CHAPTER 21 MCQ PDF

694 | CHAPTER 21 The Electric Field I: Discrete Charge Distributions Today, the study and use of electricity continue. Electrical engineers improve ex- isting electrical technologies, increasing performance and efficiency in devices such as hybrid cars and electric power plants...

694 | CHAPTER 21 The Electric Field I: Discrete Charge Distributions Today, the study and use of electricity continue. Electrical engineers improve ex- isting electrical technologies, increasing performance and efficiency in devices such as hybrid cars and electric power plants. Electrostatic paints are used in the auto industry for engine parts and for car frames and bodies. This painting process creates a more durable coat than does liquid paint, and is easier on the environ- ment because no solvents are used. In this chapter, we begin our study of electricity with electrostatics, the study of charges at rest. After introducing the concept of charge, we briefly look at conductors and insulators and how conductors can be given a net charge. We then study Coulomb’s law, which describes the force exerted by one charge on another. Next, we introduce the electric field and show how it can be visualized by electric field lines that indicate the magnitude and direction of the field, just as we visualized the velocity field of a flowing fluid using streamlines (Chapter 13). Finally, we discuss the behavior of point charges and dipoles in electric fields. 21-1 CHARGE Suppose we rub a hard-rubber rod with fur and then suspend the rod from a string so that it is free to rotate. Now we bring a second hard-rubber rod that has been rubbed with fur near it. The rods repel each other (Figure 21-1a). Two glass rods that have been rubbed with silk (Figure 21-1b) also repel each other. But, when we place a hard-rubber rod rubbed with fur near a glass rod rubbed with silk (Figure 21-1c) they attract each other. Rubbing a rod causes the rod to become electrically charged. If we repeat the experiment with various materials, we find that all charged objects fall into one of just two groups—those like the hard-rubber rod rubbed with fur and those like the glass rod rubbed with silk. Objects from the same group A cat and a balloon. (Roger Ressmeyer/CORBIS.) repel each other, while objects from different groups attract each other. Benjamin Franklin explained this by proposing a model in which every object has a normal amount of electricity that can be transferred from one object to the other when two objects are in close con- tact, as when they are rubbed together. One object would have an excess charge hard rubber – + and the other object would have a defi- –– ––– ++ ++ + –– – ––– ++ + + ++ ciency of charge, and the excess charge – glass + glass hard rubber equals the deficiency of charge. Franklin described the resulting charges as posi- (a) (b) tive (plus sign) or negative (minus sign). He also chose positive to be the charge acquired by a glass rod when it is rubbed with a piece of silk. The piece of silk then acquires a negative charge of equal mag- nitude during the procedure. Based on Franklin’s convention, if hard rubber and fur are rubbed together, the hard F I G U R E 2 1 - 1 (a) Two hard-rubber rubber acquires a negative charge and rods that have been rubbed with fur repel – each other. (b) Two glass rods that have the fur acquires a positive charge. ++ ––– ++ + ––– been rubbed with silk repel each other. Two objects that have the same sign glass + hard rubber (c) A hard-rubber rod that has been rubbed (both  or both ) repel each other, with fur attracts a glass rod that has been and two objects that have oppositely (c) rubbed with silk.  Charge S E C T I O N 2 1- 1 | 695 signed charges attract each other (Figure 21-1). An object that is neither positively nor negatively charged is said to be electrically neutral. Table 21-1 The Triboelectric Today, we know that when glass is rubbed with silk, electrons are transferred Series from the glass to the silk. Because the silk is negatively charged (according to Franklin’s convention, which we still use) electrons are said to have a negative  Positive End of Series charge. Table 21-1 is a short version of the triboelectric series. (In Greek tribos Asbestos means “a rubbing.”) The farther down the series a material is, the greater its affin- ity for electrons. If two of the materials are brought in contact, electrons are trans- Glass ferred from the material higher in the table to the one farther down the table. For Nylon example, if Teflon is rubbed with nylon, electrons are transferred from the nylon to Wool the Teflon. Lead Silk CHARGE QUANTIZATION Aluminum Matter consists of atoms that are neutral. Each atom has a tiny but massive nucleus Paper that is composed of protons and neutrons. Protons are positively charged, whereas Cotton neutrons are neutral. The number of protons that an atom of a particular element Steel has is the atomic number Z of that element. Surrounding the nucleus is an equal number of negatively charged electrons, leaving the atom with zero net charge. An Hard rubber electron is about 2000 times less massive than a proton, yet the charges of these two Nickel and copper particles are exactly equal in magnitude. The charge of the proton is e and that of Brass and silver the electron is e, where e is called the fundamental unit of charge. The charge of Synthetic rubber an electron or proton is an intrinsic property of the particle, just as mass and spin are intrinsic properties of these particles. Orlon All observable charges occur in integral amounts of the fundamental unit of Saran charge e; that is, charge is quantized. Any observable charge Q occurring in nature Polyethylene can be written Q  Ne, where N is an integer.* For ordinary objects, however, N Teflon is usually very large and charge appears to be continuous, just as air appears to be continuous even though air consists of many discrete particles (molecules, atoms, Silicone rubber and ions). To give an everyday example of N, charging a plastic rod by rubbing it with a piece of fur typically transfers 1010 or more electrons to the rod.  