* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download THE ORIGIN OF ELECTRICITY
Survey
Document related concepts
Electron mobility wikipedia , lookup
Weightlessness wikipedia , lookup
Elementary particle wikipedia , lookup
Introduction to gauge theory wikipedia , lookup
Newton's laws of motion wikipedia , lookup
Magnetic monopole wikipedia , lookup
History of electromagnetic theory wikipedia , lookup
Electrical resistivity and conductivity wikipedia , lookup
Speed of gravity wikipedia , lookup
Work (physics) wikipedia , lookup
Anti-gravity wikipedia , lookup
Aharonov–Bohm effect wikipedia , lookup
Fundamental interaction wikipedia , lookup
Maxwell's equations wikipedia , lookup
Electromagnetism wikipedia , lookup
Field (physics) wikipedia , lookup
Lorentz force wikipedia , lookup
Transcript
Lightning,which is the flow of electric charge in the atmosphere, is nature'smost spectacular display of electricity. Here,multiple lightning strikes occur at night over an urban skyline. (RalphWetmore/Stone/Getty Images) THE ORIGIN OF ELECTRICITY The electrical nature of matter is inherent in atomic structure. An atom consists of a small,relatively massive nucleus that contains particles called protons and neutrons. A protonhas a mass of 1.673 X 10-27 kg, and a neutron has a slightly greater mass of 1675X 10-27kg. Surrounding the nucleus is a diffuse cloud of orbiting particles called electrons,as Figure 18.1 suggests. An electron has a mass of 9.11 X 10-31 kg. Like mass, electriccharge is an intrinsic property of protons and electrons, and only two types of chargehave been discovered, positive and negative. A proton has a positive charge, and anelectron has a negative charge. A neutron has no net electric charge. electron proton neutron Figure 18.1 An atom contains a small, positively charged nucleus, about which the negatively charged electrons move. The closed-loop paths shown here are symbolic only. In reality, the electrons do not follow discrete paths, as Section 30.5 discusses. 537 538 | CHAPTER 18 ELECTRIC FORCES AND ELECTRIC FIELDS equals the the charge on the proton exactly Experiment reveals that the magnitude of electron the and the proton carries a charge +e, magnitude of the charge on the electron; is the charge the magnitude of an electric carries a charge —e.The SI unit for measuring experimentally to have the value coulomb* (C), and e has been determined e = 1.60 X 10-19C electron and of the charge on a proton or an The symbol e represents only the magnitude whether the charge is positive or negadoes not include the algebraic sign that indicates equal numbers of protons and electrons. tive. In nature, atoms are normally found withbecause the algebraic sum of the positive Usually, then, an atom carries no net charge the electrons is zero. When an atom, or charge of charge of the nucleus and the negative is said to be electrically neutral. The neutrons object any object, carries no net charge, the particles. in the nucleus are electrically neutral an electron or on a proton are built up on Charges of larger magnitude than that on Thus, any charge of magnitude q is an an object by adding or removing electrons. is an integer. Because any electric charge q integer multiple of e; that is, q = Ne, where N indivisible charges of magnitude e, electric occurs in integer multiples of elementary, emphasizes the quantized nature of electric charge is said to be quantized. Example I charge. Example 1 | A Lot of Electrons How many electrons are there in one coulomb of negative charge? electrons, since they carry Reasoning The negativecharge is due to the presence of excess is e = 1.60 x 10-19c, magnitude negative charge. Because an electron has a charge whose on each electron. e charge the by divided the number of electrons is equal to the charge q Solution The number N of electrons is q 1.00 c 1.60 x 10-19c Animal fur Ebonite rod Figure 18.2 When an ebonite rod is rubbed against animal fur, electrons from atoms of the fur are transferred to the rod. This transfer gives the rod a negative charge (—) and leaves a positive charge (+) on the fur. 6.25 x 1018 CHARGED OBJECTS AND THE ELECTRIC FORCE Electricity has many useful applications, and they are related to the fact that it is posSibleto transfer electric charge from one object to another. Usually electrons are transferred, and the body that gains electrons acquires an excess of negative charge. The body that loses electrons has an excess of positive charge. Such separation of charge occurs often when two unlike materials are rubbed together. For example, when an ebonite (hard, black rubber) rod is rubbed against animal fur, some of the electrons from atoms of the fur are transfe1Tedto the rod. The ebonite becomes negatively charged, and the fur becomes positively charged, as Figure 18.2 indicates. Similarly, if a glass rod is rubbed with a silk cloth, some of the electrons are removed from the atoms of the glass and deposited on the silk, leaving the silk negatively charged and the glass positively charged. There are many familiar examples of charge separation,as when you walk across a nylon rug or run a comb through dry hair. In each case, objects become "electrified" as surfaces rub against one another. When an ebonite rod is rubbed with animal fur, the rubbing process serves only to separate electrons and protons already present in the materials. No electrons or protons are created or destroyed. Whenever an electron is transferred to the rod, a proton is left behind on the fur. Since the charges on the electron and proton have identical magnitudes but opposite signs, the algebraic sum of the two charges is zero, and the transfer does not change the net charge of the fur/rod system. If each material contains an equal number Of protons and electrons to begin with, the net charge of the system is zero initially and remains zero at all times during the rubbing process. *The definition of the coulomb depends on electric currents and magnetic fields, concepts that will be discussed later. Therefore, we postpone its definition until Section 21.7. 18.2 | CHARGED OBJECTS AND THE ELECTRIC FORCE | 539 Electriccharges play a role in many situations other than rubbing two surfaces together.They are involved, for instance, in chemical reactions, electric circuits, and radecay.A great number of experimentshave verified that in any situation, the dioactive of electric charge is obeyed. lawofconservation LAWOF CONSERVATION OF ELECTRIC CHARGE Duringany process, the net electric charge of an isolated system remains constant (isconserved). (a) It is easy to demonstrate that two electrically charged objects exert a force on one another.Consider Figure 18.3a, which shows two small balls that have been oppositely chargedand are light and free to move. The balls attract each other. On the other hand, ballswith the same type of charge, either both positive or both negative, repel each other, aspartsb and c of the drawing indicate. The behavior depicted in Figure 18.3 illustrates thefollowingfundamental characteristic of electric charges: Like charges repel and unlike charges attract each other. ATA GLANCE CONCEPTS Like other forces that we have encountered, the electric force(alsocalled the electrostatic force) can alter the motion of an object. It can do so by contributingto the net external force EF that acts on the object. Newton's second law, EF = ma, specifies the acceleration ä that arises because of the net external force. The Concepts-at-a-Glancechart in Figure 18.4 is an expanded version of the charts in Figures 4.9,10.4,and 11.5, and emphasizes that any external electric force that acts on an object mustbe included when determining the net external force to be used in the second law. < Thephysicsof electronic ink. A new technology based on the electric force may revolutionizethe way books and other printed matter are made. This technology, called electronicink, allows letters and graphics on a page to be changed instantly, much like the symbolsdisplayed on a computer monitor. Figure 18.5a illustrates the essential features ofelectronicink. It consists of millions of clear microcapsules, each having the diameter of a human hair and filled with a dark inky liquid. Inside each microcapsule are several dozenextremely tiny white beads that carry a slightly negative charge. The microcapsules aresandwichedbetween two sheets, an opaque base layer and a transparent top layer, at (b) (c) Figure 18.3 (a) A positive charge (+) and a negative charge (—) attract each other. (b) Two negative charges repel each other. (c) Two positive charges repel each other. CONCEPTSAT A GLANCE External Forces 1. Gravitational Force (Section4.7) 2. Normal Force (Section 4.8) 3. Frictional Forces (Section 4.9) 4. Tension Force (Section 4.10) 5. Restoring Force of a Spring (Section 10.l) 6. CollisionalForces (Section 11.2) 7. Electrostatic Force Figure18.4 CONCEPTS AT A GLANCE External electrostatic forces that act on an object must included in the net external force when using Newton's second law to determine acceleration. In a DNA analysis using electrophoresis, an electrostatic force accelerates electrically charged DNAfragmentsto different terminal speeds in a gel. The different fragments are represented by the glowing bands. (Eurelios/Phototake) Newton's Second Law 540 | CHAPTER 18 ELECTRIC FORCES AND ELECTRIC FIELDS See your sales person for the right fit Base layer Beads Microcapsules (a) Dark liquid Base layer Dark pixel Light pixel Figure 18.5 (a) Electronic ink consists of microcapsules filled with a dark, inky liquid and dozens of white beads. (b) Dark and light pixels are formed when positive and negative charges are placed in the base layer by electronic circuitry. (b) which the reader looks. When a positive charge is applied to a small region of thebase layer, as shown in part b of the drawing, the negatively charged white beads are drawnto it, leaving dark ink at the top layer. Thus, a viewer sees only the dark liquid. When a neg- ative charge is applied to a region of the base layer, the negatively charged whitebeads are repelled from it and are forced to the top of the microcapsules; now a viewerseesa white area due to the beads. Thus, electronic ink is based on the principle thatlike charges repel and unlike charges attract each other; a positive charge causes one colorto appear, and a negative charge causes another color to appear. Each small region, whether dark or light, is known as a pixel (short for "picture element"). Computerchipsprovide the instructions to produce the negative and positive charges on the base layer of each pixel. Letters and graphics are produced by the patterns generated with the two colors• CONDUCTORS AND INSULATORS Electric charge can not only exist on an object, but it can also move throughan object. However, materials differ vastly in their abilities to allow electric charge to move or be conducted through them. To help illustrate such differences in conductivity,Figure 18.6a recalls the conduction of heat through a bar of material whose ends are maintained at different temperatures. As Section 13.2 discusses, metals conduct heat readilyand' therefore, are known as thermal conductors. On the other hand, substances that conduct heat poorly are referred to as thermal insulators. 