Chapter 15 Electric Forces and Fields PDF

44920_15_p497-530 12/29/04 9:18 AM Page 497 This nighttime view of multiple bolts...

44920_15_p497-530 12/29/04 9:18 AM Page 497 This nighttime view of multiple bolts of lightning was photographed in Tucson, Arizona. During a thunder- storm, a high concentration of electrical charge in a thundercloud creates a higher-than-normal electric field between the thundercloud and the negatively charged Earth’s surface. This strong electric field creates an electric discharge between the charged cloud and the ground — an enormous spark. Other discharges that are observed in the sky include cloud-to-cloud discharges and the more frequent intracloud © Keith Kent/Science Photo Library/Photo Researchers, Inc. discharges. 15 CHAPTER Electric Forces and 15.1 O U T L I N E Properties of Electric Electric Fields Charges 15.2 Insulators and Conductors 15.3 Coulomb’s Law 15.4 The Electric Field Electricity is the lifeblood of technological civilization and modern society. Without it, we 15.5 Electric Field Lines revert to the mid-nineteenth century: no telephones, no television, none of the household 15.6 Conductors in Electrostatic appliances that we take for granted. Modern medicine would be a fantasy, and due to the Equilibrium lack of sophisticated experimental equipment and fast computers — and especially the slow 15.7 Millikan’s Oil-Drop dissemination of information — science and technology would grow at a glacial pace. Experiment Instead, with the discovery and harnessing of electric forces and fields, we can view 15.8 The Van de Graaff arrangements of atoms, probe the inner workings of the cell, and send spacecraft beyond the Generator limits of the solar system. All this has become possible in just the last few generations of 15.9 Electric Flux and human life, a blink of the eye compared to the million years our kind spent foraging the Gauss’s Law savannahs of Africa. Around 700 B.C. the ancient Greeks conducted the earliest known study of electricity. It all began when someone noticed that a fossil material called amber would attract small objects after being rubbed with wool. Since then we have learned that this phenomenon is not re- stricted to amber and wool, but occurs (to some degree) when almost any two nonconduct- ing substances are rubbed together. In the current chapter we use the effect of charging by friction to begin an investigation of electric forces. We then discuss Coulomb’s law, which is the fundamental law of force be- tween any two stationary charged particles. The concept of an electric field associated with charges is introduced and its effects on other charged particles described. We end with dis- cussions of the Van de Graaff generator and Gauss’s law. 15.1 PROPERTIES OF ELECTRIC CHARGES After running a plastic comb through your hair, you will find that the comb attracts bits of paper. The attractive force is often strong enough to suspend the 497 44920_15_p497-530 12/29/04 9:18 AM Page 498 498 Chapter 15 Electric Forces and Electric Fields Figure 15.1 (a) A negatively charged rubber rod, suspended by a thread, is attracted to a positively charged glass rod. (b) A negatively charged rubber rod is repelled by another negatively charged rubber rod. Rubber Rubber F – – – – –– – –– – F F + + Glass – – Rubber + + + + – –– – + – F (a) (b) paper from the comb, defying the gravitational pull of the entire Earth. The same effect occurs with other rubbed materials, such as glass and hard rubber. © American Philosophical Society/AIP Image not Available Another simple experiment is to rub an inflated balloon against wool (or across your hair). On a dry day, the rubbed balloon will then stick to the wall of a room, often for hours. These materials have become electrically charged. You can give your body an electric charge by vigorously rubbing your shoes on a wool rug or by sliding across a car seat. You can then surprise and annoy a friend or co-worker with a light touch on the arm, delivering a slight shock to both yourself and your victim. (If the co-worker is your boss, don’t expect a promotion!) These experi- ments work best on a dry day because excessive moisture can facilitate a leaking BENJAMIN FRANKLIN away of the charge. (1706–1790) Experiments also demonstrate that there are two kinds of electric charge, which Franklin was a printer, author, physical Benjamin Franklin (1706 – 1790) named positive and negative. Figure 15.1 illus- scientist, inventor, diplomat, and a found- trates the interaction of the two charges. A hard rubber (or plastic) rod that has ing father of the United States. His work been rubbed with fur is suspended by a piece of string. When a glass rod that has on electricity in the late 1740s changed a jumbled, unrelated set of observations into been rubbed with silk is brought near the rubber rod, the rubber rod is attracted a coherent science. toward the glass rod (Fig. 15.1a). If two charged rubber rods (or two charged glass rods) are brought near each other, as in Figure 15.1b, the force between them is repulsive. These observations may be explained by assuming that the rubber and glass rods have acquired different kinds of excess charge. We use the convention suggested by Franklin, where the excess electric charge on the glass rod is called positive and that on the rubber rod is called negative. On the basis of observations Like charges repel such as these, we conclude that like charges repel one another and unlike charges Unlike charges attract ! attract one another. Objects usually contain equal amounts of positive and nega- tive charge — electrical forces between objects arise when those objects have net negative or positive charges. Nature’s basic carriers of positive charge are protons, which, along with neu- trons, are located in the nuclei of atoms. The nucleus, about 10!15 m in radius, is surrounded by a cloud of negatively charged electrons about ten thousand times larger in extent. An electron has the same magnitude charge as a proton, but the opposite sign. In a gram of matter there are approximately 1023 positively charged protons and just as many negatively charged electrons, so the net charge is zero. Because the nucleus of an atom is held firmly in place inside a solid, protons never move from one material to another. Electrons are far lighter than protons and hence more easily accelerated by forces. Furthermore, they occupy the outer re- gions of the atom. Consequently, objects become charged by gaining or losing electrons. Charge transfers readily from one type of material to another. Rubbing the two materials together serves to increase the area of contact, facilitating the transfer process. 