Negative End of Series CHARGE CONSERVATION When objects are rubbed together, one object is left with an excess of electrons and is therefore negatively charged; the other object is left with a deficit of electrons and is therefore positively charged. The net charge of the two objects remains con- stant; that is, charge is conserved. The law of conservation of charge is a fundamen- tal law of nature. In certain interactions among elementary particles, charged par- ticles such as electrons are created or annihilated. However, during these processes, equal amounts of positive and negative charge are produced or de- stroyed, so the net charge of the universe is unchanged. The SI unit of charge is the coulomb, which is defined in terms of the unit of electric current, the ampere (A).† The coulomb (C) is the amount of charge flowing through a cross section of wire in one second when the current in the wire is one ampere. (The cross section of a solid object is the intersection of the object and a plane. Here we consider a plane that cuts across the wire.) The fundamental unit of electric charge e is related to the coulomb by e  1.602177  1019 C  1.60  1019 C 21-1 F U N DA M E N TA L U N IT O F C H A RG E * In the standard model of elementary particles, protons, neutrons, and some other elementary particles are made up of more fundamental particles called quarks that have charges of  13 e or  23 e. Only combinations that result in a net charge of Ne, where N is an integer, are observed. † The ampere (A) is the unit of current used in everyday electrical work. 696 | CHAPTER 21 The Electric Field I: Discrete Charge Distributions Charging by contact. A piece of plastic about 0.02 mm wide was charged by contact with a piece of nickel. Although the plastic carries a net positive charge, regions of negative charge (dark blue) as well as regions of positive charge (yellow) are indicated. The photograph was taken by sweeping a charged needle of width 107 m over the sample and recording the electrostatic force on the needle. (Bruce Terris/IBM Almaden Research Center.) PRACTICE PROBLEM 21-1 A charge of magnitude 50 nC (1.0 nC  109 C) can be produced in the laboratory by simply rubbing two objects together. How many electrons must be transferred to produce this charge? Example 21-1 How Many in a Penny? A copper penny* (Z  29) has a mass of 3.10 grams. What is the total charge of all the elec- trons in the penny? PICTURE The electrons have a total charge given by the number of electrons in the penny, Ne , multiplied by the charge of an electron, e. The number of electrons in a copper atom is 29 (the atomic number of copper). So, the total charge of the electrons is 29 electrons multi- plied by the number of copper atoms Nat in a penny. To find Nat , we use the fact that one mole of any substance has Avogadro’s number (NA  6.02  1023) of particles (molecules, atoms, or ions), and the number of grams in one mole of copper is the molar mass M, which is 63.5 g>mol for copper. SOLVE 1. The total charge Q is the number of electrons multiplied by Q  Ne(e) the charge: 2. The number of electrons is Z multiplied by the number of Ne  ZNat copper atoms Nat: 6.02  1023 atoms>mol 3. Compute the number of copper atoms in 3.10 g of copper: Nat  (3.10 g)  2.94  1022 atoms 63.5 g>mol 4. Compute the number of electrons Ne: Ne  ZNat  (29 electrons>atom)(2.94  1022 atoms)  8.53  1023 electrons 5. Use this value of Ne to find the total charge: Q  Ne  (e)  (8.53  1023 electrons)(1.60  1019 C>electron)  1.37  105 C CHECK There are 29  (6.02  1023) electrons in 63.5 g of copper, so in 3.10 g of copper there are (3.10>63.5)  29  (6.02  1023)  8.53  1023 electrons—in agreement with our step-4 result. PRACTICE PROBLEM 21-2 If one million electrons were given to each person in the United States (about 300 million people), what percentage of the number of electrons in a penny would this represent? * The penny was composed of 100 percent copper from 1793 to 1837. In 1982, the composition changed from 95 percent copper and 5 percent zinc to 2.5 percent copper and 97.5 percent zinc.  Conductors and Insulators S E C T I O N 2 1- 2 | 697 – – – F I G U R E 2 1 - 2 An electroscope. Two gold leaves are attached to a conducting post that has a conducting ball on top. The ball, post, and leaves are insulated from the container. When uncharged, the – – leaves hang together vertically. When the ball is touched by a – – negatively charged plastic rod, some of the negative charge from the rod is transferred to the ball and moves to the gold leaves, which then spread apart because of electrical repulsion between their negative charges. (Touching the ball with a positively charged glass rod would also cause the leaves to spread apart. In this case, the positively charged glass rod would remove electrons from the metal ball, leaving a net positive charge on the ball, rod, and leaves.) 21-2 CONDUCTORS AND INSULATORS In many materials, such as copper and other metals, some of the electrons are free to move about the entire material. Such materials are called conductors. In other ✓ CONCEPT CHECK 21-1 materials, such as wood or glass, all the electrons are bound to nearby atoms and Two identical conducting spheres, none can move freely. These materials are called insulators. one that has an initial charge Q In a single atom of copper, 29 electrons are bound to the nucleus by the elec- and the other is initially unchar- trostatic attraction between the negatively charged electrons and the positively ged, are brought into contact. charged nucleus. The outer (valence) electrons are more weakly bound to a nu- (a) What is the new charge on each cleus than the inner (core) electrons. When a large number of copper atoms are sphere? (b) While the spheres are combined in a piece of metallic copper, the strength of the attractions of electrons in contact, a positively charged to a nucleus of an atom is reduced by their interactions with the electrons and nu- rod is moved close to one sphere, clei of neighboring atoms. One or more of the valence electrons in each atom is no causing a redistribution of the longer bound to the atom but is free