1 541 18.4 | CHARGING BY CONTACT AND BY INDUCTION Hotter object Cooler object Heat Negatively charged object (a) + Positively charged object (b) to the conduction of heat arises when a metal bar is placed beA situationanalogous objects, as in Figure 18.6b. Electrons are conducted through the bar from twocharged tween charged object toward the positively charged object. Substances that readily thenegatively electrical conductors. Although there are exceptions, good electriccharge are called conduct good electrical conductors. Metals such as copper, alugenerally conductors are thermal electrical conductors and, therefore, are used in electrisilver,and gold are excellent minum, charge poorly are known as electrical insulators. electric conduct that Materials calWiling. electrical also insulators. Common electrical insulators are insulators Inmanycases,thermal such Insulators, as the rubber or plastic that coats elecwood. and arerubber,manyplastics, from going where it is not wanted. wiring,prevent electric charge trical atomic strucThedifferencebetween electrical conductors and insulators is related to force of atthe outer weaker in orbits a those experience nucleus, the orbit Aselectrons ture. elecoutermost the inner orbits. Consequently, in the do those than to the nucleus traction In ones. the inner be dislodged more easily than can electrons) (alsocalled the valence Irons wander and atom become detached from a parent agoodconductor,some valence electrons or less freely throughout the material, belonging to no one atom in particular. The exmore actnumberof electrons detached from each atom depends on the nature of the material, but isusuallybetweenone and three. When one end of a conducting bar is placed in contact witha negativelycharged object and the other end in contact with a positively charged object,asin Figure 18.6b, the "free" electrons are able to move readily away from the negative endandtowardthe positive end. The ready movement of electrons is the hallmark of a good In an insulator the situation is different, for there are very few electrons free to conductor. move throughoutthe material. Virtually every electron remains bound to its parent atom. Witlmtthe "free" electrons, there is very little flow of charge when the material is placed between two oppositely charged bodies, so the material is an electrical insulator. from Figure 18.6 (a) Heat is conducted the to bar the hotter end of the metal conducted cooler end. (b) Electrons are of the end from the negatively charged end. charged metal bar to the positively Ebonite rod CHARGING BY CONTACT AND BY INDUCTION Whena negatively charged ebonite rod is rubbed on a metal object, such as the sphere in Figure 18.7a, some of the excess electrons from the rod are transferred to the Object. Oncethe electrons are on the metal sphere, where they can move readily, they repeloneanother and spread out over the sphere's surface. The insulated stand prevents themfromflowing to the earth, where they could spread out even more. When the rod is removed, as in part b of the picture, the sphere is left with a negative charge distributed overits surface.In a similar manner, the sphere would be left with a positive charge after being rubbedwith a positively charged rod. In this case, electrons from the sphere would betransferredto the rod. The process of giving one object a net electric charge by placing itincontactwith another object that is already charged is known as charging by contact. It is also possible to charge a conductor in a way that does not involve contact. In Figure 18.8,a negatively charged rod is brought close to, but does not touch, a metal Metal sphere Insulated stand (a) (b) Figure 18.7 (a) Electrons are transferred by rubbing the negatively charged rod on the metal sphere. (b) When the rod is removed, the electrons distributethemselves over the surface of the sphere. Ebonite rod Metal sphere Grounding wire Insulated stand -z Connection to ground (a) (b) (c) Figure 18.8 (a) When a charged rod is brought near the metal sphere without touching it, some of the positive and negative charges in the sphere are separated. (b) Some of the electrons leave the sphere through the grounding wire, with the result (c) that the sphere acquires a positive net charge. 542 | CHAPTER 18 I ELECTRIC FORCES AND ELECTRIC FIELDS Ebomte rod Positive surface charge Plastic Figure 18.9 The negatively charged rod induces a slight positive surface charge on the plastic. sphere. In the sphere, the free electrons closest to the rod move to the other side, aspana of the drawing indicates. As a result, the part of the sphere nearest the rod becomesposi. tively charged and the part farthest away becomes negatively charged. These positively and negatively charged regions have been "induced" or "persuaded" to form because of the repulsive force between the negative rod and the free electrons in the sphere.Ifthe rod were removed, the free electrons would return to their original places, andthe charged regions would disappear. Under most conditions the earth is a good electrical conductor. So when a metalwire is attached between the sphere and the ground, as in Figure 18.8b, some of the freeelec. trons leave the sphere and distribute themselves over the much larger earth. If the ground. ing wire is then removed, followed by the ebonite rod, the sphere is left with a positive net charge, as part c of the picture shows. The process of giving one object a net electm charge without touching the object to a second charged object is called charging byin. duction. The process could also be used to give the sphere a negative net charge,if a posi. tively charged rod were used. Then, electrons would be drawn up from the ground through the grounding wire and onto the sphere. If the sphere in Figure 18.8 were made from an insulating material like plastic,in. stead of metal, the method of producing a net charge by induction would not work,be. cause very little charge would flow through the insulating material and down the ground. ing wire. However, the electric force of the charged rod would have some effect, Figure 18.9 illustrates. The electric force would cause the positive and negative chargesin the molecules of the insulating material to separate slightly, with the negative chargesbe. ing "pushed" away from the negative rod. Although no net charge is created, the surface of the plastic does acquire a slight induced positive charge and is attracted to the negative rod. For a similar reason, one piece of cloth can stick to another in the phenomenon known as "static cling," which occurs when an article of clothing has acquired an electric charge while being tumbled about in a clothes dryer. CHECK YOUR UNDERSTANDING 1 Two metal spheres are identical. They are electrically neutral and are touching. An electrically charged ebonite rod is then brought near the spheres without touching them, as the drawing shows. After a while, with the rod held in place, the spheres are separated,andthe rod is then removed. The following statements refer to the masses rnAand rnBof the spheres after they are separated and the rod is removed. Which one or more of thestatements is true? (a) rnB,(b) 'DA> if the rod is positive, (c) rnA< rnBif the rod is posi• tive, (d) rnA> if the rod is negative, (e) tm < if the rod is negative. [The answer(s)is (are) given at the end of the book.] Background: Mass and electric charge are both properties of electrons and protons.Ina metal, which move, electrons or protons? An object can be given an electric charge by contact with another charged object or by induction. For similar questions (including calculational counterparts), consult Self-Assessment Test 18.1,which is described at the end of Section 18.5. COULOMB'S LAW THE FORCE THAT POINT CHARGES EXERT ON EACH OTHER The electrostatic force that stationary charged objects exert on each other depenb on the amount of charge on the objects and the distance between them. ' reveal that the greater the charge and the closer together they are, the greater is theforce 18.5 I COULOMB'S LAW | 543 for explaining these features in more detail, Figure 18.10 shows two Toset the stage objects are so small, compared to the distance r between them, that bodies. These If the charges have unlike signs, as in part a of the picture, each object is atat the other by a force th is directed along the line between them; +F is the to tracted on object I by and object 2 and —F is the electric force exerted exerted on obelectricforce in as part b, the charges have the same sign (both positive If, l. object or both ject2 by each object is repelled from the other. The repulsive forces, like the attractive negative), the line between the charges. Whether attractive or repulsive, the two forces,act along magnitude but opposite in direction. These forces always exist as a in forcesare equal on a different object, in accord with Newton's action—reactionlaw. pair,eachone acting Charles Augustin de Coulomb (1736—1806)carried out a numphysicist TheFrench determine how the electric force that one point charge applies to anto berofexperiments amount of each charge and the separation between them. His result, otherdependson the law, is stated below. Coulomb's nowknownas (a) (b) Figure18.10 Each point charge exerts a force on the other Regardlessof whether the forces are (a) attractive or (b) repulsive, they are directed along the line between the charges and have equal magnitudes. COULOMB'SLAW Themagnitude F of the electrostatic force exerted by one point charge qi on another pointcharge is directly proportional to the