44920_15_p497-530 12/29/04 9:18 AM Page 499 15.2 Insulators and Conductors 499 An important characteristic of charge is that electric charge is always conserved. " Charge is conserved; Charge isn’t created when two neutral objects are rubbed together; rather, the ob- charge is quantized jects become charged because negative charge is transferred from one object to the other. One object gains a negative charge while the other loses an equal amount of negative charge and hence is left with a net positive charge. When a glass rod is rubbed with silk, as in Figure 15.2, electrons are transferred from the – – + rod to the silk. As a result, the glass rod carries a net positive charge, the silk a net – ++ + + + – negative charge. Likewise, when rubber is rubbed with fur, electrons are trans- – ferred from the fur to the rubber. – In 1909 Robert Millikan (1886 – 1953) discovered that if an object is charged, its charge is always a multiple of a fundamental unit of charge, designated by the sym- bol e. In modern terms, the charge is said to be quantized, meaning that charge occurs in discrete chunks that can’t be further subdivided. An object may have a charge of " e, " 2e, " 3e, and so on, but never 1 a fractional charge of " 0.5e or Figure 15.2 When a glass rod is " 0.22e. Other experiments in Millikan’s time showed that the electron has a rubbed with silk, electrons are charge of ! e and the proton has an equal and opposite charge of #e. Some parti- transferred from the glass to the silk. Because of conservation of charge, cles, such as a neutron, have no net charge. A neutral atom (an atom with no net each electron adds negative charge to charge) contains as many protons as electrons. The value of e is now known to be the silk, and an equal positive charge 1.602 19 $ 10!19 C. (The SI unit of electric charge is the coulomb [C].) is left behind on the rod. Also, because the charges are transferred in discrete bundles, the charges on the two objects are " e, " 2e, " 3e, and so on. 15.2 INSULATORS AND CONDUCTORS Substances can be classified in terms of their ability to conduct electric charge. In conductors, electric charges move freely in response to an electric force. All other materials are called insulators. + – Rubber + – – – – – + + – – Glass and rubber are insulators. When such materials are charged by rubbing, only –– –– –– –– + + – – the rubbed area becomes charged, and there is no tendency for the charge to + – + – move into other regions of the material. In contrast, materials such as copper, alu- minum, and silver are good conductors. When such materials are charged in some (a) Before small region, the charge readily distributes itself over the entire surface of the ma- terial. If you hold a copper rod in your hand and rub the rod with wool or fur, it – will not attract a piece of paper. This might suggest that a metal can’t be charged. – – – However, if you hold the copper rod with an insulator and then rub it with wool or – – – – fur, the rod remains charged and attracts the paper. In the first case, the electric – – – charges produced by rubbing readily move from the copper through your body – and finally to ground. In the second case, the insulating handle prevents the flow (b) Contact of charge to ground. Semiconductors are a third class of materials, and their electrical properties are somewhere between those of insulators and those of conductors. Silicon and ger- – manium are well-known semiconductors that are widely used in the fabrication of – – – – a variety of electronic devices. – – – – – – – Charging by Conduction (c) After breaking contact Consider a negatively charged rubber rod brought into contact with an insulated Figure 15.3 Charging a metallic neutral conducting sphere. The excess electrons on the rod repel electrons on the object by conduction. (a) Just before sphere, creating local positive charges on the neutral sphere. On contact, some contact, the negative rod repels the sphere’s electrons, inducing a local- electrons on the rod are now able to move onto the sphere, as in Figure 15.3, neu- ized positive charge. (b) After con- tralizing the positive charges. When the rod is removed, the sphere is left with a net tact, electrons from the rod flow onto negative charge. This process is referred to as charging by conduction. The object the sphere, neutralizing the local pos- itive charges. (c) When the rod is being charged in such a process (the sphere) is always left with a charge having removed, the sphere is left with a the same sign as the object doing the charging (the rubber rod). negative charge. 1There is strong evidence for the existence of fundamental particles called quarks that have charges of " e/3 or " 2e/3. The charge is still quantized, but in units of " e/3 rather than " e. A more complete discussion of quarks and their properties is presented in Chapter 30. 