to move about the whole piece of metal, charges on the two spheres so the much as an air molecule is free to move about in a room. The number of these free charge on the sphere closest to electrons depends on the particular metal, but it is typically about one per atom. the rod has a charge of Q. What (The free electrons are also referred to as conduction electrons or delocalized elec- is the charge on the other sphere? trons.) An atom that has an electron removed or added, resulting in a net charge on the atom, is called an ion. In metallic copper, the copper ions are arranged in a regular array called a lattice. A conductor is neutral if for each lattice ion hav- ing a positive charge e there is a free electron having a negative charge e. The net charge of the conductor can be changed by adding or removing electrons. A conductor that has a negative net charge + – ++ has a surplus of free electrons, while a con- + –– ++ + –– ++ ductor that has a positive net charge has a + +++ –– –– + deficit of free electrons. + + (a) CHARGING BY INDUCTION The conservation of charge is illustrated by a + – ++ + –– ++ simple method of charging a conductor called + –– + –– +++ charging by induction, as shown in Figure + –– ++ 21-3. Two uncharged metal spheres touch each + + F I G U R E 2 1 - 3 Charging by induction. other. When a positively charged rod (Figure (a) Neutral conductors in contact become 21-3a) is brought near one of the spheres, con- (b) oppositely charged when a charged rod duction electrons flow from one sphere to the attracts electrons to the left sphere. (b) If the other, toward the positively charged rod. The spheres are separated before the rod is – + removed, they will retain their equal and positively charged rod in Figure 21-3a attracts – – + + –– – –– ++ + ++ opposite charges. (c) When the rod is removed the negatively charged electrons, and the – + and the spheres are far apart, the distribution sphere nearest the rod acquires electrons from of charge on each sphere approaches the sphere farther away. This leaves the near (c) uniformity. 698 | CHAPTER 21 The Electric Field I: Discrete Charge Distributions + + + + – + + – + – – + –– +++ + –– –– – – + –––– +++ + –––– + + –– –– – –– ++ –– + –– + –– + –– – + + + + + Symbol for + ground (a) (b) (c) (d) FIGURE 21-4 Induction via grounding. (a) The free charge from the ground neutralize the positive charge on the far face. The on the single neutral conducting sphere is polarized by the conductor is then negatively charged. (c) The negative charge positively charged rod, which attracts negative charges on the remains if the connection to the ground is broken before the rod is sphere. (b) When the conductor is grounded by connecting it removed. (d) After the rod is removed, the sphere has a uniform with a wire to a very large conductor, such as Earth, electrons negative charge. sphere with a net negative charge and the far sphere with an equal net positive charge. A conductor that has separated equal and opposite charges is said to be po- ✓ CONCEPT CHECK 21-2 larized. If the spheres are separated before the rod is removed, they will be left Two identical conducting spheres with equal amounts of opposite charges (Figure 21-3b). A similar result would be are charged by induction and then obtained with a negatively charged rod, which would drive electrons from the near separated by a large distance; sphere to the far sphere. sphere 1 has charge Q and For many purposes, Earth itself can be modeled as an infinitely large conductor sphere 2 has charge Q. A third that has an infinite supply of charged particles. If a conductor is electrically con- identical sphere is initially un- nected to Earth it is said to be grounded. Grounding a metal sphere is indicated charged. If sphere 3 is touched to schematically in Figure 21-4b by a connecting wire ending in parallel horizontal sphere 1 and separated, then lines. Figure 21-4 demonstrates how we can induce a charge in a single conductor touched to sphere 2 and sepa- by transferring charge from Earth through a ground wire and then breaking the rated, what is the final charge on connection to the ground. (In practice, a person standing on the floor and touching each of the three spheres? the sphere with his hand provides an adequate ground for electrostatic demon- strations such the one described here.) The lightning rod on this building is These fashionable ladies are wearing hats with grounded so that it can conduct electrons metal chains that drag along the ground, from the ground to the positively which were supposed to protect them from charged clouds, thus neutralizing them. lightning. (Ann Roman Picture Library.) (© Grant Heilman.)  Coulomb’s Law S E C T I O N 2 1- 3 | 699 Duplex wall outlet Ground conectors Grounding terminal Grounding rod Ground F I G U R E 2 1 - 5 The two round ground connectors of a duplex 120-volt wall outlet are connected to an 8-ft-long metal grounding rod by a copper wire. The grounding rod is driven into the ground. 21-3 COULOMB’S LAW Charles Coulomb (1736–1806) studied the force exerted by one charge on another using a torsion balance of his own invention.* In Coulomb’s experiment, the charged spheres were much smaller than the distance between them so that the charges could be treated as point charges. Coulomb used the method of charging by induction to produce equally charged spheres and to vary the amount of charge on the spheres. For example, beginning with charge q0 on each sphere, he could reduce the charge to 12 q0 by temporarily grounding one sphere to discharge it, dis- connecting it from ground, and then placing the two spheres in contact. The results of the experiments of Coulomb and others are summarized in Coulomb’s law: The force exerted by one point charge on another acts along the line be- tween the charges. It varies inversely as the square of the distance separat- ing the charges and is proportional to the product of the charges. The force is repulsive if the charges have the same sign and attractive if the charges have opposite signs. C OU L O M B ’ S L AW Coulomb’s torsion balance. (Bundy Library, Norwalk, CT.) * Coulomb’s experimental apparatus was essentially the same as that described for the Cavendish experiment in Chapter 11, with the masses replaced by small charged spheres. For the magnitudes of charges easily transferred by rubbing, the gravitational attraction of the spheres is completely negligible compared with their electric attraction or repulsion. 