leaf and some to the left leaf. The negative charge on thealuminum leaves repel each other, thereby separating the leaves, as shown in figure


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20.5(b). The amount of separation of the leaves becomes a measure of the quantityof charge. If instead of the negative rod, a positive rod were touched to theelectroscope, the leaves would separate in the same way, only this time positivecharge would be on each leaf. Electrons on the electroscope would flow to thepositive rod leaving a deficiency of negative charges on the leaves of theelectroscope. Hence, the leaves would have positive charges on them.

Conductors and InsulatorsTwo electroscopes are set up, as shown in figure 20.6. When the negatively chargedrod is touched to the metal bulb between the two electroscopes, the electroscope to

Negativelycharged rod

Copper wireRubberband

Metalball

Glassrod

Figure 20.6 A conductor and an insulator.

the right shows a charge but the one to the left does not. The reason for thisphenomenon can be explained by the connecting material between the metal bulband the electroscope. The copper wire is called a conductor because it allows thecharge deposited on the bulb to be conducted to the electroscope. The rubber band iscalled an insulator because it does not permit the charge to reach the electroscope.(The rubber band insulates the electroscope from the charged bulb.)

In general, most substances fall into either one of these categories. Materialsthat permit the free flow of electric charge through them are called conductors.Materials that do not permit the free flow of electric charge through themare called insulators or dielectrics. Most metals are good conductors of electriccharge, whereas most nonmetals are insulators. (There are a few materials calledsemiconductors, whose characteristics lie between those of conductors and those ofinsulators.)

An interesting characteristic of all conductors is that whenever an electriccharge is placed on a conducting body, that charge will redistribute itself until all ofthe charge is on the outside of the body. For example, if electric charges are placedon a solid metallic sphere, the charges exert forces of repulsion on one another and


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the charges try to move as far apart as they can. The greatest separation they canachieve is when they are on the outside of the sphere.

Charging by InductionMost bodies are electrically neutral, that is, they contain the same number ofpositive charges as negative charges. If a negatively charged rod is brought into thevicinity of an uncharged conducting sphere the negative rod repels the electrons onthe sphere, as shown in figures 20.7(a) and 20.7(b). The sphere still contains equal

(a) Negative rod far away from metallic sphere

(b) Negative rod brought closeto, but not touching, metal sphere

(c) Grounding the metal sphere. (d) The sphere is charged positively.

Figure 20.7 Charging by induction.

numbers of positive and negative charges but they are now redistributed so that theright side of the sphere has a negative distribution of charge whereas the left sidehas a positive one. (If the rod is removed, the charge would redistribute itself to itsinitial neutral configuration.) If you were to touch your finger to the sphere youwould provide a path for the electrons to escape to the ground, as shown in figure20.7(c).

When you remove your finger from the sphere and remove the negative rod,the sphere is charged positively. A positive charge has been induced on the spherewithout having touched it with a positive charge.

20.5 Coulomb’s LawAs has just been seen, electric charges exert forces on each other. But what

are the magnitudes of these forces? Charles Augustin de Coulomb (1736-1806)invented a torsion balance in 1777 in which a quantity of force could be measuredby the amount of twist it produced in a thin wire. In 1785 he used this torsionbalance to measure the force between electrical charges. If very small sphericalcharges, q1 and q2, called point charges, are separated by a distance r between their


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centers, figure 20.8, then Coulomb found that the force between the charges couldbe stated as: The force between the point charges q1 and q2 is directly proportional

rq

1q

2

Figure 20.8 Coulomb’s Law.

to the product of their charges and inversely proportional to the square of thedistance separating them. The direction of this force lies along the line separatingthe charges. This result is known as Coulomb’s law. Coulomb’s law can be statedmathematically as

(20.1)F = kq1q2

r2

where k is a constant depending upon the units employed and on the medium inwhich the charges are located. For a vacuum

k = 8.9876 × 109 N m2 /C2

For air the value of k is so close to the value of k for a vacuum that the samevalue will be used for both. To simplify the solution of problems in this text thevalue of k used will be rounded off to the value

k = 9.00 × 109 N m2/C2

To simplify more advanced theories of electromagnetism, Coulomb’s law is alsowritten in the form

(20.2)F = 14✜✒o

q1q2

r2

where (20.3)k = 1

4✜✒o

and εo = 8.854 × 10−12 C2/N m2 and is called the permittivity of free space. If thecharges are placed in a medium other than air or vacuum, then there will be adifferent value for the permittivity ε for that medium and hence a different value ofk. In this text only the simple form, equation 20.1, of Coulomb’s law will be used.

If the charges are much larger than point charges, Coulomb’s law can still beused if the distance separating the charges is quite large compared to the size of theelectric charge. Under these circumstances the charges approximate point charges.Let us consider some examples of the use of Coulomb’s law.