magnitudes I and Iq21of the charges andinverselyproportional to the square of the distance r between them: Iqlllq21 (18.1) 2 wherek is a proportionality constant: k = 8.99 X 109N • m2/C2in SI units. The electrostaticforce is directed along the line joining the charges, and it is attractive if the chargeshave unlike signs and repulsive if the charges have like signs. It is common practice to express k in terms of another constant €0, by writing k — €0is called the permittivity of free space and has a value of €0= 1/(47k) = 8.85X 10-12 C2/(N • m2). Equation 18.1 gives the magnitude of the electrostatic force thateachpoint charge exerts on the other. When using this equation, then, it is important torememberto substitute only the charge magnitudes (without algebraic signs) for Iql I andIq21, as Example 2 illustrates. V Needmorepractice? Example 2 | A Large Attractive Force Interactive LearningWare 18.1 An Twoobjects,whose charges are +1.0 and —1.0C, are separated by 1.0 km. Compared to 1.0km,the sizes of the objects are small. Find the magnitude of the attractive force that either chargeexertson the other. ReasoningConsidering that the sizes of the objects are small compared to the separation distance,we can treat the charges as point charges. Coulomb's law may then be used to find the magnitude of the attractive force, provided that only the magnitudesof the charges are used object of mass 215 kg is located just above the surface of the earth (radius = 6.38 x 106 m). Suppose that the object and the earth have a. identical charge q. Therefore, two forces—the gravitational force and the electrostatic force— SolutionThe magnitude of the force is act on the object. Assuming that the earth's charge is located at the center of the earth, determine q forthesymbols Iqi I and Iq21that appear in the law. Iql llq21 r2 (8.99 x 109 (1.0 >< 10 3 C) (18.1) Theforce calculated in Example 2 corresponds to about 2000 pounds and is so large bec ausecharges of ± 1.0 only in the C are enormous. Such large charges are encountered mostsevere transferred C be 25 can as conditions, as in a lightning bolt, where as much e magnitudeof a by a vertical bar to the variable is sometimes called the absolute value and is symbolized leftandto the of the variable q, which is value thevalue right of the variable. Thus, Iq I denotes the magnitude or absolute of q without • then Iql = 2.0 c. c, = -2.0 q If its algebraic plus or minus sign. For example, (magnitude and sign) when the object is in equilibrium. Related Homework: Problem 61 For an interactive solution, go to www.wiley.com/college/cutnell 544 | CHAPTER 18 I ELECTRIC FORCES AND ELECTRIC FIELDS between the cloud and the ground. The typical charges produced in the much smaller and are measured convenientlyin microcoulombs(l microcoulombs 1 = 10-6 C). Coulomb's law has a form that is remarkably similar to Newton's lawof gravitas tion (F = GmlR12/r2).The force in both laws depends on the inverse square(1/r2)of the distance between the two objects and is directed along the line betweenthem. In addition, the force is proportional to the product of an intrinsic propertyof each in Coulomb's law and the objects, the magnitudes of the charges 1and difference between the twolaws major a is there But ml and in the gravitation law. attractive, depending on The electrostatic force can be either repulsive or not the charges have the same sign; in contrast, the gravitational force is alwaysanat. tractive force. Section 5.5 discusses how the gravitational attraction between the earth anda lite provides the centripetal force that keeps a satellite in orbit. Example 3 illustrates that famous in a role similar model a plays of the atomcr. the electrostatic force of attraction ated by the Danish physicist Niels Bohr (1885—1962). ANALYZING MULTIPLE-CONCEPT PROBLEMS Example 3 | A Model of the Hydrogen Atom In the Bohr model of the hydrogen atom, the electron (charge = —e)is in a circular orbit about the nuclear proton (charge +e) at a radius of 5.29 X 10-11m, as Figure 18.11 shows. The mass of the electron is 9.11 ><10-31kg. Determine the speed of the electron. Reasoning Recall from Section 5.3 that a net force is required to keep an object such as an electron moving on a circular path. This net force is called the centripetal force and always points toward the center of the circle. It has a magnitude given by Fc = mv2/r, where m and v are, respectively, the mass and speed of the electron and r is Figure 18.11 In the Bohr modelof the the radius of the orbit. This equation can be solved for the speed of the electron. Since hydrogen atom, the electron (—e)orbits the mass and orbital radius are known, we can calculate the electron's speed provided the proton (+e) at a distanceof that a value for the centripetal force can be found. For the electron in the hydrogen r = 5.29 X 10-11 m. The velocityofthe atom, the centripetal force is provided almost exclusively by the electrostatic force that electron is v. the proton exerts on the electron. This attractive force points toward the center of the circle and its magnitude is given by Coulomb's law. The electron is also pulled toward the proton by the gravitationalforce• However, the gravitational force is negligible in comparison to the electrostatic force. Knowns and Unknowns Description Electron charge Electron mass Proton charge Radius of orbit Unknown Variable Orbital speed of electron The data for this problem are: Symbol Value -1.60 x 10-19c m 9.11 X 10-31 kg +1.60 x 10-19c 5.29 x 10-11 m 9 Modeling the Problem STEP 1 Centripetal Force An electron of mass m that moves with a constant speed v on a circular path of radius r experiences a net force, called the centripetal force. The magnitude Fc of this force is given by FC = mv2/r (Equation 5.3). By solving this equation for the speed, we obtain Equation 1 at the right. The mass and radius in this expres- sion are known. However, the magnitude of the centripetal force is not known, so we will evaluate it in Step 2. Coulomb's Law As the electron orbits the proton in the hydrogen atom, it is proton by the electrostatic force. The magnitude F of the electrostatic attractedto the Coulomb's law as F = klqlllq21/r2 (Equation18.1),where and Iq21 forceis givenby the charges, r is the of orbital radius, and k = 8.99 X 109N • m2/C2. arethemagnitudes force is provided almost entirely by centripetal the electrostaticforce, it follows Sincethe = I—el and Iq21 = l+el. Furthermore, With these substitutions,Equation F. thatFC= 18.1becomes Fc=k —ell + el (1) Fc=k 2 Allthevariables on the right side of this expression are known, so we substitute it into 1, as indicated in the right column. Equation solution Algebraically combining the results of the modeling steps, we have STEP 1 STEP 2 I—ell + el r2 m m mr Thespeedof the orbiting electron is kl —ell + el mr (8.99 x 109 N. m2/c 2)l -1.60 x 10-19Cll +1.60 x 10-19cl (9.11 X Related Homework: 10 -31 X 10 -11 m) - 2.19x Problems 16, 20 Sincethe electrostatic force depends on the inverse square of the distance between The physics of thecharges,it becomes larger for smaller distances, such as those involved when a strip adhesion. Ofadhesivetape is stuck to a smooth surface. Electrons shift over the small distances betweenthe tape and the surface. As a result, the materials become oppositely charged. Sincethe distance between the charges is relatively small, the electrostatic force of attractionis large enough to contribute to the adhesive bond. Figure 18.12 shows an image of thestickysurface of a piece of tape after it has been pulled off a metal surface. The image wasObtainedusing an atomic-force microscope and reveals the tiny pits left behind when microscopicportions of the adhesive remain stuck to the metal because of the strong adhesivebonding forces. gm 1.5 THE FORCE ON A POINT CHARGE DUETo TWO OR MORE OTHER POINT CHARGES Up to now, we have been discussing the electrostatic force on a point charge (magnitudeIqi l) due Suppose that a third point to another point charge (magnitude11/21). charge(magnitude Iq31) is also present. What would be the net force on due to both andq3?It is convenient to deal with such a problem in parts. First, find the magnitude and directionOfthe force exerted on by (ignoring (13).Then, determine the force exerted onqi by (ignoring The net force on is the vector sum of these forces. Examples 4 and 5 illustrate this approach when the charges lie along a straight line and on a plane, respectively. 1.0 0.5 Figure 18.12 After a strip of adhesive tape has been pulled off a metal surface, there are tiny pits (approximately one ten-millionth of a meter in diameter) in the stick surface of the tape, as this image shows. It was obtained using an atomic-force microscope. (CourtesyLouis Scudiero and J. Thomas Dickinson, Washington State University.) 546 | CHAPTER 18 I ELECTRIC FORCES AND ELECTRIC FIELDS nA, 0.20 m -4.0 PC 0.15m +3.0 q3 -7.0 pc (a) (b) Free-body diagram for Figure 18.13 (a) Three charges lying along the x axis. (b) The force exerted on qi by is F12,while the force exertedon by Example 4 | Three Charges on a Line Figure 18.13a shows three point charges that lie along the x axis in a vacuum. Determine the magnitude and direction of the net electrostatic force on Cll. Reasoning Part b of the drawing shows a free-body diagram of the forces that act onq. Since and have opposite signs, they attract one another. Thus, the force exertedonq, by is F12, and it points to the left. Similarly, the force exerted on by is F13andisalso an attractive force. It points to the right in Figure 18.13b.The magnitudes of these forcescan be obtained from Coulomb's law. The net force is the vector sum of F12and Fn. Solution The magnitudes of the forces are is Fn. llq21 109 r 12 Problem solving insight When using Coulomb's law (F = k Iql llq21/r2), rememberthat the symbols qtl and Iq21stand for the charge magnitudes. Do not substitute negative numbers for these symbols. (8.99 x IL/lIIq31 x X 10-6 C) X 10-6 (8.99 X 109 r 13 10 -6 (0.20 m) 2 X 10-6 C) (0.15 = 2.7 N = 8.4N Since F12points in the negative x direction, and F13points in the positive x direction,thenet force F is (-2.7 N) + (8.4 N) The plus sign in the answer indicates that the net force points to the right in the drawing. Example 5 | Three Charges in a Plane Figure 18.14a shows three point charges that lie in the x, y plane in a vacuum. Find the magnitude and direction of the net electrostatic force