44920_15_p497-530 12/29/04 9:18 AM Page 500 500 Chapter 15 Electric Forces and Electric Fields Charging by Induction An object connected to a conducting wire or copper pipe buried in the Earth is said to be grounded. The Earth can be considered an infinite reservoir for elec- trons; in effect, it can accept or supply an unlimited number of electrons. With this idea in mind, we can understand the charging of a conductor by a process (a) known as induction. Consider a negatively charged rubber rod brought near a neutral (uncharged) conducting sphere that is insulated, so there is no conducting path to ground (Fig. + + – – 15.4). Initially the sphere is electrically neutral (Fig. 15.4a). When the negatively + – + – charged rod is brought close to the sphere, the repulsive force between the elec- – – – – + trons in the rod and those in the sphere causes some electrons to move to the side + – ++ – – of the sphere farthest away from the rod (Fig. 15.4b). The region of the sphere (b) nearest the negatively charged rod has an excess of positive charge because of the migration of electrons away from that location. If a grounded conducting wire is then connected to the sphere, as in Figure 15.4c, some of the electrons leave the sphere and travel to ground. If the wire to ground is then removed (Fig. 15.4d), the + + conducting sphere is left with an excess of induced positive charge. Finally, when + + the rubber rod is removed from the vicinity of the sphere (Fig. 15.4e), the induced – – – – – – – + – positive charge remains on the ungrounded sphere. Even though the positively + ++ charged atomic nuclei remain fixed, this excess positive charge becomes uniformly (c) distributed over the surface of the ungrounded sphere because of the repulsive forces among the like charges and the high mobility of electrons in a metal. In the process of inducing a charge on the sphere, the charged rubber rod doesn’t lose any of its negative charge because it never comes in contact with the + + sphere. Furthermore, the sphere is left with a charge opposite that of the rubber + + rod. Charging an object by induction requires no contact with the object inducing – – – + the charge. + ++ A process similar to charging by induction in conductors also takes place in in- (d) sulators. In most neutral atoms or molecules, the center of positive charge coin- cides with the center of negative charge. However, in the presence of a charged object, these centers may separate slightly, resulting in more positive charge on + one side of the molecule than on the other. This effect is known as polarization. + + The realignment of charge within individual molecules produces an induced + + charge on the surface of the insulator, as shown in Figure 15.5a. This explains why a balloon charged through rubbing will stick to an electrically neutral wall, or the + + + comb you just used on your hair attracts tiny bits of neutral paper. (e) Figure 15.4 Charging a metallic object by induction. (a) A neutral Quick Quiz 15.1 metallic sphere with equal numbers of positive and negative charges. A suspended object A is attracted to a neutral wall. It’s also attracted to a positively (b) The charge on a neutral metal charged object B. Which of the following is true about object A? (a) It is un- sphere is redistributed when a charged rubber rod is placed near charged. (b) It has a negative charge. (c) It has a positive charge. (d) It may be ei- the sphere. (c) When the sphere is ther charged or uncharged. grounded, some of the electrons leave it through the ground wire. (d) When the ground connection is removed, the nonuniformly charged 15.3 COULOMB’S LAW sphere is left with excess positive charge. (e) When the rubber rod is In 1785 Charles Coulomb (1736 – 1806) experimentally established the fundamen- moved away, the charges on the tal law of electric force between two stationary charged particles. sphere redistribute themselves until the sphere’s surface becomes uni- formly charged. An electric force has the following properties: 1. It is directed along a line joining the two particles and is inversely propor- tional to the square of the separation distance r, between them. 2. It is proportional to the product of the magnitudes of the charges, !q 1! and !q 2 !, of the two particles. 3. It is attractive if the charges are of opposite sign and repulsive if the charges have the same sign. 44920_15_p497-530 12/29/04 9:18 AM Page 501 15.3 Coulomb’s Law 501 Figure 15.5 (a) The charged Insulator object on the left induces charges on + – + the surface of an insulator. (b) A charged comb attracts bits of paper + – + because charges are displaced in the paper. + – + + – + ©1968 Fundamental Photographs + – + + – + Charged Induced object charges (a) (b) From these observations, Coulomb proposed the following mathematical form for the electric force between two charges: The magnitude of the electric force F between charges q 1 and q 2 separated " Coulomb’s law by a distance r is given by !q 1!!q2 ! F % ke [15.1] r2 where ke is a constant called the Coulomb constant. Equation 15.1, known as Coulomb’s law, applies exactly only to point charges and to spherical distributions of charges, in which case r is the distance between the two centers of charge. Electric forces between unmoving charges are called electro- static forces. Moving charges, in addition, create magnetic forces, studied in Chapter 19. The value of the Coulomb constant in Equation 15.1 depends on the choice of units. The SI unit of charge is the coulomb (C). From experiment, we know that the Coulomb constant in SI units has the value ke % 8.9875 $ 10 9 N & m2/C 2 [15.2] Photo courtesy of AIP Niels Bohr Library, E. Scott Barr Collection This number can be rounded, depending on the accuracy of other quantities in a given problem. We’ll use either two or three significant digits, as usual. The charge on the proton has a magnitude of e % 1.6 $ 10!19 C. Therefore, it Image not Available would take 1/e % 6.3 $ 1018 protons to create a total charge of # 1.0 C. Likewise, 6.3 $ 1018 electrons would have a total charge of ! 