Notice the similarity between the form of Coulomb’s law of electrostatics andNewton’s law of universal gravitation:


20-10

(6.37)F = Gm1m2r2

Let us compare the electric and gravitational forces between an electron and aproton in a hydrogen atom. The mass of the proton is mp = 1.67 × 10−27 kg, whereasthe mass of the electron is me = 9.11 × 10−31 kg. The radius of the lowest energyorbit for the electron is r = 5.29 × 10−11 m. The charge on the electron and proton is1.60 × 10−19 C. The gravitational force between the electron and the proton is foundfrom Newton’s law of universal gravitation, equation 6.37, as

Fg =Gmpme

r2

=(6.67 % 10−11N m2/kg2)(1.67 % 10−27 kg)(9.11 % 10−31 kg)

(5.29 % 10−11 m)2

= 3.63 × 10−47 N

The electric force between the electron and the proton is found fromCoulomb’s law of electrostatics, equation 20.1, as

Fe =kqpqe

r2

=(9.00 % 109N m2/C2)(1.60 % 10−19 C)2

(5.29 % 10−11 m)2

= 8.23 × 10−8 N

Although both forces seem quite small, let us compare the relative magnitudeof these forces by taking the ratio of the electric force to the gravitational force, thatis,

FeFg

= 8.23 % 10−8N3.63 % 10−47 N = 2.27 % 1039

orFe = (2.27 × 1039)Fg

That is, for the electron-proton system discussed here, the electric force is 1039 timesgreater than the gravitational force. Because magnitudes are sometimes hard tovisualize in scientific notation for the beginning student, we can also write thisnumber as

Fe = 2,270,000,000,000,000,000,000,000,000,000,000,000,000 Fg

Therefore, on the atomic level the gravitational force is an extremely weak force,whereas the electric force is very large. Hence, in the solution of electrostaticsproblems the gravitational force can be ignored when compared to the electric force.

On an atomic level, the gravitational force is indeed very weak. By contrast,however, when extremely large masses are involved, such as in a large dying star,the gravitational force is so great, that electrons are driven right into the nuclei of


20-11

atoms, converting the nuclear protons into neutrons, and creating a neutron star.The density of the matter in such a neutron star is so great that one tablespoon ofthat matter would weigh 10 billion tons if it were located in the gravitational fieldat the surface of the earth.

Let us now consider the electrostatic force between the two protons in ahelium nucleus. The protons are separated by a distance of about 2.40 × 10−15 m.The force between the two protons is

Fpp =kqpqp

r2

=(9.00 % 109N m2/C2)(1.60 % 10−19 C)2

(2.40 % 10−15 m)2

= 40.0 N

If we compare this repulsive force between the two protons in the helium nucleus tothe attractive electrostatic force between the electron and the proton in thehydrogen atom, we see that their ratio is

Fpp

Fep= 40.0 N

8.23 % 10−8 N = 4.86 % 108

orFpp = 4.86 × 108 Fep

The force between the two protons in the helium nucleus is an enormous forceand hence the helium nucleus should blow itself apart. The fact that it does not isan indication of the existence of another force, the strong nuclear force, that holdsthe protons together within the nucleus. From the calculation, the nuclear forceholding the nucleus together is at least 108 times greater than the electric forceholding the atom together, that is,

FN = 108 FA

Let us consider some examples of the use of Coulomb’s law.

Example 20.1

Coulomb’s law for two point charges. A point charge, q1 = 2.00 µC is placed 0.500 mfrom another point charge q2 = −5.00 µC. Calculate the magnitude and direction ofthe force on each charge.

r+q1 -q

2

F12 F21

Diagram for example 20.1


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Solution

The force acting on charge q1, is a force of attraction caused by the negative chargeof q2. This force will be called F12 (force on charge 1 caused by charge 2). Since thisforce is in the positive x-direction it can be written as

F12 = iF12

The magnitude of the force F12 is found by equation 20.1 as

F12 =kq1q2

r2

F12 = 9.00 % 109 N m2

C2(2.00 % 10−6 C)(5.00 % 10−6 C)

(0.500 m)2

F12 = 0.360 Nand the force is

F12 = (0.360 N)i

The force acting on charge q2, F21, is a force of attraction caused by charge q1. Sincethis force points in the negative x-direction, it can be written as

F21 = − F21i

The magnitude of the force F21 is found from Coulomb’s law and is given by

F21 =kq2q1

r2

F21 = 9.00 % 109 N m2

C2(5.00 % 10−6 C)(2.00 % 10−6 C)

(0.500 m)2

F21 = 0.360 N and the force is

F21 = − (0.360 N)i

Note that the magnitudes of the forces on q1 and q2 are identical, but theirdirections are opposite. This could have been deduced immediately from Newton’sthird law, for if charge 1 exerts a force on charge 2, then charge 2 must exert anequal but opposite force on charge 1.