on Cll. Need more practice? Interactive LearningWare 18.2 A positive charge is placed on the x axis (q = +7.00 gc, x = 0.600m), and another positive charge is placed on the y axis (q = +9.00 y = 0.400 m). A third charge (q — -6.00 gc, m = 5.00 x 10-8 kg) is placed at the coordinate origin. If the charge at the origin were free to move, what would be the magnitude of its acceleration? Related Homework: Problem 17 For an interactive solution, go to www.wiley.com/college/cutnell Reasoning The force exerted on by is F12and is an attractive force becausethetwo charges have opposite signs. It points along the line between the charges. The force exertedon qi by is F13 and is also an attractive force. It points along the line between qi and(13. Coulomb's law specifies the magnitudes of these forces. Since the forces point in differentdirections (see Figure 18.14b), we will use vector components to find the net force. Solution The magnitudes of the forces are Iql Il q21 (8.99 x 109 r 12 F 13= k Iqlllq31 X 10-6 X 10-6 C) x 10-6 X 10-6 C) (0.15 (8.99 X 109 r 13 (0.10 F12 sin 730 730 COS 73 0 is F = + F13.The angle that F makes with the +x axis is 0. = 18N The net force F is the vector sum of F12and F13,as part b of the drawing shows. The comop nents of F that lie in the x and y directions are Fr and Fy, respectively. Our approachto findin g F is the same as that used in Chapters I and 4. The forces F12and F13are resolved intox andy components. Then, the x components are combined to give Fx, and the y componentsare -6.0pc Figure 18.14 (a) Three charges lying in a plane. (b) The net force acting on = 9.6 N +4.0{IC = 0.10 m (a) -5.0 pc (b) Free-body diagram for qi 18.6 | THE ELECTRIC FIELD | 547 Fy. Once Fx and Fy are known, the magnitude and direction of F can be decombinedto give using trigonometry. termined x component y component Force F12 +(9.6 N) cos 730 = +2.8 N +(9.6 N) sin 730 = +9.2 N ON F13 +21 N = +9.2 N O (21 tan -1 tan -l The electrostatic force is a vector and has a direction as well as a magnitude. When adding electrostatic forces, take into account the directions of all forces, using vector components as needed. F and the angle 0 of the net force are Themagnitude F = Fx2+ F 2 Problem solving insight + (9.2 = 9.2 N 21 N CHECK YOUR UNDERSTANDING 2 Thedrawing shows three point charges arranged in differentways. The charges are +q, -q, and —q;each has the same magnitude, one being positive and the other two negative. Ineach part of the drawing the distance dis the same. Rank the arrangements in descendingorder (largest first) according to the magnitude of the net electrostaticforce that acts on thepositive charge. (The answer is given at the end of the book.) d -q d (a) —q —q d 900 d (b) —q (c) Background:The electrostatic force that one point charge exerts on another is described by Coulomb'slaw. Like any force, this force is a vector quantity, and the rules of vector additionapply. Forsimilar questions (including calculational counterparts), consult Self-Assessment Test 18.1,which is described next. ELF-ASSESSMENT TEST 18.1 www.wiley.com/college/cutnell Testyour understanding of the material in Sections 18.1-18.5: • The Origin of Electricity • Charged Objects and the Electric Force • Conductors and Insulators • Charging by Contact and by Induction • Coulomb's Law THE ELECTRIC FIELD DEFINITION Ebonite rod As We know, a charge can experience an electrostatic force due to the presence of Othercharges. For instance, the positive charge qo in Figure 18.15 experiences a force F, Which is the vector forces exerted by the charges on the rod and the two spheres.It is useful sum of the to think of CIO as a test charge for determining the extent to which the sum)undingcharges to select generate a force. However, in using a test charge, we must be careful one with a very small magnitude, so that it does not alter the locations of the othercharges. The next example illustrates how the concept of a test charge is applied. e Example 6 | A Test Charge positive test forceF charge shown in Figure 18.15 is = +3.0 X 10-8 C and experiences a X 10-8 N in the direction shown in the drawing.(a) Find theforce per coulomb Figure 18.15 A positive charge qo experiences an electrostatic force F due to the surrounding charges on the ebonite rod and the two spheres. 548 CHAPTER 18 ELECTRIC FORCES AND ELECTRIC FIELDS Using the result of part (a), predict the forcethatac that the test charge experiences. (b) hogt if it replaced qo. of +12 ><10-8 C would experience apply a force F to the test charge Reasoning The charges in the environment is F/qo. If is replaced by a newc test charge force per coulomb experienced by the force per coulomb times q. then the force on this new charge is the charge is Solution (a) The force per coulomb of 6.0 x 10-8 N = 2.0 N/C F 3.0 x 10-8 c that the surrounding charges can exert 2.0 (b) The result from part (a) indicates a fore charge of +12 ><10-8 C would experience force per coulomb of charge. Thus, a whose magnitude is as that experienced by the test charge,since The direction of this force would be the same have the same positive sign. in Example 6(a) is one illustrationof The electric force per coulomb, F/qo, calculated The idea is calledtheelectric an idea that is very important in the study of electricity. field. electric the of field. Equation 18.2 presents the definition DEFINITION OF THE ELECTRIC FIELD bya The electric field E that exists at a point is the electrostatic force F experienced itself: charge the by small test charge* qo placed at that point divided (182) The electric field is a vector, and its direction is the same as the directionof theforce F on a positive test charge. SI Unit of Electric Field: newton per coulomb (NIC) Equation 18.2 indicates that the unit for the electric field is that of forcedivided by charge, which is a newton/coulomb (NIC) in SI units. It is the surrounding charges that create an electric field at a givenpoint.Any itive or negative charge placed at the point interacts with the field and, as a result,expen• ences a force, as the next example indicates. Example 7 | An Electric Field Leads to a Force In Figgre 18.16a the charges on the two metal spheres and the ebonite rod createanelectnc field E at the spot indicated. This field has a magnitude of 2.0 N/C and is directedasinthe Of(a) drawing. Determine the force on a charge placed at that spot, if the charge has a value = +18 X 10-8 C and (b) = -24 X 10-8 C. 011 the Reasoning The electric field at a given spot can exert a variety of forces, depending magnitude and sign of the charge placed there. The charge is assumed to be smallenough it does not alter the locations of the surrounding charges that create the field. Solution (a) The magnitude of the force is the product of the magnitudesof qoandE' F = IqolE = (18 x 10-8 N/C) = Since qo is positive, the force points in the same direction as the electric field,as part drawing indicates. *As long as the test charge is small enough that it does not disturb the surrounding charges,it maybe either positive or negative. Compared to a positive test charge, a negative test charge Of the same magnitude experiences a force of the same magnitude that points in the opposite direction. However, the sameelectnc is given by Equation 18.2, in which F is replaced by —F and qo is replaced by —CIO. 18.6 | THE ELECTRIC FIELD | 549 (a) Figure 18.16 The electric field E that exists at a given spot can exert a variety of forces. The force exerted depends on the magnitude and sign of the charge placed at that spot. (a) The force on a positive charge points in the same direction as E, while (b) the force on a negative charge points opposite to E. (b) (b)Inthiscase, the magnitude of the force is F = IqolE (24 X 10-8 N/C) (18.2) Theforceon the negative charge points in the direction opposite to the force on the positive charge—thatis, opposite to the electric field (see part b of the drawing). At a particular point in space, each of the surroundingcharges contributes to the net electricfield that exists there. To determine the net field, it is necessary to obtain the variouscontributionsseparately and then find the vector sum of them all. Such an approach is anillustrationof the principle of linear superposition, as applied to electric fields. (This principleis introduced in Section 17.1, in connection with waves.) Example 8 emphasizes thevectornature of the electric field, and Example 9 illustrates that a charged particle accelerates in an electric field. V Example 8 | Electric Fields Add as Vectors Do Figure18.17shows two charged objects, A and B. Each contributes as follows to the net electricfieldat point P: EA = 3.00 N/C directed to the right, and EB = 2.00 N/C directed downward.Thus, EA and EB are perpendicular. What is the net electric field at P? ReasoningThe net electric field E is the vector sum of EA and EB•.E = EA + EB. As illustrated in Figure 18.17, EA and EB are perpendicular, so E is the diagonal of the rectangle shownin the drawing. Thus, we can use the Pythagorean theorem to find the magnitude of E andtrigonometryto find the directional angle 0. Solution The magnitude of the net electric field is = (3.00 N/C)2 + (2.00 N/C)2 = Thedirectionof E is given by the angle 0 in the drawing: 0 = tan -1 tan -1 2.00 N/C 3.00 N/C Figure 18.17 The electric field contributions EA and EB, which come from the two charge distributions. are added vectorially to obtain the net field E at point P. ANALYZING MULTIPLE-CONCEPT PROBLEMS Example9 | A Proton Accelerating in an Electric Field In a a vacuum, a proton (charge = +e, mass = 1.67 X 10-27kg) is moving parallel to a uniformelectric field that is directed along the +x axis (see Figure 18.18). The proton startsWitha velocity Of +2.5 X 104m/s and accelerates in the same direction as the electricfield, which has a value of +2.3 X 103NIC. Find the velocity of the proton Whenits displacement is +2.0 mm from the starting point. +2.00 mm Figure 18.18 A proton, moving to the right, accelerates in the presence of the electric field Ex. 550 | CHAPTER 18 I ELECTRIC FORCES AND ELECTRIC FIELDS Reasoning Since we know the initial velocity and displacement of the proton, we can determine its final velocityfrom tion of kinematics, provided the proton's acceleration can be found. The acceleration is given by Newton's secondlawasan the force acting on the proton divided by its mass. The net force is the electrostatic force, since the proton is