700 | CHAPTER 21 The Electric Field I: Discrete Charge Distributions The magnitude of the electric force exerted by a point charge q1 on another point charge q2 a distance r away is thus given by q1 q2 r12 = r2 – r1 r1 k ƒ q1 q2 ƒ r2 F 21-2 O r2 C OU L O M B ’ S L AW F O R T H E M AG N IT U D E O F T H E F O RC E E X E RT E D BY q1 O N q2 (a) rˆ12 q1 q2 where k is an experimentally determined positive constant called the Coulomb constant, which has the value kq1q2 r12 F12 = rˆ12 k  8.99  109 N # m 2>C 2 2 r12 21-3 S S S (b) If q1 is at position r1 and q2 is at r2 (Figure 21-6), the force F 1 2 exerted by q1 on q2 is S FIGURE 21-6 (a) Charge q1 at position r1 S S kq1 q2 and charge q2 at r relative to the origin O. F1 2  rn1 2 21-4 S 2 (b) The force F1 2 exerted by q1 on q2 is in the r21 2 S S S direction of the vector r1 2  r2  r1 if both C OU L O M B ’ S L AW ( V E C TO R F O R M ) charges have the same sign, and in the opposite direction if they have opposite signs. where r1 2  r2  r1 is the vector pointing from q1 to q2 , and rn1 2  r1 2 >r1 2 is a unit The unit vector rn 1 2  r1 2 >r1 2 is in the direction S S S S S vector in the same direction. from q1 to q2. S In accord with Newton’s third law, the electrostatic force F 2 1 exerted by q2 on q1 S is the negative of F1 2. Note the similarity between Coulomb’s law and Newton’s law of gravity. (See Equation 11-3.) Both are inverse-square laws. But the gravita- tional force between two particles is proportional to the masses of the particles and ! Equation 21-4 gives the correct direction for the force, whether or not the two charges are both positive, is always attractive, whereas the electric force is proportional to the charges of the both negative, or one positive and one particles and is repulsive if the charges have the same sign and attractive if they negative. have opposite signs. Example 21-2 Electric Force in Hydrogen In a hydrogen atom, the electron is separated from the proton by an average distance of about 5.3  1011 m. Calculate the magnitude of the electrostatic force of attraction exerted by the proton on the electron. PICTURE Assign the proton as q1 and the electron as q2. Use Coulomb’s law to determine the magnitude of the electrostatic force of attraction exerted by the proton on the electron. SOLVE 1. Sketch the electron and the proton and label the Proton Electron sketch with the suitable symbols (Figure 21-7): q1 q2 F r q1 = +e q2 = −e FIGURE 21-7 k ƒ q1q2 ƒ ke2 (8.99  109 N # m2>C2)(1.60  1019 C)2 2. Use the given information and Equation 21-2 F   r 2 r 2 (5.3  1011 m)2 (Coulomb’s law) to calculate the electrostatic force:  8.2  108 N CHECK The order of magnitude is plausible. The powers of ten in the numerator combined are 109  1038  1029, the power of ten in the denominator is 1022, and 1029>1022  107. In comparison, 8.2  108  107.  Coulomb’s Law S E C T I O N 2 1- 3 | 701 TAKING IT FURTHER Compared with macroscopic interactions, this is a very small force. However, because the mass of the electron is only about 1030 kg, this force produces an ac- celeration of F>m  9  1022 m>s2. The proton is almost 2000 times more massive than the electron, so the acceleration of the proton is about 4  1019 m>s2. To put these accelerations in perspective, the acceleration due to gravity g is a mere 101 m>s2. PRACTICE PROBLEM 21-3 Two point charges of 0.0500 mC each are separated by 10.0 cm. Find the magnitude of the force exerted by one point charge on the other. Because the electrical force and the gravitational force between any two parti- cles both vary inversely with the square of the distance between the particles, the ratio of these forces is independent of that distance. We can therefore compare the relative strengths of the electrical and gravitational forces for elementary particles such as the electron and proton. Example 21-3 Ratio of Electric and Gravitational Forces Compute the ratio of the electric force to the gravitational force exerted by a proton on an electron in a hydrogen atom. PICTURE Use Coulomb’s law and q1  e and q2  e to find the electric force. Use Newton’s law of gravity, the mass of the proton, mp  1.67  1027 kg, and the mass of the electron, me  9.11  1031 kg, to find the gravitational force. SOLVE ke2 Gmpme 1. Express the magnitudes of the electric force Fe and the Fe  Fg  r2 r2 gravitational force Fg in terms of the charges, masses, separation distance r, and electrical and gravitational constants: Fe ke2 2. Determine the ratio. Note that the separation distance r cancels:  Fg Gmpme Fe (8.99  109 N # m2>C 2)(1.60  1019 C)2 (6.67  1011 N # m2>kg 2)(1.67  1027 kg)(9.11  1031 kg) 3. Substitute numerical values:  Fg  2.27  1039 CHECK In the numerator of the fraction in step 3, the coulomb units cancel out. In the de- nominator of the fraction, the kilogram units cancel out. The result is that both numerator and denominator have units of N # m2. The fraction has no units, as expected for a ratio of two forces. TAKING IT FURTHER The fact that the ratio (step 3) is so large reveals why the effects of gravity are not considered when discussing atomic or molecular interactions. Although the gravitational force is incredibly weak compared with the electric force and plays essentially no role at the atomic level, it is the dominant force be- tween large objects such as planets and stars. Because large objects contain almost equal numbers of positive and negative charges, the attractive and repulsive elec- trical forces cancel. The net force between astronomical objects is therefore essen- tially the force of gravitational attraction alone. 