To go to this Interactive Example click on this sentence.


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http://www.farmingdale.edu/faculty/peter-nolan/pdf/univphys/UPEX20-01.XLS

Example 20.2

The effect of touching two charged spheres. Two identical metal spheres are placed0.200 m apart. A charge q1 of 9.00 µC is placed on one sphere while a charge q2 of−3.00 µC is placed upon the other. (a) What is the force on each of the spheres? (b) Ifthe two spheres are brought together and touched and then returned to theiroriginal positions, what will be the force on each sphere?

r+q1 -q

2

F12 F21

r+q1 +q

2

F12F21

(a) (b)Diagram for example 20.2

Solution

a. The magnitude of the attractive force on sphere 1 is found from Coulomb’s law as

F12 =kq1q2

r2

F12 = 9.00 % 109 N m2

C2(9.00 % 10−6 C)(3.00 % 10−6 C)

(0.200 m)2

F12 = 6.08 Nand the force is found as

F12 = (6.08 N)i

The force on sphere 2 is equal and opposite to this.

b. When the spheres are touched together, the −3.00 µC of charge q2, neutralizes+3.00 µC of charge q1, leaving a net charge of

9.00 × 10−6 C − 3.00 × 10−6 C = 6.00 × 10−6 C

This charge of 6.00 × 10−6 C is then equally distributed between sphere one andsphere two, giving q1’ = q2’ = 3.00 × 10−6 C. Since the charges on both spheres arenow positive, the force on each sphere is now repulsive. When the spheres areremoved to the original distance the magnitude of the force on each sphere is now

F =kq2∏ q1∏

r2

= 9.00 % 109 N m2

C2(3.00 % 10−6 C)(3.00 % 10−6 C)

(0.200 m)2


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F = 2.03 N


20.6 Multiple Discrete ChargesIf there are three or more charges present, then the force on any one charge

is found by the vector addition of the forces associated with the other charges. Thatis, the resultant force on any one charge is given by

(20.4)F = F1 + F2 + F3 + F4 + ....

Example 20.3

Multiple charges in a line. Three charges are placed on the line as shown. Theseparation of the charges are r12, = 0.500 m and r23 = 0.500 m. If q1 = 1.00 µC, q2 =−2.00 µC, and q3 = 3.00 µC, find: (a) the resultant force on charge q1, (b) theresultant force on charge q2, and (c) the resultant force on charge q3.

q1

q2

q3

r12r23

r13

Diagram for example 20.3.

Solution

The resultant force on each charge is equal to the vector sum of all the forces actingon that charge. a. As can be seen in diagram 20.3(a), the force on charge 1 is

F1 = F12 + F13

where F12 is the force on charge 1 caused by charge 2, and F13 is the force on charge1 caused by charge 3. Because q2 is negative, F12 is a force of attraction to the rightand is given by

F12 = iF12


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q1 q2q3

r12 r23

r13

F12F13

Diagram for example 20.3(a).

Since q3 is positive, F13 is a force of repulsion to the left and is given by

F13 = − iF13

The total force on charge 1 is therefore

F1 = iF12 − iF13

where

F12 =kq1q2

r122

F12 = 9.00 % 109 N m2

C2(1.00 % 10−6 C)(2.00 % 10−6 C)

(0.500 m)2

F12 = 0.0720 N while

F13 =kq1q3

r132

F13 = 9.00 % 109 N m2

C2(1.00 % 10−6 C)(3.00 % 10−6 C)

(1.00 m)2

F13 = 0.0270 N Therefore,

F1 = iF12 − iF13 = (0.0720 N)i − (0.0270 N)iF1 = (0.0450 N)i

That is, the resultant force on charge 1 is a force of 0.0450 N to the right.

b. The resultant force on charge 2 is F2 = F21 + F23

or as can be seen from diagram 20.3b

F2 = iF23 − iF21

From Newton’s third lawF21 = − F12 = − (0.0720 N)i

and


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q1 q2q3

r12 r23

r13

F21 F23

Diagram for example 20.3(b).

F23 =kq2q3

r232

F23 = 9.00 % 109 N m2

C2(2.00 % 10−6 C)(3.00 % 10−6 C)

(0.500 m)2

F23 = 0.216 N

Therefore, the net force on charge 2 is

F2 = iF23 − iF21 = (0.216 N)i − (0.0720 N)iF2 = (0.144 N)i

to the right.

c. The resultant force on charge 3 is F3 = F31 + F32

and as can be seen in diagram 20.3(c) this can be written as

F3 = iF31 − iF32

q1 q2q3

r12 r23

r13

F32 F31

Diagram for example 20.3(c).