moving in an electric field. The electrostatic force depends on the proton's charge and the electric field, both of which are known. Knowns and Unknowns The data for this problem are listed as follows: Description Proton charge Proton mass Initial velocity of proton Electric field Displacement of proton Symbol Value Comment +1.60 x 10-19c m vox 1.67 X 10-27 kg +2.5 X 104 m/s x +2.3 X 103 N/C +2.0 mm 2.0 mm = 2.0 X 10-3 m Unknown Variable Final velocity of proton Modeling the Problem STEP 1 Kinematics To obtain the final velocity vx of the proton we employ Equation 3.6a from the equations of kinematics: vx2 = v0x2 + 2axx. We have chosen this equation because two of the variables, the initial velocity vot and the displacement x, are known. Taking the square root of each side of this relation and choosing the + sign, since the proton is moving in the +x direction (see Figure 18.18), we arrive at Equation 1 in the right column. Although the acceleration ax is not known, we will obtain an expression for it in Step 2. STEP 2 Newton's Second Law Newton's second law, as given in Equation 4.2a, states that the acceleration ax of the proton is equal to the net force EFx acting on it divided by the proton's mass m: ax = EFx/m. Only the electrostatic force Fx acts on the proton, so it is the net force. Setting SFx = Fx in Newton's second law gives ax — m m This expression can be substituted into Equation 1, as indicated at the right. The electrostatic force is not known, so we proceed to Step 3 to evaluate it using the concept of the electric field. STEP 3 The Electric Field Since the proton is moving in a uniform electric field E it experiences an electrostatic force Fx given by Fx = qoEx(Equation 18.2), where qo is the charge. Setting = e for the proton, we have Fx = eEx + 2ax X ax = All the variables on the right side of this equation are known, so we substitute it into Equation 2, as shown in the right column. Solution Vx= + Algebraically combining the results of the three steps, we have m eEx m x (l) (2) m 18.6 THE ELECTRIC FIELD 551 velocity of the proton is Thefinal + Vox+ 2 m x vs -+ (2.5 X 104m/s)2 +2 (1.60 X 10-19 X 103N/C) (2.0 X 10-3 m) 1.67 X 10-27kg +3.9 X 104m/s denotes that the final velocity points along the +x axis. wherethe + sign Problems 38, 40, 66 Related Homework: POINT CHARGES A more complete understanding of the electric field concept can be gained by consideringthe field created by a point charge, as in the following example. Example 10 | The Electric Field of a Point Charge (a) Thereis an isolated point charge of q +15 'LC in a vacuum at the left in Figure 18.19a. Using a testcharge of qo = +0.80 /-LC,determine the electric field at point P, which is 0.20 m away. Reasoning Following the definition of the electric field, we place the test charge qo at point P, determine the force acting on the test charge, and then divide the force by the test charge. Solution Coulomb's law (Equation 18.1), gives the magnitude of the force: Iqollql r2 (8.99 x 109 x 10-6 (0.20 x 10-6 C) q (b) q = 2.7 N Equation18.2 gives the magnitude of the electric field: 2.7 N 0.80 x 10-6 c 3.4 x 106 NIC Iqol Theelectric field E points in the same direction as the force F on the positive test charge. Sincethe test charge experiences a force of repulsion directed to the right, the electric field vectoralso points to the right, as Figure 18.19b shows. (c) Figure 18.19 (a) At location P, a positive test charge qo experiences a repulsive force F due to the positive point charge q. (b) At P the electric field E is directed to the right. (c) If the charge q were negative rather than positive, the electric field would have the same magnitude as in (b) but would point to the left. The electric field produced by a point charge q can be obtained in general terms from Coulomb'slaw. First, note that the magnitude of the force exerted by the charge q on a testcharge is F klqllqol/r2. Then, divide this value by Iqol to obtain the magnitude Ofthe field. Since Iqol is eliminated algebraically from the result, the electric field does notdependon the test charge: Pointcharge q klql 2 (18.3) As in Coulomb's law, the symbol Iql denotes the magnitude of q in Equation 18.3, withOutregard to whether q is positive or negative. If q is positive,then E is directed away from as in Figure 18.19b. On the other hand, if q is negative,then E is directed toward q, since a negativecharge attracts a positive test charge. For instance, Figure 18.19c shows the electric fieldthat Would exist at P if there were a charge of —qinstead of +q at the left of the drawIng•Example I I reemphasizes the fact that all the surroundingcharges make a contribution tothe electric field that exists at a given place. d V Example 11 | The Electric Fields from SeparateCharges May Cancel TwoPOSitivepoint charges, +4.0 pc, are separated in a vacuum by a +16 'LC and distanceof 3.0 m, as Figure 18.20 illustrates. Find the spot on the line between the charges Wherethe net electric field is zero. 3.0 m Figure 18.20 The two point charges q, and create electric fields El and E2 that cancel at a location P on the line between the charges. 552 CHAPTER 18 ELECTRIC FORCES AND ELECTRIC FIELDS Reasoning Between the charges the two field contributions have opposite directions, and the net electric field is zero at the place where the magnitude of El equals that of E2. However, since is smaller than (Il, this location must be closer to (12,in order that the field of the smaller charge can balance the field of the larger charge. In the drawing, the cancellationspot is labeled P, and its distance from is d. Solution At P, El = E2, and using the expression E = klql/r 2, we have Problem solving insight Equation 18.3gives only the magnitude of the electric field produced by a point charge. Therefore, do not use negative numbers for the symbol Iql in this equation. —q 1 k(4.o X 10-6 C) (3.0 m — Rearranging this expression shows that 4.0(3.0 m — d) 2 = d 2. Taking the square root of each side of this equation reveals that 2.0(3.Om - d) The plus and minus signs on the right occur because either the positive or negative root can be taken. Therefore, there are two possiblevalues for d: +2.0 m and +6.0 m. The value +6.0 m corresponds to a location off to the right of both charges, where the magnitudes of El and E2 are equal, but where the directions are the same. Thus, El and E2 do not cancel at this spot. The other value for d corresponds to the location shown in the drawing and is the zero-fieldIocation: d = +2.0 m When point charges are arranged in a symmetrical fashion, it is often possible to deduce useful information about the magnitude and direction of the electric field by taking advantage of the symmetry. Conceptual Example 12 illustrates the use of this technique. c 4 k(16 x 10-6 c) 3 (a) V Conceptual Example 12 | Symmetry and the Electric Field Figure 18.21 shows point charges fixed to the corners of a rectangle in two different ways. The charges all have the same magnitudes, but they have different signs. Consider the net electric field at the center C of the rectangle in each case. Which field is larger? Reasoning and Solution In Figure 18.21a, the charges at corners 1 and 3 are both +q. The positive charge at corner I produces an electric field at C that points toward corner 3. In contrast, the positive charge at corner 3 produces an electric field at C that points toward corner l. Thus, the two fields have opposite directions. The magnitudes of the fields are identicalbecause the charges have the same magnitude and are equally far from the center. Therefore, the fields from the two positive charges cancel. Now, let's look at the electric field produced by the charges on corners 2 and 4 in Figure 18.2la. The electric field due to the negative charge at corner 2 points toward corner 2, and the field due to the positive charge at corner 4 points the same way. Furthermore, the magnitudes (b) Figure 18.21 Charges of identical magnitude, but different signs, are placed at the corners of a rectangle. The charges give rise to different electric fields at the center C of the rectangle, depending on the signs of the charges. of these fields are equal because each charge has the same magnitude and is located at the same distance from the center of the rectangle. As a result, the two fields combine to give the net electric field Eu shown in the drawing. In Figure 18.21b, the charges on corners 2 and 4 are identical to those in part a Ofthe drawing, so this pair gives rise to the electric field labeled E24.The charges on corners I and 3 are identical to those on corners 2 and 4, so they give rise to an electric field labeled E13, which has the same magnitudeas E24.The net electric field E is the vector sum of E24and E13 and is also shown in the drawing. Clearly, this sum is greater than Eu alone. Therefore' the netfield in part b is larger than that in part a. Related Homework: Conceptual Question 12, Problem 34 Need more practice? nteractive LearningWare 18.3 A constant electric field exists in a region of space. The field has a magnitude of 1600 N/C and points due north. A point charge of +4.0 x 10-9 C is then placed in this electric field. (a) What is the magnitude and direction of the net electric field at a spot 13 cm due east of the charge? Specify the direction as an angle relative to due east. (b) If a —7.00'LC charge were placed at this spot, what would be the electrostatic force (magnitude and direction) exerted on it? Related Homework: Problem 59 For 4. an interactive solution, go to www.wiley.com/college/cutnell 18.7 THE PARALLEL ELECTRIC FIELD LINES 553 PLATE CAPACITOR which gives the electric field of a point charge, is a very useful reEquation 18.3, calculus, this equation can be applied in a variety of situaintegral aid of stilt.Withthe distributed over one or more surfaces. One such example are point charges tionswhere importance is the parallel plate capacitor. As Figure 18.22 practical thathasconsiderable device consists of two parallel metal plates, each with area A. A charge +q is shows,this over one plate, while a charge —qis spread uniformly over the other spreaduniformly between the plates and away from the edges, the electric field points plate.In the region plate toward the negative plate and is perpendicular to both. It can be fromthe positive 16 in Section 18.9) that the electric field has a magnitude of shown(see Example (18.4) parallelplate capacitor Area = A —q