1.0 C. Compare this with the number of free electrons in 1 cm3 of copper, which is on the order of 1023. Even so, 1.0 C is a very large amount of charge. In typical electrostatic experiments in which a rubber or glass rod is charged by friction, there is a net charge on the or- der of 10!6 C ( % 1 ' C). Only a very small fraction of the total available charge is transferred between the rod and the rubbing material. Table 15.1 lists the charges and masses of the electron, proton, and neutron. CHARLES COULOMB TABLE 15.1 (1736 –1806) Charge and Mass of the Electron, Proton, and Neutron Coulomb’s major contribution to science was in the field of electrostatics and Particle Charge (C) Mass (kg) magnetism. During his lifetime, he also Electron ! 1.60 $ 10!19 9.11 $ 10!31 investigated the strengths of materials and Proton # 1.60 $ 10!19 1.67 $ 10!27 identified the forces that affect objects on Neutron 0 1.67 $ 10!27 beams, thereby contributing to the field of structural mechanics. 44920_15_p497-530 12/29/04 9:18 AM Page 502 502 Chapter 15 Electric Forces and Electric Fields When using Coulomb’s force law, remember that force is a vector quantity and r + F12 must be treated accordingly. Active Figure 15.6a shows the electric force of repul- q2 sion between two positively charged particles. Like other forces, electric forces : : + obey Newton’s third law; hence, the forces F12 and F21 are equal in magnitude but : q1 opposite in direction. (The notation F 12 denotes the force exerted by particle 1 on F21 (a) : particle 2; likewise, F21 is the force exerted by particle 2 on particle 1.) From New- ton’s third law, F 12 and F 21 are always equal regardless of whether q 1 and q 2 have – the same magnitude. q2 F12 Quick Quiz 15.2 F21 + Object A has a charge of # 2 ' C, and object B has a charge of # 6 'C. Which state- q1 ment is true? (b) : : : : : : ACTIVE FIGURE 15.6 (a) FAB % ! 3F BA (b) FAB % ! FBA (c) 3FAB % ! FBA Two point charges separated by a distance r exert a force on each other given by Coulomb’s law. The force The Coulomb force is similar to the gravitational force. Both act at a distance on q 1 is equal in magnitude and without direct contact. Both are inversely proportional to the distance squared, opposite in direction to the force on q 2. (a) When the charges are of with the force directed along a line connecting the two bodies. The mathematical the same sign, the force is repulsive. form is the same, with the masses m 1 and m 2 in Newton’s law replaced by q 1 and q 2 (b) When the charges are of opposite in Coulomb’s law and with Newton’s constant G replaced by Coulomb’s constant sign, the force is attractive. k e. There are two important differences: (1) electric forces can be either attractive or repulsive, but gravitational forces are always attractive, and (2) the electric force Log into PhysicsNow at www.cp7e.com between charged elementary particles is far stronger than the gravitational force and go to Active Figure 15.6, where you between the same particles, as the next example shows. can move the charges to any position in two-dimensional space and observe the electric forces acting on them. EXAMPLE 15.1 Forces in a Hydrogen Atom Goal Contrast the magnitudes of an electric force and a gravitational force. Problem The electron and proton of a hydrogen atom are separated (on the average) by a distance of about 5.3 $ 10!11 m. Find the magnitudes of the electric force and the gravitational force that each particle exerts on the other, and the ratio of the electric force Fe to the gravitational force Fg. Strategy Solving this problem is just a matter of substituting known quantities into the two force laws and then finding the ratio. Solution " # !e !2 9 N&m 2 (1.6 $ 10!19 C)2 Substitute !q 1! % !q 2! % e and the distance into Coulomb’s Fe % k e % 8.99 $ 10 law to find the electric force: r2 C2 (5.3 $ 10!11 m)2 % 8.2 $ 10!8 N mem p Substitute the masses and distance into Newton’s law of Fg % G gravity to find the gravitational force: r2 " N&m2 # (9.11 $ 10(5.3 $kg)(1.67 !31 $ 10 !27 kg) % 6.67 $ 10!11 kg2 !11 10 m)2 % 3.6 $ 10!47 N Fe Find the ratio of the two forces: % 2.27 $ 1039 Fg Remarks The gravitational force between the charged constituents of the atom is negligible compared with the electric force between them. The electric force is so strong, however, that any net charge on an object quickly attracts nearby opposite charges, neutralizing the object. As a result, gravity plays a greater role in the mechanics of moving objects in everyday life. 44920_15_p497-530 12/29/04 9:18 AM Page 503 15.3 Coulomb’s Law 503 Exercise 15.1 Find the magnitude of the electric force between two protons separated by one femtometer (10!15 m), approxi- mately the distance between two protons in the nucleus of a helium atom. The answer may not appear large, but if not for the strong nuclear force, the two protons would fly apart at an initial acceleration of nearly 7 $ 10 28 m/s 2 ! Answers 2.30 $ 102 N The Superposition Principle When a number of separate charges act on the charge of interest, each exerts an electric force. These electric forces can all be computed separately, one at a time, then added as vectors. This is another example of the superposition principle. The following example illustrates this procedure in one dimension. INTERACTIVE EXAMPLE 15.2 May the Force Be Zero Goal Apply Coulomb’s law in one dimension. 