But by Newton’s third law F31 = −F13 = (0.0270 N)i

and


20-17

F32 = −F23 = − (0.2160 N)i

The net force on charge 3 is therefore

F3 = (0.0270 N)i − (0.2160 N)iF3 = −(0.189 N)i

to the left.


Example 20.4

Multiple charges not on a line. Find the resultant force on charge q2 in the diagramif q1 = 13.0 µC, q2 = 4.00 µC, q3 = 5.00 µC. r13 = 0.500 m, r23 = 0.800 m,

q1

q2

r12

r13

r23

q3

F21

F2

x

y

21yF

21xF

23F

θ

θ


Solution

The resultant force on charge q2 is found as

F2 = F21 + F23


20-18


where F21 is the force on charge 2 caused by charge 1, and F23 is the force on charge2 caused by charge 3. The distance r12 is found from the diagram and thePythagorean theorem as

r12 = r132 + r23

2

r12 = (0.500 m)2 + (0.800 m)2

r12 = 0.940 m

The magnitude F21 is found from Coulomb’s law as

F21 =kq2q1

r122

F21 = 9.00 % 109 N m2

C2(4.00 % 10−6 C)(13.00 % 10−6 C)

(0.940 m)2

F21 = 0.530 Nwhile the magnitude of F23 is

F23 =kq2q3

r232

F23 = 9.00 % 109 N m2

C2(4.00 % 10−6 C)(5.00 % 10−6 C)

(0.800 m)2

F23 = 0.281 N

The addition of the two vectors is an example of the addition of vectors discussed inchapter 3. The resultant vector is given by

F2 = iF2x + jF2y

The magnitude of the resultant vector is found, as before, as

F2 = (F2x )2 + (F2y )2

The force F21 is given byF21 = iF21x + jF21y

The vector F21 has an x-component given by

F21x = F21 cosθ

The angle θ is found from the geometry of the diagram as

✕ = tan−1 0.8000.500 = 58.00

Therefore, the x-component of F21 is

F21x = F21 cosθ = (0.530 N)cos58.00 = 0.281 N


20-19

Similarly, the y-component of F21 is

F21y = F21 sinθ = (0.530 N)sin58.00 = 0.449 N

The vector F23 is in the y direction and is given by

F23 = jF23

and has no x-component. Therefore, the x-component of the resultant vector is

F2x = F21x = 0.281 N

and as can be seen from the diagram the y-component of the resultant vector is

F2y = F23 + F21y = 0.281 N + 0.449 N = 0.730 N

The resultant force on charge 2 is

F2 = iF2x + jF2y = (0.281 N)i + (0.730 N)j

The magnitude of the resultant force on charge 2 is therefore

F2 = (F2x )2 + (F2y )2 = (0.281 N)2 + (0.730 N)2

F2 = 0.782 N

The angle φ that F2 makes with the x-axis is found from

$ = tan−1 F2y

F2x= tan−1 0.730

0.281φ = 68.90


Example 20.5

More multiple charges. Find the resultant force on charge q2 in the diagram if q1 =3.00 µC, q2 = 5.00 µC, q3 = 4.00 µC, r12 = 0.500 m and r23 = 0.500.

Solution

The resultant force on charge 2 is found from the vector sum


20-20


q1

q3

r12

r13r23

q2

F21

F223F

θ


F2 = F21 + F23

Because F21 is perpendicular to F23, the magnitude of the resultant force is

F2 = (F21 )2 + (F23 )2

where

F21 =kq2q1

r212

F21 = 9.00 % 109 N m2

C2(5.00 % 10−6 C)(3.00 % 10−6 C)

(0.500 m)2

F21 = 0.540 N and

F23 =kq2q3

r232

F23 = 9.00 % 109 N m2

C2(5.00 % 10−6 C)(4.00 % 10−6 C)

(0.500 m)2

F23 = 0.720 N

As can be seen from the diagram, the resultant force on charge 2 is

F2 = iF21 − jF23 F3 = (0.540 N)i − (0.720)j

The magnitude of the resultant force on charge 2 is therefore

F2 = (F21 )2 + F232 = (0.540 N)2 + (0.720 N)2

F2 = 0.900 N


20-21

The direction of the resultant force is determined by

✕ = tan−1 F23F21

= tan−1 −0.7200.540

θ = −53.10


Example 20.6

Charged pith balls. Two equally charged pith balls are separated by 0.100 m asshown in figure 20.9(a). Find the charge on each ball and the tension in the stringif the mass of each ball is 5.00 × 10−3 kg, and the length l of the string is 0.250 m.

r

l

θ

(a)

θ

T

w

FeTx

Ty

(b)

Figure 20.9 Charged pith balls.