Figure 18.22 A parallel plate capacitor. of free space. In this expression the Greek symbol sigma (c) where€0is the permittivity the charge density. denotesthe charge per unit area (u = q/A) and is sometimes called has the same value at all places between the Exceptin the region near the edges, the field distinct contrast to plates.The field does not depend on the distance from the charges, in point charge. thefieldcreated by an isolated CHECK YOUR UNDERSTANDING 3 A positivepoint charge +q is fixed in position at the center of a square, as the drawing shows.A second point charge is fixed to either corner B, corner C, or corner D. The net electricfieldat corner A is zero. (a) At which corner is the second charge located? (b) Is the secondcharge positive or negative? (c) Does the second charge have a greater, a smaller, or thesamemagnitude as the charge at the center? (The answers are given at the end of the book.) Background: The electric field is a vector quantity. Its magnitudeand direction depend on themagnitudes and algebraic signs of the charges that create it. Forsimilarquestions (including calculational counterparts), consult Self-Assessment Test 18.2,whichis described at the end of Section 18.9. c ELECTRIC FIELD LINES As we have seen, electric charges create an electric field in the space surrounding them.It is useful to have a kind of "map" that gives the direction and indicates the strengthof the field at various places. The great English physicist Michael Faraday (1791 1867)proposed an idea that provides such a "map," the idea of electric field lines. Sincethe electric field is the electric force per unit charge, the electric field lines are sometimes called lines offorce. TOintroduce the electric field line concept, Figure 18.23a shows a positive point charge+q. At the locations numbered 1—8,a positive test charge would experience a 1 2 8 5 (a) (b) Figure 18.23 (a) At any of the eight marked spots around a positive point charge +q, a positive test charge would experience a repulsive force directed radially outward. (b) The electric field lines are directed radially outward from a positive point charge +q. 554 CHAPTER 18 ELECTRIC FORCES AND ELECTRIC FIELDS —q Figure 18.24 The electric field lines are directed radially inward toward a negative point charge —q. Edge view Figure 18.25 In the central region of a parallel plate capacitor, the electric field lines are parallel and evenly spaced, indicating that the electric field there has the same magnitudeand direction at all points. repulsive force, as the arrows in the drawing indicate. Therefore, the electricfieldc by the charge +q is directed radially outward. The electric field lines are lines show this direction, as part b of the drawing illustrates. They begin on the charge+qand point radially outward. Figure 18.24 shows the field lines in the vicinityof a negative charge —q.In this case they are directed radially inward because the force on a positive test charge is one of attraction, indicating that the electric field points inward.In general, electric field lines are always directed away from positive charges and towardnegaåve charges. The electric field lines in Figures 18.23 and 18.24 are drawn in only two dimensions as a matter of convenience. Field lines radiate from the charges in three dimensions, and an infinite number of lines could be drawn. However, for clarity only a small ever included in pictures. The number is chosen to be proportional to the magnitude of the charge; thus, five times as many lines would emerge from a +5q chargeas froma+q charge. The pattern of electric field lines also provides information about the magnitude strength of the field. Notice in Figures 18.23 and 18.24 that near the charges,where the electric field is stronger, the lines are closer together. At distances far from thecharges where the electric field is weaker, the lines are more spread out. It is true in general that the electric field is stronger in regions where the field lines are closer together.In factno matter how many charges are present, the number of lines per unit area passingperpen. dicularly through a surface is proportional to the magnitude of the electric field. In regions where the electric field lines are equally spaced, there is the samenumber of lines per unit area everywhere, and the electric field has the same strengthat allpoints For example, Figure 18.25 shows that the field lines between the plates of a parallelplate capacitor are parallel and equally spaced, except near the edges where theybulgeout. ward. The equally spaced, parallel lines indicate that the electric field has the samemagnitude and direction at all points in the central region of the capacitor. Often, electric field lines are curved, as in the case of an electric dipole. Anelectric dipole consists of two separated point charges that have the same magnitude butopposite signs. The electric field of a dipole is proportional to the product of the magnitudeofone of the charges and the distance between the charges. This product is called the dipolemo• ment. Many molecules, such as 1--120 and HCI, have dipole moments. Figure 18.26depicts the field lines in the vicinity of a dipole. For a curved field line, the electric fieldvectorat a point is tangent to the line at that point (see points 1, 2, and 3 in the drawing).Thepat• be tern of the lines for the dipole indicates that the electric field is greatest in the region tween and immediately surrounding the two charges, since the lines are closesttogether there. Notice in Figure 18.26 that any given field line starts on the positive chargeand on the negative charge. In general, electric field lines always begin on a positivecharge 1 Figure 18.26 The electric field lines of an electric dipole are curved and extend from the positive to the negative charge. At any point, such as l, 2, or 3, the field created by the dipole is tangent to the line through the point. 2 3 18.7 ELECTRIC FIELD LINES 555 Figure 18.27 The electric field lines for two identical positive point charges. If the charges were both negative, the directions of the lines would be reversed. andend on a negative charge and do not start or stop in midspace.Furthermore, the numberof lines leaving a positive charge or entering a negativecharge is proportional to the magnitude of the charge. This means, for example, that if 100 lines are drawn leavinga +4 /JC charge, then 75 lines would have to end on a —3kLCcharge and 25 lines ona —1/.LCcharge. Thus, 100 lines leave the charge of +4 'LC and end on a total charge of —4'LC, so the lines begin and end on equal amounts of total charge. The electric field lines are also curved in the vicinity of two identical charges. Figure 18.27shows the pattern associated with two positive point charges and reveals that there is an absence of lines in the region between the charges. The absence of lines indicates thatthe electric field is relatively weak between the charges. Some of the important properties of electric field lines are considered in Conceptual Example 13. V Conceptual Example 13 | Drawing Electric Field Lines Figure18.28a shows three negative point charges ( —q, —q,and —2q)and one positive point charge(+4q), along with some electric field lines drawn between the charges. There are three thingswrong with this drawing. What are they? Reasoningand Solution One aspect of Figure 18.28athat is incorrect is that electric field linescross at point P. Field lines can never cross, and here's why.An electric charge placed at P experiencesa single net force due to the presence of the other charges in its environment. Therefore,there is only one value for the electric field (which is the force per unit charge) at thatpoint. If two field lines intersected, there would be two electric fields at the point of intersection,one associated with each line. Since there can be only one value of the electric field at anypoint, there can be only one electric field line passing through that point. Another mistake in Figure 18.28a is the number of electric field lines that end on the negativecharges. Remember that the number of field lines leaving a positive charge or entering a (a) negativecharge is proportional to the magnitudeof the charge.The —2qcharge has half the magnitudeof the +4q charge. Therefore, since 8 lines leave the +4q charge, 4 of them (one- half Of them) must enter the —2q charge. Of the remaining 4 lines that leave the positive charge,2 enter each of the —qcharges, according to a similar line of reasoning. The third error in Figure 18.28a is the way in which the electric field lines are drawn between the+4q charge and the —q charge at the left of the drawing.As drawn, the lines are parallel and evenlySpaced.This would indicate that the electric field everywhere in this region has a constant magnitudeand direction, as is the case in the central region of a parallel plate capacitor. But the electricfield between the +4q and —qcharges is not constant everywhere. It certainly is stronger InPlacesclose to the +4q or —qcharge than it is midway between them. The field lines, therefore, shouldbe drawn with a curved nature, similar (but not identical) to those that surround a dipole. Igure18.28b shows more nearly correct representations of the field lines for the four charges. Related Homework: Problems 25, 26 -q —W (b) Figure 18.28 (a) Incorrectly and (b) correctly drawn electric field lines. 