2.0 m Problem Three charges lie along the x -axis as in Figure 15.7. The positive charge x 2.0 m – x q 1 % 15 ' C is at x % 2.0 m, and the positive charge q 2 % 6.0 'C is at the origin. Where must a negative charge q 3 be placed on the x-axis so that the resultant elec- tric force on it is zero? + – + x q2 F23 q 3 F13 q1 Strategy If q 3 is to the right or left of the other two charges, then the net force on Figure 15.7 (Example 15.2) Three : : q 3 can’t be zero, because then F 13 and F 23 act in the same direction. Consequently, point charges are placed along the : : x -axis. The charge q 3 is negative, q3 must lie between the two other charges. Write F 13 and F 23 in terms of the un- whereas q1 and q 2 are positive. If the known coordinate position x, sum them and set them equal to zero, solving for the resultant : force on q 3 is zero, then the unknown. The solution can be obtained with the quadratic formula. force F 13 exerted by q 1 on q 3 must be equal : in magnitude and opposite the force F 23 exerted by q 2 on q 3. Solution : (15 $ 10!6 C)!q 3! Write the x-component of F 13 : F13x % # k e (2.0 m ! x)2 : (6.0 $ 10!6 C)!q3 ! Write the x-component of F 23 : F23x % !k e x2 (15 $ 10!6 C)!q 3! (6.0 $ 10!6 C)!q 3! Set the sum equal to zero: ke 2 !k e %0 (2.0 m ! x) x2 Cancel k e , 10!6 and q 3 from the equation, and 6(2 ! x)2 % 15x 2 rearrange terms (explicit significant figures and units are temporarily suspended for clarity): Put this equation into standard quadratic form, 6(4 ! 4x # x 2) % 15x 2 : 2(4 ! 4x # x 2) % 5x 2 ax 2 # bx # c % 0: 3x 2 # 8x ! 8 % 0 ! 8 " √64 ! (4)(3)(! 8) ! 4 " 2√10 Apply the quadratic formula: x% % 2&3 3 Only the positive root makes sense: x % 0.77 m Remarks Notice that it was necessary to use physical reasoning to choose between the two possible answers for x. This is nearly always the case when quadratic equations are involved. 44920_15_p497-530 12/29/04 9:18 AM Page 504 504 Chapter 15 Electric Forces and Electric Fields Exercise 15.2 Three charges lie along the x -axis. The positive charge q 1 % 10.0 ' C is at x % 1.00 m, and the negative charge q 2 % ! 2.00 'C is at the origin. Where must a positive charge q 3 be placed on the x-axis so that the resultant force on it is zero? Answer x % ! 0.809 m You can predict where on the x-axis the electric force is zero for random values of of q 1 and q 2 by logging into PhysicsNow at www.cp7e.com and going to Interactive Example 15.2. EXAMPLE 15.3 A Charge Triangle Goal Apply Coulomb’s law in two dimensions. y F13 F 13 sin 37° Problem Consider three point charges at the corners of a 4.00 m F23 37.0° q2 – + triangle, as shown in Figure 15.8, where q 1 % 6.00 $ 10!9 C, F 13 cos 37° q 2 % ! 2.00 $ 10!9 C, and q 3 % 5.00 $ 10!9 C. (a) Find the q3 : components of the force F 23 exerted by q 2 on q 3. : (b) Find the components of the force F 13 exerted by q 1 on 3.00 m 5.00 m q 3. (c) Find the resultant force on q 3 , in terms of compo- nents and also in terms of magnitude and direction. Strategy Coulomb’s law gives the magnitude of each q1 + x force, which can be split with right-triangle trigonometry into x- and y-components. Sum the vectors component- Figure : 15.8 (Example 15.3) The force : exerted by q 1 on q 3 is wise, and then find the magnitude and direction of the F. The force exerted by q 2 on q 3:is F 23.:The resultant force :13 F 3 exerted on q 3 is the vector sum F 13 # F 23. resultant vector. Solution (a) Find the components of the force exerted by q 2 on q 3. : !q 2 !!q3 ! Find the magnitude of F 23 with Coulomb’s law: F23 % k e r2 (2.00 $ 10!9 C)(5.00 $ 10!9 C) %(8.99 $ 109 N&m 2/C 2) (4.00 m)2 F23 % 5.62 $ 10!9 N : Because F 23 is horizontal and points in the negative F23x % ! 5.62 $ 10!9 N x-direction, the negative of the magnitude gives the x-component, and the y-component is zero: F23y % 0 (b) Find the components of the force exerted by q 1 on q 3. : !q 1!!q 3 ! Find the magnitude of F 13 : F13 % k e r3 (6.00 $ 10 !9 C)(5.00 $ 10!9 C) %(8.99 $ 10 9 N&m 2/C 2) (5.00 m)2 F13 % 1.08 $ 10!8 N Use the given triangle to find the components: F13x % F 13 cos ( % (1.08 $ 10!8 N )cos(37)) % 8.63 $ 10!9 N F 13y % F 13 sin ( % (1.08 $ 10!8 N)sin(37o) % 6.50 $ 10!9 N 44920_15_p497-530 12/29/04 9:18 AM Page 505 15.4 The Electric Field 505 (c) Find the components of the resultant vector. Sum the x-components to find the resultant Fx : Fx % ! 5.62 $ 10!9 N # 8.63 $ 10!9 N % 3.01 $ 10!9 N Sum the y-components to find the resultant Fy : Fy % 0 # 6.50 $ 10!9 N % 6.50 $ 10!9 N Find the magnitude of the resultant force on the charge ! F ! % √F x 2 # F y 2 : q 3, using the Pythagorean theorem: % √(3.01 $ 10!9 N)2 # (6.50 $ 10!9 N)2 % 7.16 $ 10!9 N " F # % tan " 6.50 # Fy $ 10 !9 N Find the angle the force vector makes with respect to ( % tan!1 !1 !9 % 65.2) the positive x-axis: x 3.01 $ 10 N Remarks The methods used here are just like those used with Newton’s law of gravity in two dimensions. Exercise 15.3 Using the same triangle, find the vector components of the electric force on q 1 and the vector’s magnitude and direc- tion. Answers F x % ! 8.63 $ 10!9 N, Fy % 5.50 $ 10!9 N, F % 1.02 $ 10!8 N, ( % 147° 15.4 THE ELECTRIC FIELD The gravitational force and the electrostatic force are both capable of acting through space, producing an effect even when there isn’t any physical contact be- tween the objects involved. Field forces can be discussed in a variety of ways, but an approach developed by Michael Faraday (1791 – 1867) is the most practical. In this approach, an electric field is said to exist in the region of space around a charged object. The electric field exerts an electric force on any other charged object within the field. This differs from the Coulomb’s law concept of a force exerted at a distance, in that the force is now exerted by something — the field — that is in the same location as the charged object. Figure 15.9 shows an object with a small positive charge q 0 placed near a second object with a much larger positive charge Q. : The electric field E produced by a charge Q at the location of a small “test” : charge q0 is defined as the electric force F exerted by Q on q 0, divided by the test charge q 0 : : : F E $ [15.4] Q q0 + + + + q0 + + SI Unit: newton per coulomb (N/C) + + + + + + + + + Conceptually and experimentally, the test charge q 0 is required to be very small Test charge (arbitrarily small, in fact), so it doesn’t cause any significant rearrangement of the : charge creating the electric field E. Mathematically, however, the size of the test Figure 15.9 A small object with a positive charge q 0 placed near charge makes no difference: the calculation comes out the same, regardless. In an object with a larger positive : charge view of this, using q 0 % 1 C in Equation 15.4 can be convenient if not rigorous. Q is subject to an electric field E When a positive test charge is used, the electric field always has the same direc- directed as shown. The magnitude of the electric field at the location of tion as the electric force on the test charge. This follows from Equation 15.4. q 0 is defined as the electric force on Hence in Figure 15.9, the direction of the electric field is horizontal and to the q 0 divided by the charge q 0. 