Solution

Let us consider the forces acting on the ball at the right. The forces acting are thetension T in the string, the weight w of the ball, and the electric force of repulsionon the ball Fe as shown in figure 20.9(b).

Since the ball is in equilibrium the first condition of equilibrium is applied as

Σ Fy = 0, Σ Fx = 0Ty − w = 0, Fe − Tx = 0

T sin θ = w, Fe = T cos θ Fe = T cos θ = 1

w T sin θ tan θ


20-22


Fe = w tan θkq2 = mg r2 tan θ

Solving for the charge q on each ball we get

q =r2mg

k tan ✕

The angle θ, found from the geometry of the figure 20.9(a), is

θ = cos−1 r/2 l

= cos−1 5.00 cm 25.0 cm

= 78.50

Therefore, the charge on each ball is

q =r2mg

k tan ✕

=(0.100 m)2(5.00 % 10−3 kg)(9.80 m/s2)

(9.00 % 109 N m2/C2 ) tan 78.50

= 1.05 × 10−7 CThe tension in the string is

T sin θ = wT = w sin θ = mg sin θ

= (5.00 × 10−3 kg)(9.80 m/s2)sin 78.50

= 5.00 × 10−2 N


In section 20.5, and in particular in equation 20.1, we wrote Coulomb’s law interms of its magnitude only. In this section we have been using the unit vectors iand j to describe the direction of the force vector. Coulomb’s law can be written in amore general form as follows. Let q1 be the primary charge that we are interestedin. We will introduce a unit vector ro that points everywhere radially away from thecharge q1 as shown in figure 20.10(a).


20-23


+q1

+q2

F12

F21

rro

-q1

+q2

F12

F21

rro

(a) Force on + charge (b) Force on − charge Figure 20.10 Coulomb’s law in vector form.

If a charge + q2 is brought into the vicinity of charge q1, it will experience theforce

(20.5)F21 = kq2q1

r2 ro

where F21 is the force on charge q2 caused by charge q1. If q1 and q2 are of like signthen the force on the secondary charge q2 is in the same direction as the unit vectorro, and the force is one of repulsion as expected and is shown in figure 20.10(a). Ifthe primary charge is negative, that is, −q1, then the charges are of opposite signand the force on charge q2 is in the opposite direction of the unit vector ro, and theforce is one of attraction as seen in figure 20.10(b). By Newton’s third law, the forceon charge q1 is equal and opposite to the force on charge q2 as expected. That is,

F12 = − F21

20.7 Forces Caused by a Continuous Distribution ofCharge

As we have seen in the last section, when there are multiple discrete chargesin a region, then the force on any one of those charges is found by the vector sum ofthe forces associated with the other charges. That is, the resultant force on any onecharge was given by equation 20.4 as

F = F1 + F2 + F3 + F4 + .…

A shorthand notation for equation 20.4 can be written as

(20.6)F=✟i=1

NFi


20-24

where, again, Σ, the Greek letter sigma, means “the sum of” and the sum goes from i= 1 to i = N.

Besides the forces caused by a discrete distribution of charge, the force on asingle charge qo caused by a continuous distribution of charge can be handled in asimilar way. The uniform distribution of charge can be broken up into a largenumber of infinitesimal elements of charge, dq, and each element of charge willproduce an element of force dF on the discrete charge qo. Coulomb’s Law can then bewritten as

(20.7)dF= kqodq

r2 ro

where r is the distance from the element of charge dq to the single discrete chargeqo. ro is a unit vector that points from the element of charge dq, and points towardthe discrete charge qo. The total force F on the discrete charge qo, caused by the forcesfrom the entire distribution of all the dq’s is again a sum, but since the elements ofcharge dq, are infinitesimal the sum becomes the integral of all the elements of forcedF. That is, the force is found as

(20.8)F= ¶ dF= ¶ kqodq

r2 ro

Remember that qo is a constant in this integration. As an example of the force caused by a continuous distribution, let us find the

force acting on a charge qo located at the origin of a semicircular ring of charge asshown in figure 20.11. The ring has a radius r and carries a continuous distributionof charge q. A small element of this charge dq is shown in the right quadrant of thediagram. This element of charge dq exerts the force dF on the discrete charge qo,given by equation 20.7, and is shown in the diagram. The total force F is found fromequation 20.8, which is a vector integration.

dq dqr r

x

y

dFdFθ θ

θθ

dF cosθ

θdF sindF

dF cosθ

θdF sindF

qo

Figure 20.11 Force on a point charge caused by a continuous distribution ofcharge.