566 | CHAPTER 18 | ELECTRIC FORCES AND ELECTRIC FIELDS Learning Aids Discussion Topic Electric field of a point charge PROBLEMS directionof the force on a positive test charge. The direction of the electric field is the same as the (NIC). The source of the electric field coulomb per newton the is field electric The SI unit for the spot. that surrounding at any spot is the charged objects charge q is The magnitude of the electric field created by a point Ex.10-u18 klql (18.3) Excess charge carried by a conductor at equilibrium IS 1841 10.There is an electric field at point P. A very small charge is placedat this point and experiences a force. Another very small chargeis then placed at this point and experiencesa force that differsin both magnitude and direction from that experienced by the firstcharge.How can these two differentforces result from the singleelectricfield that exists at point P? Il. Ihree point charges are fixed to the comers of a square,one to a comer.in such a way that the net electric field at the empty comer is zero.Do these charges all have (a) the same sign and (b) the same magnitude(but, possibly, different signs)? Justify your answers. (18.4) Electric field of a parallel plate Direction of electric field Strength of electric field foryour answer. nonconducting rod, positive charges are spread evenly, 9. On a thin, the same amount of charge per unit length at every is there that so pint. On another identical rod, positive charges are spread evenly overonly the left half, and the same amount of negative charges are spreadevenly over the right half. For each rod, deduce the direction of theelectric field at a point that is located directly above the midpointof the rod. Give your reasoning. ILW 18.3 and r is the distance from the where Iql is the magnitude of the charge and has no algebraic sign a negative charge. toward and charge positive a from charge. The electric field points away plate, the magnitude of each on For a parallel plate capacitorthat has a charge per unit area of is plates the between the electric field Electric field lines simple harmonic motion on being released, as will the notexhibit spring. Explain why not. the on panicle point charges are fixed to opposite corners of a Identical 8. ssm does a third point charge experience the greater force, square.Where empty corners or at the center of the square?Account atone of the FIELDLINES Electricfield lines are lines that can be thought of as a "map," inso18.7 ELECTRIC far as the lines provide informationabout the directionand strength of the electric field. The lines are directed away from positive charges and toward negative charges. The direction of the lines gives the direction of the electric field, since the electric field vector at a point is tangent to the line at that point. The electric field is strongest in regions where the number of lines per unit area C 18.2 passing perpendicularlythrough a surface is the greatest—that is, where the lines are packed together most tightly. 12. Review Conceptual Example 12 as an aid in answering this ques- 18.8 THE ELECTRIC FIELD INSIDE A CONDUCTOR: SHIELDING Excess negativeor positive EL 14 tion.Supposein figure 18.21 that charges +q are placed on corners I and3 of the rectangle, and charges —qare placed on comers 2 and 4. Whatis the net electric field at the center C of the rectangle? 13. In figure 18.26 there is no place on the line through the charges wherethe electric field is zero, neither to the left of the positive charge,nor between the charges, nor to the right of the negative charge.Now, suppose the magnitude of the negative charge were geater than the magnitude of the positive charge. Is there any place onthe line through the charges where the electric field is zero? Jus- charge resides on the surface of a conductorat equilibrium under electrostatic conditions. In such a situation, the electric field at any point within the conductingmaterial is zero, and the electric field just outside the surface of the conductoris perpendicular to the surface. through a surface is related to the magnitude E of the 18.9 GAUSS' LAW The electric flux electric field, the area A of the surface, and the angle that specifies the direction of the field relative to the normal to the surface: = Electric flux (18.6) cos Gauss' law states that the electric flux througha closed surface (a Gaussian surface) is equal to the net charge Q enclosed by the surface divided by €0, the permittivity of free space: = Gauss' law tify your answer. ELL 16 cos 4)AA 14. ssm Drawings I and Il showtwo examples of (18.7) IS 18.51 electric fieldlines.Decide whichof the following statementsare true and use Self-AssessmentTest 18.2to evaluate your understanding of Sections 18.6—18.9. whichare false, defending CONCEPTUAL QUESTIONS SSm Solutionis in the StudentSolutionsManual. 5. Blow up a balloon and rub it against your shirt a left with a charge? Account for your answer. remain stuck to the ceiling. Why? 6. A proton and an electron are held in place on the x axis. fie copper sphere is neutral. The rod and the sphere do not touch. Is it possible for the rod and the sphere to (a) attract one another and (b) repel one another? Explain. 4. On a dry day, Just after washing your hair to remove naturaloils and drying it thoroughly.run a plastic comb through it. Small bits of paper will be attracted to the comb. Explain why the electric field becomesstronger. (c) The electric field in I is the same everywherebut becomesstronger in Il as you move from left to right. (d) The electric fields in both I and Il could be created by negativecharges located somewhere on the left and positive charges somewhereon the right. (e) Both I and Il arise from a single positive point charge located somewhere on the left. 15. A positively charged particle is moving horizontally when it enters the region between the plates of a capacitor. as the drawing illustrates. (a) Draw the traJectory that the particle follows in moving through the capacitor. (b) When the particle is within the capacitor, which of the following four vectors, if any, are parallel to the electric field inside the capacitor: the particle's displacement,its velocity, its linear momentum,its acceleration?For each vector explain why the vector is, or is not, parallel to the electric field of the capacitor. 16. Refer to figure 18.27. Imagine a plane that is perpendicular to a line betweenthe charges, midway between them, and is half into and half out of the paper.The electric flux through this plane is zero. Explain why 17. ssm Two charges, +q and —q,are inside a Gaussian surface. Since the net charge inside the Gaussian surface is zero, Gauss' law states that the electric flux throughthe surface is also zero: that is. = O. Does the fact that = O imply that the electric field É at any point on the Gaussian surface is also zero? Justify your answer. (Hint: Imagine a Gaussian surface that encloses the two charges in Figure 18.26.) 18. The drawing shows three charges, labeled q, q:, and '13. A Gaussian surface is drawn around q, and '12. (a) Which charges determine the electric flux throughthe Gaussiansurface? (b) Which charges produce the electric field at the point P? Justify your answers. Gaussian surface 19. A charge +q is placed inside a spherical Gaussian surface. The charge is not located at the center of the sphere. (a) Can Gauss' law tell us exactly where the charge is located inside the sphere? Justify your answer. (b) Can Gauss' law tell us about the magnitude of the electric flux through the Gaussian surface? Why? PROBLEMS 1. In Figure 18.8 the grounding wire is removed first, followed by the rod. and the sphere is left with a positive charge. If the rod were removed first. followed by the grounding wire, would the sphere be 2. A metallic object is given a positive charge by the process Of induction. as illustrated in Figure 18.8. (a) Does the mass of the object increase. decrease, or remain the same? Why? (b) What happensto the mass of the object if it is given a negative charge by induction? Explain. 3. A rod made from insulating material carries a net charge, while a 11 yourchoicein each case. (a) In both I and Il the electric field is the sameeverywhere.(b) As you move from left to right in each case, | 567 Of times. In so doing you give the balloon a net electric charge. touch the balloon to the ceiling. On being released, the balloonwill ton is at x = d, while the electron is at x = +d. Theyarereleased simultaneously, and the only force that affects their motionsis the electrostaticforce of attractionthat each applies to the other. particle reaches the origin first? Give your reasoning. 7. A particle is attached to a spring and is pushed so that the spnng is compressed more and more. As a result, the spring exerts a greater and greater force on the particle. Similarly, a charged partw experiences a greater and greater force when pushed closer of closer to another particle that is fixed in position and has a charge the same polarity In spite of the similarity. the charged particle Problemsthat are not marked with a star are consideredthe easiest to solve. Problems that are marked witha single star C) are more difficult,while thosemarkedwith a double star ( •e) are the most diffcult. Noteto Instructors: Most of the homeworkproblemsin this chapterare availablefor assignment via an online homeworkmanagementprogram such as WileyPLUS or WebAssign,and those marked withthe icon GO are presented in a guided tutorialformat that provides enhanced interactivity. See Pmface for addittonal details. ssm Solutionis in the Student Solutions Manual. This icon represents a biomedical application. www Solution is available on the World Wide Web at www.wiIey.com/college/cutnelJ Section18.1The Origin of Electricity, 18.2Charged Objects and the Electric Force, Section183 Conductors and Insulators, Section18.4 Charging by Contact and by Induction 1. ssm How many electrons must be removed from an electrically neutralsilver dollar to give it a charge of +2.4 gC? 2. A plate carries a charge of —3.0/LC,while a rod carries a charge of +2.0 gC. How many electrons must be transferredfrom the plate to the rod, so that both objects have the same charge? 3. A metal sphere has a charge of +8.0 /.cc. What is the net charge after 6.0 X 1013electrons have been placed on it? 4. Four identicalmetallic objects carry the following charges: + 1.6. +6.2, —4.8.and —9.4pc. ne objects are brought simultaneously 570 | CHAPTER 18 I ELECTRIC FORCES ELECTRIC AND FIELDS CONCEPTS electric field E in the drawing is uniform and has a magnitude 42. The drawing shows an electron entering the lower left side of a paral- of 250 NIC. Find the electric flux through (a) surface I and (b) surface 2. lel plate capacitor and ex- 0.150 cm iting at the upper right side. The initial speed of the electron is 7.00 x 106 m/s. The capacitor is 48. A rectangular surface (0.16 m ><0.38 m) is oriented in a uni. form electric field of 580 NIC. What is the maximum possibleelec. tric flux through the surface? 49. A vertical wall (5.9 m X 2.5 m) in a house faces due east.A 2.00 cm long, and its plates are separated by O.150 cm. Assume that the electric field between the plates is uniform everywhere and find its magnitude. 