44920_15_p497-530 12/29/04 9:19 AM Page 506 506 Chapter 15 Electric Forces and Electric Fields Figure 15.10 (a) The electric field A at A due to the negatively charged E sphere is downward, toward the nega- + tive charge. (b) The electric field at P P q0 P due to the positively charged con- E ducting sphere is upward, away from the positive charge. (c) A test charge – – + + q 0 placed at P will cause a rearrange- + + – – – + + + ment of charge on the sphere, unless + + q 0 is very small compared with the – – – + + + + + charge on the sphere. – – + + + + (a) (b) (c) right. The electric field at point A in Figure 15.10a is vertical and downward be- q0 E cause at that point a positive test charge would be attracted toward the negatively P charged sphere. r Once the electric field due to a given arrangement of charges is known at some q point, the force on any particle with charge q placed at that point can be calcu- + lated from a rearrangement of Equation 15.4: (a) : : F % qE [15.5] q0 Here q 0 has been replaced by q, which need not be a mere test charge. : As shown in Active Figure 15.11, the direction of E is the direction of the force P that acts on a positive test charge q 0 placed in the field. We say that an electric field q E exists at a point if a test charge at that point is subject to an electric force there. – Consider a point charge q located a distance r from a test charge q 0. According to (b) Coulomb’s law, the magnitude of the electric force of the charge q on the test charge is ACTIVE FIGURE 15.11 A test charge q 0 at P is a distance r !q!!q 0 ! F % ke from a point charge q. (a) If q is r2 positive, the electric field at P points radially outwards from q. (b) If q is Because the magnitude of the electric field at the position of the test charge is de- negative, the electric field at P points radially inwards toward q. fined as E % F/q 0, we see that the magnitude of the electric field due to the charge q at the position of q 0 is Log into PhysicsNow at www.cp7e.com !q! and go to Active Figure 15.11, where E % ke [15.6] you can move point P to any position r2 in two-dimensional space and observe the electric field due to q. Equation 15.6 points out an important property of electric fields that makes them useful quantities for describing electrical phenomena. As the equation indicates, an electric field at a given point depends only on the charge q on the object set- ting up the field and the distance r from that object to a specific point in space. As a result, we can say that an electric field exists at point P in Active Figure 15.11 whether or not there is a test charge at P. The principle of superposition holds when the electric field due to a group of point charges is calculated. We first use Equation 15.6 to calculate the electric field produced by each charge individually at a point and then add the electric fields to- gether as vectors. It’s also important to exploit any symmetry of the charge distribution. For ex- ample, if equal charges are placed at x % a and at x % ! a, the electric field is zero at the origin, by symmetry. Similarly, if the x -axis has a uniform distribution of pos- itive charge, it can be guessed by symmetry that the electric field points away from the x -axis and is zero parallel to that axis. Quick Quiz 15.3 A test charge of #3 ' C is at a point P where the electric field due to other charges is directed to the right and has a magnitude of 4 $ 106 N/C. If the test charge is re- placed with a charge of ! 3 ' C, the electric field at P (a) has the same magnitude as before, but changes direction, (b) increases in magnitude and changes direction, (c) remains the same, or (d) decreases in magnitude and changes direction. 44920_15_p497-530 12/29/04 9:19 AM Page 507 15.4 The Electric Field 507 Quick Quiz 15.4 A circular ring of charge of radius b has a total charge q uniformly distributed around it. The magnitude of the electric field at the center of the ring is (a) 0 (b) k e q/b 2 (c) k e q 2/b 2 (d) k e q 2/b (e) none of these. Quick Quiz 15.5 A “free” electron and a “free” proton are placed in an identical electric field. Which of the following statements are true? (a) Each particle is acted upon by the same electric force and has the same acceleration. (b) The electric force on the proton is greater in magnitude than the force on the electron, but in the opposite direction. (c) The electric force on the proton is equal in magnitude to the force on the electron, but in the opposite direction. (d) The magnitude of the accelera- tion of the electron is greater than that of the proton. (e) Both particles have the same acceleration. EXAMPLE 15.4 Electrified Oil Goal Use electric forces and fields together with Newton’s second law in a one-dimensional problem. Problem Tiny droplets of oil acquire a small negative charge while dropping through a vacuum (pressure % 0) in an experiment. An electric field of magnitude 5.92 $ 10 4 N/C points straight down. (a) One particular droplet is observed to remain suspended against gravity. If the mass of the droplet is 2.93 $ 10!15 kg, find the charge carried by the droplet. (b) Another droplet of the same mass falls 10.3 cm from rest in 0.250 s, again moving through a vac- uum. Find the charge carried by the droplet. : Strategy We use Newton’s second law with both gravitational and electric forces. In both parts, the electric field E is pointing down, taken as the negative direction, as usual. In part (a), the acceleration is equal to zero. In part (b), the acceleration is uniform, so the kinematic equations yield the acceleration. Newton’s law can then be solved for q. Solution (a) Find the charge on the suspended droplet. Apply Newton’s second law to the droplet in the vertical (1) ma % * F % ! mg # Eq direction: mg (2.93 $ 10!15 kg)(9.80 m/s2) E points downward, hence is negative. q% % E ! 