The vector integration can be simplified into a scalar integration by notingthat the element of charge dq has a mirror image of charge of the same magnitudeon the other half of the semicircular ring as shown in the second quadrant of thediagram. Each charge produces an element of force dF. This element of force can bebroken up into two components, one, dF cosθ, which lies along the x-axis, while the


20-25

second component, dF sinθ, lies along the negative y-axis. When we add up(integrate) the effect of all the charges dq we see that half the components dF cosθ,are in the positive x-direction, while the other half are in the negative x-direction.Thus, the sum of all the x-components of the forces will be zero, and we need onlyconsider the y-components. Both y-components, dF sinθ, are in the negativey-direction, and the total force can be written as the scalar integration

(20.9)F = ¶ 2dF sin ✕ = ¶ 2kqodq

r2 sin ✕

Notice that, in general, k and qo, are constants and, for this particularproblem, r, the radius of the ring is also a constant. Hence, they can all be takenoutside the integral sign. The integral is now given as

(20.10)F =2kqo

r2 ¶ sin ✕ dq

Notice that the integration is over dq. Rather than integrating over a charge, it iseasier to integrate over a geometrical figure. We define a linear charge density λλλλas the amount of charge per unit length, that is,

(20.11)✘ =qs

where q is the total charge on the semicircular ring and s is the total length of thesemicircular ring and also represents the arc of the semicircular ring. Solvingequation 20.11 for q gives

q = λs

and upon differentiating for the element of charge, dq, we get

dq = λ ds (20.12)

Replacing equation 20.12 into equation 20.10 we get

(20.13)F =2kqo

r2 ¶ sin ✕ ✘ds

Now the sinθ and ds are not independent and the relation between them must bestated before the integration can begin. The angle θ is related to the radius r andthe arc s of a circle by

s = rθDifferentiating this equation we get

ds = r dθ (20.14)


20-26

where ds is an element of an arc of the semicircular ring, r is the radius of the ring,and dθ is the small angle subtended by the arc ds. Replacing equation 20.14 intoequation 20.13 gives

F =2kqo

r2 ¶ sin ✕ ✘rd✕

Assuming λ is a constant, it can be taken out of the integral to yield

(20.15)F =2kqo✘

r ¶0✜/2

sin✕ d✕

Notice that the integration is now over the angle θ, and the integration is from θ = 0to θ = π/2. Integrating equation 20.15 gives

F =2kqo✘

r (−cos ✕)|0✜/2

F = −2kqo✘

r cos ✜2 − cos 0

F = −2kqo✘

r (0 − 1)

(20.16)F =2kqo✘

r

Equation 20.16 gives the force acting on the discrete charge qo in terms of thelinear charge density λ. It can be expressed in terms of the total charge q on thesemicircular ring by writing equation 20.11 as

(20.17)✘ =qs =

q✜r

Notice that the arc s of the semicircular ring is equal to half the circumference of acircle, namely, πr. Substituting equation 20.17 into equation 20.16 we get

F =2kqo

rq✜r

(20.18)F =2kqoq✜r2

Equation 20.18 gives the force on a discrete charge qo by a continuous distribution ofcharge q along the surface of a semicircular ring of radius r.

The force on a discrete charge qo caused by any other continuous distributionof charge can be found in the same way.

The Language of Physics

ElectrostaticsThe study of electric charges at rest under the action of electric forces (p. ).


20-27

The Fundamental Principle of ElectrostaticsLike electric charges repel each other while unlike electric charges attract eachother (p. ).

QuarksElementary particles of matter. There are six quarks. They are: up, down, strange,charm, bottom, and top. The proton and neutron are made of quarks, but theelectron is not (p. ).

ConductorsMaterials that permit the free flow of electric charge through them (p. ).

Insulators or DielectricsMaterials that do not permit the free flow of electric charge through them (p. ).

Coulomb’s lawThe force between point charges q1 and q2 is directly proportional to the product oftheir charges and inversely proportional to the square of the distance separatingthem. The direction of the force lies along the line separating the charges (p. ).

Linear charge density λλλλ is defined as the amount of charge per unit length (p. ).

Summary of Important Equations

Coulomb’s law (20.1)F = kq1q2

r2

Force caused by multiple discrete charges F = F1 + F2 + F3 + F4 + … (20.4)

Coulomb’s law in vector form (20.5)F21 = kq2q1

r2 ro

Coulomb’s law in differential form (20.7)dF= kqodq

r2 ro

Forced caused by a continuous charge distribution (20.8)F= ¶ dF= ¶ kqodq

r2 ro

Linear charge density (20.11)✘ =qs

Linear element of charge dq = λ ds (20.12)


20-28

Questions for Chapter 20

1. Describe the process of charging by induction.2. What did Ben Franklin have to do with the classification of electrical

charge?3. The difference between one chemical element and another is the number of

protons that each contains. Is it possible for one chemical element to have the samenumber of neutrons as another chemical element?