702 | CHAPTER 21 The Electric Field I: Discrete Charge Distributions FORCE EXERTED BY A SYSTEM OF CHARGES In a system of charges, each charge exerts a force, given by Equation 21-4, on every other charge. The net force on any charge is the vector sum of the individual forces See exerted on that charge by all the other charges in the system. This result follows Math Tutorial for more from the principle of superposition of forces. information on Trigonometry Example 21-4 Electric Force on a Charge Three point charges lie on the x axis; q1 is at the origin, q2 is at x  2.0 m, and q0 is at posi- tion x (x 2.0 m). (a) Find the total electric force on q0 due to q1 and q2 if q1  25 nC, q2  10 nC, q0  20 nC, and x  3.5 m. (b) Find an expression for the total electric force on q0 due to q1 and q2 throughout the region 2.0 m x. S PICTURE The total electric force on q0 is the vector sum of the force F1 0 exerted by q1 S and the force F2 0 exerted by q2. The individual forces are found using Coulomb’s law and the principle of superposition. Note that rn1 0  rn2 0  in because both rn1 0 and rn2 0 are in the x direction. SOLVE y, m (a) 1. Draw a sketch of the system of charges (Figure 21-8a). Identify the distances r1 0 and r2 0 on the graph: q2 = –10 nC + – + x, m 1 2 3 4 q1 = +25 nC q0 = +20 nC FIGURE 21-8a k ƒ q1q0 ƒ 2. Find the force exerted by q1 on q0. These F1 0  r21 0 charges have the same sign, so they repel. The force is in the x direction: S k ƒ q1q0 ƒ (8.99  109 N # m2>C 2)(25  109 C)(20  109 C) F1 0  F1 0 in   in  in r21 0 (3.5 m)2  (0.37  106 N)in k ƒ q2q0 ƒ 3. Find the force exerted by q2 on q0. These F2 0  r22 0 charges have opposite signs, so they attract. The force is in the x direction: S k ƒ q2q0 ƒ (8.99  109 N # m2>C 2)(10  109 C)(20  109 C) F2 0  F2 0 in   in   in r22 0 (1.5 m)2  (0.80  106 N)in S S S 4. Combine your results to obtain the net force. Fnet  F1 0  F2 0  (0.43  106 N)in (b) 1. Draw a sketch of the system of charges. y, m Label the distances r1 0 and r2 0 (Figure 21-8b): r10 = x r20 = x − 2.0 m 2.0 m + − + x, m 1 2 3 4 q1 q2 q0 FIGURE 21-8b  Coulomb’s Law S E C T I O N 2 1- 3 | 703 S k ƒ q1q0 ƒ 2. Find an expression for the force on q0 due F1 0  in x2 to q1. S k ƒ q2q0 ƒ 3. Find an expression for the force on q0 F2 0   in (x  2.0 m)2 due to q2. k ƒ q1q0 ƒ k ƒ q2 q0 ƒ Fnet  F1 0  F2 0  a b in S S S 4. Combine your results to obtain an  x2 (x  2.0 m)2 expression for the net force. CHECK In steps 2, 3, and 4 of Part (b) , both forces approach zero as x S , as expected. In addition, the magnitude of the step-3 result approaches infinity as x S 2.0 m, also as expected. TAKING IT FURTHER The charge q2 is located between charges q1 and q0 , so you might Fx , mN S think that the presence of q2 will affect the force F1 0 exerted by q1 on q0. However, this is S not the case. That is, the presence of q2 does not effect the force F1 0 exerted by q1 on q0. (That 0.1 this is so is called the principle of superposition.) Figure 21-9 shows the x component of the force on q0 as a function of the position x of q0 throughout the region 2.0 m x. Near q2 0 x, m the force due to q2 dominates, and because opposite charges attract the force on q2 is in the x 2 5 10 15 direction. For x W 2.0 m the force is in the x direction. This is because for large x the dis- −0.1 tance between q1 and q2 is negligible so the force due to the two charges is almost the same as that for a single charge of 15 nC. −0.2 −0.3 PRACTICE PROBLEM 21-4 If q0 is at x  1.0 m, find the total electric force acting on q0. −0.4 For the charges in a system to remain stationary, there must be forces, other than the electric forces the charges exert on each other, acting on the charges so that the −0.5 net force on each charge is zero. In the preceding example, and those that follow throughout the book, we assume that there are such forces so that all the charges FIGURE 21-9 remain stationary. Example 21-5 Summing Forces in Two Dimensions Charge q1  25 nC is at the origin, charge q2  15 nC is on the x axis at x  2.0 m, and charge q0  20 nC is at the point x  2.0 m, y  2.0 m as shown in Figure 21-10. Find the magnitude and direction of the resultant electric force on q0. PICTURE The resultant electric force is the vector sum of the individual forces exerted by each charge on q0. We compute each force from Coulomb’s law and write it in terms of its rectangular components. SOLVE y, m 1. Draw the coordinate axes showing the positions of the three S 3 charges. Show the resultant electric force F on charge q0 as q0 = +20 nC S S F1 0 the vector sum of the forces F1 0 due to q1 and F2 0 due to q2 (Figure 21-10a): 2 + F1 0 + F2 0 = Fnet 1 F2 0 + – x, m 1 2 3 4 q1 = +25 nC, q2 = –15 nC FIGURE 21-10a 704 | CHAPTER 21 The Electric Field I: Discrete Charge Distributions S S S S 2. The resultant force F on q0 is the sum of the individual F  F1 0  F2 0 forces: so ©Fx  F1 0 x  F2 0 x and ©Fy  F1 0 y  F2 0 y S k ƒ q1q0 ƒ (8.99  109 N # m2>C 2)(25  109 C)(20  109 C) 3. The force F1 0 is directed away from the origin along the F1 0   r21 0 (2.0 22 m)2 line from q1 to q0. Use r1 0  2.012 m for the distance between q1 and q0 to calculate its magnitude:  5.62  107 N S 4. Because F1 0 makes an angle of 45° with the x and y axes, F1 0 x  F1 0 y  F1 0 cos 45°  (5.62  107N) cos 45° its x and y components are equal to each other:  3.97  107 N S S k ƒ q2 q0 ƒ (8.99  109 N # m2>C2)(15  109 C)(20  109 C) 5. The force F2 0 exerted by q2 on q0 is attractive and in the F2 0   jn   jn r22 0 (2.0 m)2 y direction as shown in Figure 21-10a:  (6.74  107N)jn 6. Calculate the components of the resultant force: Fx  F1 0 x  F2 0 x  (3.97  107 N)  0  3.97  107 N Fy  F1 0 y  F2 0 y  (3.97  107 N)  (6.74  107 N) Fy  2.77  107 N 7. Draw the resultant force (Figure 21-10b) along with its y two components: q0 Fx = 3.97 * 10–7 N + x θ F Fy = –2.77 * 10–7 N FIGURE 21-10b 8. The magnitude of the resultant force is found from its F  3F 2x  F 2y  4(3.97  107 N)2  (2.77  107 N)2 components:  4.84  107 N  4.8  107 N Fy 2.77 9. The resultant force points to the right and downward as tan u    0.698 Fx 3.97 shown in Figure 21-10b, making an angle u with the x axis given by: u  tan1 (0.698)  34.9°  35° CHECK We expect the two forces to be approximately equal in magnitude because even though q1 is a bit larger than ƒ q2 ƒ , q2 is a bit closer