5.92 $ 104 N/C Set a % 0 and solve for q: % ! 4.85 $ 10!19 C (b) Find the charge on the falling droplet. Use the kinematic displacement equation to find the +y % 12 at 2 # v 0 t acceleration: Substitute +y % ! 0.103 m, t % 0.250 s, and v 0 % 0: ! 0.103 m % 12 a(0.250 s)2 : a % ! 3.30 m/s2 Solve Equation 1 for q and substitute: m(a # g) q% E (2.93 $ 10!15 kg)(! 3.30 m/s2 # 9.80 m/s2) % ! 5.92 $ 104 N/C % !3.22 $ 10!19 C 44920_15_p497-530 12/29/04 9:19 AM Page 508 508 Chapter 15 Electric Forces and Electric Fields Remark This example exhibits features similar to the Millikan Oil-Drop experiment discussed in Section 15.7, which determined the value of the fundamental electric charge e. Notice that in both parts of the example, the charge is very nearly a multiple of e. Exercise 15.4 Suppose a droplet of unknown mass remains suspended against gravity when E % ! 2.70 $ 10 5 N/C. What is the minimum mass of the droplet? Answer 4.41 $ 10!15 kg Problem-Solving Strategy Calculating Electric Forces and Fields The following procedure is used to calculate electric forces (the same procedure can be used to calculate an electric field, a simple matter of replacing the charge of inter- est, q, with a convenient test charge and dividing by the test charge at the end): 1. Draw a diagram of the charges in the problem. 2. Identify the charge of interest, q, and circle it. 3. Convert all units to SI, with charges in coulombs and distances in meters, so as to be consistent with the SI value of the Coulomb constant ke. 4. Apply Coulomb’s law. For each charge Q , find the electric force on the charge of interest, q. The magnitude of the force can be found using Coulomb’s law. The vector direction of the electric force is along the line of the two charges, directed away from Q if the charges have the same sign, toward Q if the charges have the opposite sign. Find the angle ( this vector makes with the positive x -axis. The x-component of the electric force exerted by Q on q will be F cos (, and the y-component will be F sin (. 5. Sum all the x-components, getting the x-component of the resultant electric force. 6. Sum all the y-components, getting the y-component of the resultant electric force. 7. Use the Pythagorean theorem and trigonometry to find the magnitude and direc- tion of the resultant force if desired. EXAMPLE 15.5 Electric Field Due to Two Point Charges Goal Use the superposition principle to calculate the electric field due to two point charges. y Problem Charge q 1 % 7.00 ' C is at the origin, and charge q 2 % ! 5.00 ' C is on the x-axis, E1 0.300 m from the origin (Fig. 15.12). (a) Find the magnitude and direction of the electric field at E point P, which has coordinates (0, 0.400) m. (b) Find the force on a charge of 2.00 $ 10!8 C f placed at P. P u E2 Strategy Follow the problem-solving strategy, finding the electric field at point P due to each 0.400 m 0.500 m individual charge in terms of x - and y-compo- nents, then adding the components of each type Figure 15.12 (Example : 15.5) to get the x- and y-components of the resultant The resultant electric field : E:at P electric field at P. The magnitude of the force in u the vector sum E 1 # E 2, equals : + – x where E 1 is the field due to the part (b) can be found by simply multiplying the 0.300 m : positive charge q 1 and E 2 is the field magnitude of the electric field by the charge. q1 q2 due to the negative charge q 2. 44920_15_p497-530 12/29/04 9:19 AM Page 509 15.4 The Electric Field 509 Solution (a) Calculate the electric field at P. : !q1! 9 N&m2/C 2) (7.00 $ 10 !6 C) Find the magnitude of E 1 with Equation 15.6: E 1 % ke % (8.99 $ 10 r 12 (0.400 m)2 % 3.93 $ 10 5 N/C : The vector E 1 is vertical, making an angle of 90° with E 1x % E 1 cos (90°) % 0 respect to the positive x -axis. Use this fact to find its E 1y % E 1 sin (90°) % 3.93 $ 10 5 N/C components: : !q 2 ! (5.00 $10!6 C) Next, find the magnitude of E 2 , again with Equation 15.6: E2 % ke 2 % (8.99 $ 109 N&m2/C 2) r2 (0.500 m)2 % 1.80 $ 10 5 N/C : adj 0.300 Obtain the x-component of E 2 , using the triangle in Fig- cos ( % % % 0.600 ure 15.12 to find cos (: hyp 0.500 E 2x % E 2 cos ( % (1.80 $ 10 5 N/C)(0.600) % 1.08 $ 10 5 N/C opp 0.400 Obtain the y-component in the same way, but a sin ( % % % 0.800 hyp 0.500 minus sign has to be provided for sin ( because this component is directed downwards: E 2y % E 2 sin ( % (1.80 $ 10 5 N/C)(! 0.800) % ! 1.44 $ 10 5 N/C Sum the x -components to get the x -component of the E x % E 1x # E 2x % 0 # 1.08 $ 10 5 N/C % 1.08 $ 10 5 N/C resultant vector: Sum the y -components to get the y-component of the E y % E 1y # E 2y % 0 # 3.93 $ 10 5 N/C ! 1.44 $ 10 5 N/C resultant vector: Ey % 2.49 $ 105 N/C Use the Pythagorean theorem to find the magnitude of E % √E x 2 # E y 2 % 2.71 $ 10 5 N/C the resultant vector: " E # % tan " 2.49 # % 66.6) Ey $ 10 5 N/C The inverse tangent function yields the direction of the , % tan!1 !1 5 x 1.08 $ 10 N/C resultant vector: (b) Find the force on a charge of 2.00 $ 10!8 C placed at P. Calculate the magnitude of the force (the direction is F % Eq % (2.71 $ 105 N/C)(2.00 $ 10!8 C) : the same as that of E because the charge is positive): % 5.42 $ 10!3 N Remarks There were numerous steps to this problem, but each was very short. When attacking such problems, it’s important to focus on one small step at a time. The solution comes not from a leap of genius, but from the assembly of a number of relatively easy parts. Exercise 15.5 (a) Place a charge of ! 