4. Discuss the planetary model of the atom and state its good points and itslimitations.

5. Is it possible for a quark to exist if it has not been seen? If it cannot beisolated does it make any sense to describe particles as though they were made upof quarks?

6. Describe the process of lightning in the atmosphere. How do the earth andthe clouds pick up electric charge? If air is an insulator, how can a lightning boltever reach the ground?

7. Can you think of a way that you could use electrostatics to measure thehumidity of the atmosphere?

8. How could you paint a metallic object with a minimum of paint using theprinciples of electrostatics?

9. Describe the phenomenon known as “St. Elmo’s Fire” in terms ofelectrostatics.

10. Why do clothes taken from a clothes dryer sometimes cling to the body?

Problems for Chapter 20

Section 20.5 Coulomb’s Law.1. A point charge of 4.00 µC is placed 25.0 cm from another point charge of

−5.00 µC. Calculate the magnitude and direction of the force on each charge. 2. If the force of repulsion between two protons is equal to the weight of the

proton, how far apart are the protons? 3. What equal positive charges would have to be placed on the earth and the

moon to neutralize the gravitational force between them? 4. What is the velocity of an electron in the hydrogen atom if the centripetal

force is supplied by the coulomb force between the electron and proton? The radiusof the electron orbit is 5.29 × 10−11 m.

5. Two identical metal spheres are placed 15.0 cm apart. A charge of 6.00 µCis placed on one sphere while a charge of −2.00 µC is placed upon the other. What isthe force on each sphere? If the two spheres are brought together and touched andthen separated to their original separation, what will be the force on each sphere?


20-29

6. Two identical metal spheres, carrying different opposite charges, attracteach other with a force of 5.00 × 10−6 N when they are 5.00 cm apart. The spheresare then touched together and then removed to the original separation where now aforce of repulsion of 1.00 × 10−6 N is observed. What is the charge on each sphereafter touching and before touching?

Section 20.6 Multiple Charges. 7. Three charges q1 = 2.00 µC, q2 = 5.00 µC, and q3 = 8.00 µC are placed along

the x-axis at 0.00, 45.0 cm, and 72.4 cm respectively. Find the force on each charge. 8. Two charges q1 = 4.00 µC and q2 = 8.00 µC are separated by the distance r

= 50.0 cm as shown. Where should a third charge be placed on the line betweenthem such that the resultant force on it will be zero? Does it matter if the thirdcharge is positive or negative?

rq

1q

2

Diagram for Problem 8.

9. Repeat problem 8 but now let charge q1 be negative, and find any positionon the line, either to the left of q1 , between q1 and q2, or to the right of q2, where athird charge can be placed that experiences a zero resultant force.

10. Three charges of 2.00 µC, −4.00 µC, and 6.00 µC are placed at the verticesof an equilateral triangle of length 10.0 cm on a side. Find the resultant force oneach charge.

11. If q1 = 5.00 µC = q2 = q3 = q4 are located on the corners of a square oflength 20.0 cm, find the resultant force on q3.

q3

q1

q2

q4

q3

q1

q2

l

Diagram for problem 11. Diagram for problem 13.

12. Charges of 2.54 µC, −7.86 µC, 5.34 µC, and −3.78 µC are placed on thecorners of a square of side 23.5 cm. Find the resultant force on the first charge.

13. Find the force on charge q3 = 1.00 µC, if q1 = + 5.00 µC and q2 = − 5.00 µC.The distance separating charges q1 and q2 is 5.00 cm, and l = 1.00 m.


20-30

14. Find the resultant force on charge q3 in the diagram if q1 = 2.00 µC, q2 =−7.00 µC, q3 = 5.00 µC, r12 = 0.750 m, r23 = 0.600 m.

q1 q2r12

r23

q3

θ

Diagram for problem 14.

Section 20.7 Force Caused by a Continuous Distribution of Charge. 15. A thin nonconducting rod of length l carries a uniform charge per unit

length λ. The rod lies on the + x-axis with one end at the point xo and the other endat x = xo + l. A point charge qo lies on the x-axis at the origin. Find the force actingon charge qo.

16. This is similar to problem 15 except the point charge qo is located on they-axis at the point yo. (a) Find the x-component of the force on the point chargecaused by the line of charge. (b) Find the y-component of the force on the pointcharge caused by the line of charge.

Additional Problems. 17. Find the force on charge q5 = 5.00 µC, located at the center of a square

25.0 cm on a side if q1 = q2 = 3.00 µC and q3 = q4 = 6.00 µC.

q3

q1

q2

q4

q5

q1 q

2

300

l = 25 cm


18. Two small, equally charged, spheres of mass 0.500 gm are suspendedfrom the same point by a silk fiber 50.0 cm long. The repulsion between them keepsthem 15.0 cm apart. What is the charge on each sphere?