to q0 than is q1. Comparing the results of steps 3 and 5 shows agreement with this expectation. PRACTICE PROBLEM 21-5 Express rn 1 0 in Example 21-5 in terms of in and jn. PRACTICE PROBLEM 21-6 In Example 21-5, is the x component of the force F1 0  (kq1q0 >r21 0)rn1 0 equal to kq1q0 >x 21 0 (where x1 0 is the x component of rn1 0 )? S 21-4 THE ELECTRIC FIELD The electric force exerted by one charge on another is an example of an action-at- a-distance force, similar to the gravitational force exerted by one mass on another. The idea of action at a distance presents a difficult conceptual challenge. What is  The Electric Field S E C T I O N 2 1- 4 | 705 the mechanism by which one particle can exert a force on another across the empty – space between the particles? Suppose that a charged particle at some point is sud- F q2 denly moved. Does the force exerted on the second particle some distance r away change instantaneously? To address the challenge of action at a distance, the con- F = F10 + F20 + F30 S F20 cept of the electric field is introduced. One charge produces an electric field E everywhere in space, and this field exerts the force on the second charge. Thus, it S F30 is the field E at the location of the second charge that exerts the force on it, not the + F10 first charge itself (which is some distance away). Changes in the field propagate q0 through space at the speed of light, c. Thus, if a charge is suddenly moved, the + force it exerts on a second charge a distance r away does not change until a time q3 + r>c later. q1 Figure 21-11a shows a set of point charges q1 , q2 , and q3 arbitrarily arranged in (a) S space. These charges produce an electric field E everywhere in space. If we place a small positive test charge q0 at some point near the three charges, there – will be a force exerted on q0 due to the other charges. The net force on q0 is the vec- q2 tor sum of the individual forces exerted on q0 by the other charges in the system. Because each of these forces is proportional to q0 , the net force will be proportional S to q0. The electric field E at a point is this force divided by q0 :* F02 S S F E (q0 is small) 21-5 + q0 q0 D E F I N IT I O N – E L E C T R I C F I E L D + F03 F01 q3 The SI unit of the electric field is the newton per coulomb (N>C). In addition, + the test charge q0 will exert a force on each of the other point charges (Figure q1 21-11b). Because these forces on the other charges might cause some of the other (b) charges to move, the charge q0 must be so small that the forces it exerts on the other charges are negligible. Thus, the electric field at the location in question is actually FIGURE 21-11 (a) A small test charge q0 defined by Equation 21-5, but in the limit that q0 approaches zero. Table 21-2 lists in the vicinity of a system of charges the magnitudes of some of the electric fields found in nature. q1 , q2 , q3 , Á , experiences a resultant S The electric field describes the condition in space set up by the system of point electric force F that is proportional to q0. The ratio F>q0 is the electric field at that point. S S charges. By moving a test charge q0 from point to point, we can find E at all points (b) The test charge q0 also exerts a force on S in space (except at any point occupied by a charge q ). The electric field E is thus a each of the surrounding charges, and each of vector function of position. The force exerted on a test charge q0 at any point is re- these forces is proportional to q0. lated to the electric field at that point by S S F  q0 E 21-6 Table 21-2 Some Electric Fields in Nature PRACTICE PROBLEM 21-7 E, N/C When a 5.0-nC test charge is placed at a certain point, it experiences a force of S 2.0  104 N in the direction of increasing x. What is the electric field E at that point? In household wires 102 PRACTICE PROBLEM 21-8 In radio waves 101 What is the force on an electron placed at a point where the electric field is In the atmosphere 102 S E  (4.0  104 N>C)in? In sunlight 103 Under a thundercloud 104 The electric field due to a single point charge can be calculated from Coulomb’s In a lightning bolt 104 law. Consider a small, positive test charge q0 at some point P a distance riP away In an X-ray tube 106 from a charge qi. The force on q0 is At the electron in a hydrogen atom 5  1011 S kqiq0 Fi0  rniP At the surface of a r2iP uranium nucleus 2  1021 * This definition is similar to that for the gravitational field of Earth, which was defined in Section 4-3 as the force per unit mass exerted by Earth on an object. 706 | CHAPTER 21 The Electric Field I: Discrete Charge Distributions The electric field at point P due to charge qi (Figure 21-12) is thus EiP rˆiP S kqi riP EiP  rniP 21-7 Field point P r2iP S C OU L O M B ’ S L AW F O R E + qi Source point i where rniP is the unit vector pointing from the source point i to the field point P. S FIGURE 21-12 The electric field E at a The resultant electric field at P due to a distribution of point charges is found by field point P due to charge qi at a source point i. summing the fields due to each charge separately: S S E P  a EiP i S 21-8 E L E C T R I C F I E L D E D U E TO A SYST E M O F P O I N T C H A RG E S ! Even though the expression for the electric field (Equation 21-7) does depend on the location of point P, it does not depend on the test charge q0. That is, electric fields follow the principle of superposition. That is, q0 itself does not appear in Equation 21-7. PROBLEM-SOLVING STRATEGY Calculating the Resultant Electric Field S PICTURE To calculate the resultant electric field E P at field point P due to a specified distribution of point charges, draw the charge configuration. Include coordinate axes and the field point on the drawing. SOLVE 1. On the drawing label the distance riP from each charge to point P. Include S an electric field vector E iP for the electric field at P due to each point charge. 