43. A small plastic ball with a mass of 6.50 X 10-3 kg and with a charge of +0.150 is suspended from an insulating thread and hangs between 1 30.00 uniform electric field has a magnitude of 150 NIC. This field is par- allel to the ground and points 350 north of east. What is theelectric flux through the wall? 50. A charge Q is located inside a rectangular box. The electricflux = +1500 N•m2/C, through each of the six surfaces of the box is: = +4600 N.m2/C, = -1800 = +2200 N.m2/C, = -3500 N. m2/C, = -5400 N.m2/C. Whatis Q? the plates of a capacitor (see the drawing). The ball is in equilibrium, with the thread making an angle of 30.00 with respect to the vertical. The area of each plate is 0.0150 m2. What is the magnitude of the charge on each plate? 51. Interactive Solution 18.51 at www.wiley.com/college/cutnell offers help with this problem in an interactive environment.A solid nonconducting sphere has a positive charge q spread uniformly 44. Two particles are in a uniform electric field whose value is = 1.4 x 10-5 kg +2500 NIC. The mass and charge of particle I are and q, = —7.0 gC, while the correspondingvalues for panicle 2 are = 2.6 ><10-5 kg and = +18 '.LC.Initially the particles are at rest. The particles are both located on the same electric field line but are separated from each other by a distance d. When released, they acceler- a point within the sphere at a radius r has a magnitude of throughout its volume. The charge density or charge per unit volume 3 ate but always remain at this same distance from each other. find d. 45. ssm Two point charges of the same magnitude but opposite signs are fixed to either end of the base of an isosceles triangle, as the drawing shows. The electric field 4 reoR 3 (Hint: For a Gaussian surface, use a sphere of radius r centered within the solid sphere. Note that the net charge within any volume is the charge density times the volume.) * 52. Refer to Concept Simulation 18.3 at www.wiley.com/college/ cutnell for a perspective that is useful in solving this problem.Two spherical shells have a common center. A —1.6 X 10-6 C chargeis spread uniformly over the inner shell, which has a radius of 0.050m. at the midpoint M between the charges has a magnitude EM.The field directly above the midpoint +q at point P has a magnitude Ep. The ratio of these two field magnitudes is EM/Ep = 9.0. Find the angle a in the drawing. Section 18.9 Gauss' Law 46. A spherical surface completely surrounds a collection of charges. Find the electric flux through the surface if the collection consists of (a) a single +3.5 X 10-6 C charge, (b) a single —2.3 X 10-6 C charge, and (c) both of the charges in (a) and (b). drawing 47. ssm The shows an edge-on view of two planar surfaces that intersect and are mutually perpendicular. Surface I has an area of 1.7 m2, while surface 2 has an area of 3.2 m2. The Use Gauss' law to show that the electric fieldat therefore, is Surface 1 350 Surface 2 A +5.1 x 10-6 C charge is spread uniformly over the outershell, of which has a radius of 0.15 m. Find the magnitude and direction the electric field at a distance (measured from the commoncenter) of (a) 0.20 m, (b) 0.10 m, and (c) 0.025 m. * 53. ssm A cube is located with one corner at the origin of anx,y z coordinate system. One of the cube's faces lies in the x, y plane, another in the y, z plane, and another in the x, z plane. In other words, the cube is in the first octant of the coordinate system.The edges of the cube are 0.20 m long. A uniform electric field is parallel to the x, y plane and points in the direction of the +y axis. magnitude of the field is 1500 NIC. (a) Find the electricflux through each of the six faces of the cube. (b) Add the six values obtained in part (a) to show that the electric flux through thecubinet cal surface is zero, as Gauss' law predicts, since there is no charge within the cube. Q 54. A long, thin, straight wire of length L has a positive charge elee distributed uniformly along it. Use Gauss' law to show thatthe tric field created by this wire at a radial distance r has a magnitude sudace• of E = A/(27T€or), where A = Q/L. (Hint: For a Gaussian use a cylinder aligned with its axis along the wire and notethat cylinder has a flat surface at either end, as well as a curved face.) ADDITIONAL PROBLEMS 55. An electric field of 260 000 NIC points due west at a certain spot. What are the magnitude and direction of the force that acts on a charge of —7.0 'LC at this spot? eachOther 56. The force of repulsion that two like charges exert on is 3.5 N. What will the force be if the distance between the is increased to five times its original value? 57. Conceptual Example 14 deals with the hollow spherical conduc- 18.31. The conductor is initially electricallyneutral, torin Figure charge +q is placed at the center of the hollowspace. a then and has a net charge of +2q instead of supposethe conductor initially What is the total charge on the interior and on the exbeingneutral. when the +q charge is placed at the center? teriorsurface from a point charge, the magnitude of the elec58. At a distance r, charge is 248 NIC. At a distance r2 from the tric field created by the a magnitude of 132 NIC. Find the ratio r2/rl. charge,the field has 59. ssm Background pertinent to this problem is available in Interactive LearningWare 18.3 at www.wiley.com/college/cutnell. A 3.0-/JCpoint charge is placed in an external uniform electric field of 1.6 X 104NIC. At what distance from the charge is the net electric field zero? 60.In a vacuum, two particles have charges of q, and 012,where = +3.5 gC. They are separated by a distance of 0.26 m, and particle I experiences an attractive force of 3.4 N. What is (magnitude and sign)? 61. ssm Interactive LearningWare 18.1 at www.wiley.com/college/ cutnelloffers some perspective on this problem. Two tiny spheres have thesame mass and carry charges of the same magnitude. The mass of eachsphereis 2.0 X 10-6 kg. The gravitational force that each sphere exertson the other is balanced by the electric force. (a) What algebraic signscan the charges have? (b) Determine the charge magnitude. 62. A charge of q = +7.50 1-CCis located in an electric field. The x andy components of the electric field are Ex = 6.00 X 103 NIC and Ey= 8.00 X 103 NIC, respectively. (a) What is the magnitude of the forceon the charge? (b) Determine the angle that the force makes withthe +x axis. 63.Two charges are located along the x axis: qi = +6.0 /LC at = +4.0 cm, and q2 = +6.0 gC at x2 = —4.0 em. Two other at = +5.0 cm, chargesare located on the y axis: = +3.0 and qg= —8.0 at = +7.0 cm. Find the net electric field (magnitudeand direction) at the origin. & CALCULATIONS | 571 64. Four point charges have equal magnitudes. Three are positive, and one is negative, as the drawing shows. They are fixed in place on the same straight line, and adjacent charges are equally separated by a distance d. Consider the net electrostatic force acting on each charge. Calculate the ratio of the largest to the smallest net force. d 65. A proton is moving parallel to a uniform electric field. The electric field accelerates the proton and thereby increases its linear mo- mentumto 5.0 X 10-23kg•m/s from 1.5 X 10-23kg•m/s in a time of 6.3 X 10-6 s. What is the magnitude of the electric field? 66. Multiple-Concept Example 9 illustrates the concepts employed in this problem. A small object, which has a charge q = 7.5 'LC and mass m = 9.0 X 10-5 kg, is placed in a constant electric field. Starting from rest, the object accelerates to a speed of 2.0 X 103 m/s in a time of 0.96 s. Determine the magnitude of the electric field. 67. ssm In the rectangle the drawing, a charge +3.0PC 4d is to be placedat the empty comer to make the net force on the charge at +3.0 pc +3.0 pc corner A point along the vertical direction. What charge (magnitude and algebraic sign) must be placed at the empty corner? 68. Two objects are identical and small enough that their sizes can be ignored relative to the distance between them, which is 0.200 m. In a vacuum, each object carries a different charge, and they attract each other with a force of 1.20 N. The objects are brought into contact, so the net charge is shared equally, and then they are returned to their initial positions. Now it is found that the objects repel one another with a force whose magnitude is equal to that of the initial attractive force. What is the initial charge on each object? Note that there are two answers. CONCEPTS & CALCULATIONS Note: Each of these problems consists of Concept Questions followedby a related quantitative Problem The Concept Questions involvelittle or no mathematics. Theyfocus on the concepts with which theproblemsdeal. Recognizing the concepts is the essential initial step in anyproblem-solving technique. 69. metal spheres have GO Concept Questions Two identical charges of q, and (12.They are brought together so they touch,and then they are separated. (a) How is the net charge on the two spheres before they touch related to the net charge after they touch?(b) After they touch and are separated, is the charge on each spherethe same? Why? Problem Four identical metal spheres have charges of -2.0 pc, qc = +5.0 gc, and = +12.0 Go magnitude should be correct relative to that of the other force. (b) Rank the magnitudes of the net electrical force on charge 2, largest first. Explain. Problem The magnitude of the charges is Iql = 8.6 LLC.and the distance between them is 3.8 mm. Determine the magnitude of the net your anforce on charge 2 for each of the three drawings. Verify that swers are consistent with your answers to the Concept Questions. 2 = -8.0 ,zc, (a) Two ofthe separated. spheresare brought together so they touch and then they are is spheres are they, if the final charge on each of the two +5.0 1..tC?(b) In are brought a similar manner, which three spheres togetherand then of the three is separated, if the final charge on each one of +3.0 1...tC? (c) How many electrons would have to be added to thespheres neutral? in pan (b) to make it electrically 70. the same. (a) Draw the electrical force that each charge exerts on charge 2. Each force should be drawn in the correct direction. and its charges Concept Questions The drawings show three signs. In all different but that have the same magnitude, and 3 is casesthe distance between 2 d between charges I and 2 and 3 Concept Questions Suppose you want to neutralize the gravitational attraction between the earth and the moon on each. (a) Should the charges by placing equal amounts of charge positive and the other negaone or negative. both positive, be both 71. GO of the charge? Why or why not? and the moon to find the magnitude