7.00 ' C at point P and find the magnitude and direction of the electric field at the location of q 2. (b) Find the magnitude and direction of the force on q 2. Answer (a) 5.84 $ 10 5 N/C, , % 20.2) (b) F % 2.92 N, , % 200.) 44920_15_p497-530 12/29/04 9:19 AM Page 510 510 Chapter 15 Electric Forces and Electric Fields 15.5 ELECTRIC FIELD LINES A convenient aid for visualizing electric field patterns is to draw lines pointing in the direction of the electric field vector at any point. These lines, introduced by Michael Faraday and called electric field lines, are related to the electric field in any region of space in the following way: : 1. The electric field vector E is tangent to the electric field lines at each point. 2. The number of lines per unit area through a surface perpendicular to the lines is proportional to the strength of the electric field in a given region. : Note that E is large when the field lines are close together and small when the lines are far apart. Figure 15.13a shows some representative electric field lines for a single posi- tive point charge. This two-dimensional drawing contains only the field lines that lie in the plane containing the point charge. The lines are actually directed radi- ally outward from the charge in all directions, somewhat like the quills of an angry porcupine. Because a positive test charge placed in this field would be re- pelled by the charge q, the lines are directed radially away from the positive charge. The electric field lines for a single negative point charge are directed to- ward the charge (Fig. 15.13b), because a positive test charge is attracted by a neg- ative charge. In either case, the lines are radial and extend all the way to infinity. Note that the lines are closer together as they get near the charge, indicating that the strength of the field is increasing. Equation 15.6 verifies that this is indeed the case. The rules for drawing electric field lines for any charge distribution follow directly from the relationship between electric field lines and electric field vectors: 1. The lines for a group of point charges must begin on positive charges and end on negative charges. In the case of an excess of charge, some lines will begin or end infinitely far away. 2. The number of lines drawn leaving a positive charge or ending on a negative charge is proportional to the magnitude of the charge. TIP 15.1 Electric Field Lines 3. No two field lines can cross each other. Aren’t Paths of Particles Figure 15.14 shows the beautifully symmetric electric field lines for two point Electric field lines are not material objects. They are used only as a charges of equal magnitude but opposite sign. This charge configuration is called pictorial representation of the an electric dipole. Note that the number of lines that begin at the positive charge electric field at various locations. must equal the number that terminate at the negative charge. At points very near Except in special cases, they do not represent the path of a charged either charge, the lines are nearly radial. The high density of lines between the particle released in an electric field. charges indicates a strong electric field in this region. Photo courtesy of Harold M. Waage, Princeton University + q – –q Image not Available (a) (b) (c) Figure 15.13 The electric field lines for a point charge. (a) For a positive point charge, the lines radiate outward. (b) For a negative point charge, the lines converge inward. Note that the figures show only those field lines which lie in the plane containing the charge. (c) The dark lines are small pieces of thread suspended in oil, which align with the electric field produced by a small charged conductor at the center. 44920_15_p497-530 12/29/04 9:19 AM Page 511 15.5 Electric Field Lines 511 Photo courtesy of Harold M. Waage, + – Image not Available Princeton University (a) (b) Figure 15.14 (a) The electric field lines for two equal and opposite point charges (an electric dipole). Note that the number of lines leaving the positive charge equals the number terminating at the negative charge. (b) The dark lines are small pieces of thread suspended in oil, which align with the electric field produced by two charged conductors. Figure 15.15 shows the electric field lines in the vicinity of two equal positive point charges. Again, close to either charge the lines are nearly radial. The same number of lines emerges from each charge because the charges are equal in mag- nitude. At great distances from the charges, the field is approximately equal to that of a single point charge of magnitude 2q. The bulging out of the electric field lines between the charges reflects the repulsive nature of the electric force between like charges. Also, the low density of field lines between the charges indi- cates a weak field in this region, unlike the dipole. Finally, Active Figure 15.16 is a sketch of the electric field lines associated with the positive charge # 2q and the negative charge ! q. In this case, the num- ber of lines leaving charge # 2q is twice the number terminating on charge ! q. Hence, only half of the lines that leave the positive charge end at the negative charge. The remaining half terminate on negative charges that we assume to be located at infinity. At great distances from the charges (great compared with the charge separation), the electric field lines are equivalent to those of a single charge # q. Quick Quiz 15.6 Rank the magnitudes of the electric field at points A, B, and C in Figure 15.15, with the largest magnitude first. (a) A, B, C (b) A, C, B (c) C, A, B (d) Can’t be determined by visual inspection + 2q + – –q B Photo courtesy of Harold M. Waage, Princeton A ACTIVE FIGURE 15.16 C Image not Available The electric field lines for a point + + charge of # 2q and a second point charge of ! q. Note