19. Two pith balls of 10.0 gm mass are hung from ends of a string 25.0 cmlong, as shown. When the balls are charged with equal amounts of charge, thethreads separate to an angle of 30.00. What is the charge on each ball?


20-31

20. Two 10.0 gm pith balls are hung from the ends of two 25.0 cm long stringsas shown. When an equal and opposite charge is placed on each ball, theirseparation is reduced from 10.0 cm to 8.00 cm. Find the tension in each string andthe charge on each ball.

q1 q

28 cm

10 cm


21. A charge of 15.0 µC is on a metallic sphere 10.0 cm radius. It is thentouched to a sphere of 5.00 cm radius, until the surface charge density is the sameon both spheres. What is the charge on each sphere after they are separated? (Thesurface charge density σ is the charge per unit area and is given by σ = q/A.)

22. Two small spheres carrying charges q1 = 7.00 µC and q2 = 5.00 µC areseparated by 20.0 cm. If q2 were free to move what would its initial acceleration be?Sphere 2 has the mass m2 = 15.0 gm.

23. Where should a fourth charge, q4 = 3.00 µC, be placed to give a net force ofzero on charge q3? Charge q1 = 2.00 µC, q2 = 4.00 µC, and q3 = 2.00 µC.

q1

q2

q3

1 m

1 m

qo

+q

-q

r

xa2 θ

qo+q r


24. Charge q1 = 3.00 µC, is located at the coordinates (0,2) and charge q2 =6.00 µC, is located at the coordinates (1,0) of a Cartesian coordinate system. Findthe coordinates of a third charge that will experience a zero net force.

25. The configuration of a positive charge q separated by a distance 2a from anegative charge −q is called an electric dipole. Show that the force exerted by anelectric dipole on a point charge qo, located as shown in the diagram varies as 1/r3


20-32

while the force between a point charge q and the point charge qo varies as 1/r2.Which force is the weaker?

26. A plastic rod 50.0 cm long, has a charge + q1 = 2.00 µC at each end. Therod is then hung from a string and placed so that each charge is only 5.00 cm fromnegative charges q2 = − 10.0 µC as shown in the diagram. Find the torque acting onthe string.

27. A charge of 5.00 µC is uniformly distributed over a copper ring 2.00 cm inradius. What force will this ring exert on a point charge of 8.00 µC that is placed3.00 m away from the ring. Indicate what assumptions you make to solve thisproblem.

28. A charge of 2.50 µC is placed at the center of a hollow sphere of charge of8.00 µC. What is the resultant force on the charge placed at the center of thesphere? Indicate what assumptions you make to solve this problem.

q1

q2

r r

q2

q1


29. A thin nonconducting rod of length l carries a nonuniform charge densityλ = Ax2. The rod lies on the x-axis with one end at the origin and the other end at x =l. A point charge qo lies on the x-axis at the point xo. Find the force acting on chargeqo.

30. Find the force acting on a point charge qo located at the origin of asemicircular ring of charge as shown in figure 20.10. The ring has a radius r andcarries a nonuniform continuous distribution of charge density λ = A sinθ.

Interactive Tutorials31. Coulomb’s law. Two charges q1 = 2.00 × 10−6 C and q2 = 3.00 × 10−6 C are

separated by a distance r = 1.00 m. Calculate the electrostatic force F of repulsionacting on charge 1 as the distance of separation is increased from r = 1 to r = 10 m.Show how the force F varies with the distance r.

32. Coulomb’s law. Two electrons of charge q1 = q2 = e = 1.60 × 10−19 C arepositioned at the coordinates (0,1) and (0,−1) (in meters) of a Cartesian coordinatesystem. Calculate the net force F3 on an electron q3 as it is moved from x = 0 to x =10.0 m along the x-axis.


20-33

33. Coulomb’s law and multiple charges. Two charges q1 = 8.32 × 10−6 C andq2 = −2.55 × 10−6 C lie on the x-axis and are separated by the distance r12 = 0.823 m.A third charge q3 = 3.87 × 10−6 C is located a distance r23 = 0.475 m from charge q2,and the line between charge 2 and 3 makes an angle φ = 60.00 with respect to thex-axis. Find the resultant force on (a) charge 3, (b) charge 2, and (c) charge 1.

34. Coulomb’s law and a continuous charge distribution. A rod of charge oflength L = 0.100 m lies on the x-axis. One end of the rod lies at the origin and theother end is on the positive x-axis. A charge q’ = 7.36 × 10−6 C is uniformlydistributed over the rod. Find the force exerted on a point charge q = 2.95 × 10−6 Cthat lies on the x-axis at distance x0 = 0.175 m from the origin of the coordinatesystem.

To go to these Interactive Tutorials click on this sentence.

To go to another chapter, return to the table of contents byclicking on this sentence.


20-34

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