2. If the field point P and the point charges are not on a single line, then S label the angle each individual electric field vector E iP makes with one of the coordinate axes. S 3. Calculate the component of each individual field vector E iP along each axis and use these to calculate the components of the resultant electric S field E P. Example 21-6 Direction of Electric Field Conceptual A positive point charge q1  q and a negative point charge of q2  2q are located on the x axis at x  a and x  a, respectively, as shown in Figure 21-13. Consider the fol- lowing regions on the x axis: region I (x a), region II (a x a), and region III (x a). In which region, or regions, is there a point at which the resultant electric field is equal to zero? S S PICTURE Let E 1 and E 2 be the electric fields due to q1 and q2, respectively. Because q1 is pos- S S itive, E 1 points away from q1 everywhere, and because q2 is negative, E 2 points toward from S q2 everywhere. The resultant electric field E is equal to the sum of the electric fields of the S S S S S two charges (E  E 1  E 2). The resultant field is zero if E 1 and E 2 are equal in magnitude and oppositely directed. The magnitude of the electric field due to a point charge approaches infinity at points close to a point charge. In addition, at points far from the charge configu- ration, the electric field approaches the electric field of a point charge equal to q1  q2 that is located at the center of charge. The electric field far from the charge configuration is that of a negative point charge because q1  q2 is negative.  The Electric Field S E C T I O N 2 1- 4 | 707 SOLVE y 1. Sketch a figure showing the two charges, the x axis, and the electric fields due the charges at points on the E1 E2 q2 = −2q E2 E1 q1 = +q E2 E1 x axis in each of regions I, II, and III. Label these x PI −a PII +a PIII points PI , PII , and PIII , respectively (Figure 21-13): 0 Region I Region II Region III FIGURE 21-13 2. Check to see if the two electric field vectors can be Throughout region I, the two electric field vectors are oppositely directed. equal in magnitude and opposite in direction However, each point in the region is closer to q2 ( 2q) than q1 ( q), anywhere in region I: so E2 is greater than E1 at each point in the region. Thus, in region I there are no points where the electric field is equal to zero. 3. Check to see if the two electric field vectors can be Throughout region II, the two electric field vectors are in the same equal in magnitude and opposite in direction direction at each point on the x axis. Thus, in region II there are no points anywhere in region II: where the electric field is equal to zero. 4. Check to see if the two electric field vectors can be Throughout region III, the two electric field vectors are oppositely equal in magnitude and opposite in direction directed. At points very close to x  a, E1 is greater than E2 (because at anywhere in region III: points close to a point charge the magnitude of the electric field due to that charge approaches infinity). However, at points where x W a, E2 is greater than E1 (because at large distances from the two charges the field direction is determined by the sign of q1  q2). Thus, there must be a point somewhere in region III where E1 is equal to E2. At that point the net electric field is zero. S CHECK The resultant electric field is zero at a point in region III, the region in which E 1 and S E 2 are oppositely directed AND in which all points are farther from q2 , the charge with the larger magnitude, than from q1. This result is as one would expect. Example 21-7 Electric Field on a Line through Two Positive Point Charges A positive point charge q1  8.0 nC is on the x axis at x  x1  1.0 m, and a second pos- itive point charge q2  12 nC is on the x axis at x  x2  3.0 m. Find the net electric field (a) at point A on the x axis at x  6.0 m, and (b) at point B on the x axis at x  2.0 m. S S PICTURE Let E 1 and E 2 be the electric fields due to q1 and q2, respectively. Because q1 is S S positive, E 1 points away from q1 everywhere, and because q2 is positive, E 2 points away S S S from q2 everywhere. We calculate the resultant field using E  E 1  E 2. SOLVE q1 = +8.0 nC q2 = +12 nC (a) 1. Draw the charge configuration and place the q1 q2 E2 E1 E1 E2 field point A on the x axis at the appropriate + + place. Draw vectors representing the electric −2 0 2 4 6 x, m field at A due to each point charge. Repeat this B A procedure for field point B (Figure 21-14): S FIGURE 21-14 Because q1 is a positive charge, E 1 points away from q1 , at S both point A and point B. Because q2 is a positive charge, E 2 points away from q2 at both point A and point B. S S S S kq1 kq2 kq1 kq2 2. Calculate E at point A, using E  E1  E2  rn  rn  in  in r21A 1A r22A 2A (xA  x1) 2 (xA  x2)2 r1A  ƒ xA  x1 ƒ  6.0 m  (1.0 m)  7.0 m and r2A  ƒ xA  x2 ƒ  6.0 m  (3.0 m)  3.0 m: (8.99  109 N # m2>C2)(8.0  109 C) (8.99  109 N # m2>C 2)(12  109 C)  2 in  in (7.0 m) (3.0 m)2  (1.47 N>C)in  (12.0 N>C)in  (13 N>C)in 708 | CHAPTER 21 The Electric Field I: Discrete Charge Distributions S S S S kq1 kq2 kq1 kq2 (b) Calculate E at point B, where E  E1  E2  rn  rn  in  (in) r21B 1B r22B 2B (xB  x1) 2 (xB  x2)2 r1B  ƒ xB  x1 ƒ  2.0 m  (1.0 m)  3.0 m and r2B  ƒ xB  x2 ƒ  ƒ 2.0 m  (3.0 m) ƒ  1.0 m: (8.99  109 N # m2>C 2)(8.0  109 C) (8.99  109 N # m2>C 2)(12  109 C)  in  in (3.0 m)2 (1.0 m)2  (7.99 N>C)in  (108 N>C)in  (100 N>C)in CHECK The Part (b) result is large and in the x direction. This result is expected because point B is close to q2 , and q2 is a large positive charge (12 nC) that produces electric field S E 2 in the x direction at B. TAKING IT FURTHER The resultant electric field at source Ex , N/C S points close to q1  8.0-nC is dominated by the field E 1 due to q1. There is one point between q1 and q2 where the re- 600 sultant electric field is zero. A test charge placed at this 400 point would experience no electric force. A sketch of Ex ver- sus x for this charge configuration is shown in Figure 21-15. 200 PRACTICE PROBLEM 21-9 Regarding Example 21-7, find the point on the x axis where the electric field is zero. –3 –2 –1 0 1 2 3 4 5 x, m – 200 – 400 – 600 FIGURE 21-15 Example 21-8 Electric Field Due to Point Charges on the x Axis Try It Yourself A point charge q1  8.0 nC is at the origin and a second point charge q2  12.0 nC is on the x axis at x  4.0 m. Find the electric field on the y axis at y  3.0 m. S S S PICTURE As in Example 21-7, E  E 1  E 2. Answers S At points on the y axis, the electric field E 1 due y, m to charge q1 is directed along the y axis, and the S field E 2 due to charge q2 is in the second quad- S rant. To find the resultant field E , we first find S E1 the x and y components of E. SOLVE θ Cover the column to the right and try these E2 P on your own before looking at the answers. 3 Steps θ 1. Sketch the two charges and the field 2 point. Include the coordinate axes. Draw the electric field due to each charge at the 5.0 m field point and label distances and angles appropriately (Figure 21-16a): 1 q2 = +12 nC + + x, m q1 = +8.0 nC 1 2 3 4 5 FIGURE 21-16a  The Electric Field S E C T I O N 2 1- 4 | 709 E1  kq1 >y2  7.99 N>C S 2. Calculate the magnitude of the field E 1 at (0, 3.0 m) S due to q1. Find the x and y components of E 1 E1x  0, E1y  E1  7.99 N>C S 3. Calculate the magnitude of the field E 2 at (0, y) E2  4.32 N>C due to q2. S 4. Write the x and y components of E 2 in terms of E2x  E2 sin u; E2y  E2 cos u the angle u. 5. Compute sin u and cos u. sin u  0.80; cos u  0.60 6. Calculate E2x and E2y. E2x  3.46 N>C; E2y  2.59 N>C 7. Sketch the components of the resultant field. y S S Include both the vector E and the angle E makes E with the x axis (Figure 21-16b): Ey θ1 Ex x P FIGURE 21-16b 8. Find the x and y components of the resultant Ex  E1x  E2x  3.46 N>C S field E. Ey  E1y  E2y  10.6 N>C S 9. Calculate the magnitude of E from its E  2E 2x  E 2y  11.2 N>C  11 N>C components. S Ey 10. Find the angle u1 made by E with the x axis. u1  tan1 ¢ ≤  108° Ex CHECK As expected, E is larger than either E1 or E2 , but less than E1  E2. (This result is S S expected because the angle between E 1 and E 2 is less than 90°.) Example 21-9 Electric Field Due to Two Equal and Opposite Charges A charge q is at x  a and a second charge q is at x  a (Figure 21-17). (a) Find the elec- tric field on the x axis at an arbitrary point x a. (b) Find the limiting form of the electric field for x W a. PICTURE We calculate the electric field at point P using the principle of superposition, S S S S E P  E 1P  E 2P. For x a, the electric field E  due to the positive charge is in the x di- S rection and the electric field E  due to the negative charge is in the x direction. The dis- tances are x  a to the positive charge and x  (a)  x  a to the negative charge. SOLVE y (a) 1. Draw the charge configuration on a coordinate axis and label the distances from each charge to the field x+a point (Figure 21-17): x a a x−a E– E+ − + P x −q +q FIGURE 21-17 710 | CHAPTER 21 The Electric Field I: Discrete Charge Distributions kq kq ain b S S S S 2. Calculate E due to the two charges for x a: (Note: The E  E  E  in  [x  a] 2 [x  (a)]2 equation on the right holds only for x a.  kq c d in 1 1  (x  a)2 (x  a)2 (x  a)2  (x  a)2 E  kq c d in  kq 2 S 4ax 3. Put the terms in square brackets under a common in x a (x  a)2(x  a)2 (x  a2)2 denominator and simplify: S 4ax 4ax 4kqa (b) In the limit x W a, we can neglect a2 compared with x 2 in the E  kq in  kq 4 in  in xWa (x  a ) 2 2 2 x x3 denominator: CHECK Both boxed answers approach zero as x approaches infinity, which is as expected. TAKING IT FURTHER Figure 21-18 shows Ex versus x for all x, for q  1.0 nC and a  1.0 m. For ƒ x ƒ W a (far from the charges), the field is given by S 4kqa E in ƒxƒ W a ƒxƒ3 Between the charges, the contribution from each charge is in the negative direction. An S expression for E is S kq k(q) E en  en a x a (x  a)2 (x  a)2 where en is a unit vector that points away from the point x  a for all values of x (except x  a) and en is a unit vector that points away from the point x  a for all values of x xa n xa n (except x  a). (Note that en   i and en  i.) ƒx  aƒ ƒx  aƒ Ex, N/C 200 100 – + x, cm –3 –2 –1 1 2 3 –100 F I G U R E 2 1 - 1 8 A plot of Ex versus x on the x axis for the charge distribution in –200 Example 21-9. ELECTRIC DIPOLES A system of two equal and opposite charges q separated by a small distance L is S called a dipole. Its strength and orientation are described by the dipole moment p, p which is a vector that points from the negative charge q toward the positive p = qL S charge q and has the magnitude qL (Figure 21-19): L S –q – + +q S p  qL 21-9 D E F I N IT I O N — D I P O L E M O M E N T F I G U R E 2 1 - 1 9 A dipole consists of a pair of equal and Sopposite charges. The dipole S S where L is the position of the positive charge relative to the negative charge. moment is p  qL , where q is the S S For the system of charges in Figure 21-17, L  2a in and the dipole moment is magnitude of one of the charges and L is the position of the negative charge relative to p  2aq in S the positive charge.  Electric Field Lines S E C T I O N 2 1- 5 | 711 S In terms of the dipole moment p, the electric field on the axis of the dipole at a S point a great distance ƒ x ƒ away is in the same direction as p and has magnitude 2kp E 3 21-10 ƒxƒ (see Example 21-9). At a point far from a dipole in any direction, the magnitude of the electric field is proportional to the magnitude of the dipole moment and de- creases with the cube of the distance. If a system has a nonzero net charge, the elec- + tric field decreases as 1>r 2 at large distances. In a system that has zero net charge, the electric field falls off more rapidly with distance. In the case of a dipole, the field falls off as 1>r 3 in all directions. 21-5 ELECTRIC FIELD LINES We can visualize the electric field by drawing a number of directed curved lines, (a) called electric field lines, to indicate both the magnitude and the direction of the S field. At any given point, the field vector E is tangent to the line through that point. (Electric field lines are also called lines of force because they show the direction of the electric force exerted on a positive test charge.) At points very near a positive point S charge, the electric field E points directly away from the charge. Consequently, the electric field lines very near a positive charge also point directly away from t