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Transcript

C|H|A|P|T|E|R 18 Electric Forces and Electric Fields Lightning, which is the flow of electric charge in the atmosphere, is nature’s most spectacular display of electricity. Here, multiple lightning strikes occur at night over an urban skyline. (Ralph Wetmore/Stone/Getty Images) Copyright © 2007 John Wiley & Sons, Inc. All rights reserved. 18.1 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 proton has a mass of , and a neutron has a slightly greater mass of . Surrounding the nucleus is a diffuse cloud of orbiting particles called electrons, as Figure 18-1 . Like mass, electric charge is an intrinsic suggests. An electron has a mass of property of protons and electrons, and only two types of charge have been discovered, positive and negative. A proton has a positive charge, and an electron has a negative charge. A neutron has no net electric charge. 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. Experiment reveals that the magnitude of the charge on the proton exactly equals the magnitude of the charge on the electron; the proton carries a charge +e, and the electron carries a charge −e. The SI unit for measuring the magnitude of an electric charge is the coulomb* (C), and e has been determined experimentally to have the value The symbol e represents only the magnitude of the charge on a proton or an electron and does not include the algebraic sign that indicates whether the charge is positive or negative. In nature, atoms are normally found with equal numbers of protons and electrons. Usually, then, an atom carries no net charge because the algebraic sum of the positive charge of the nucleus and the negative charge of the electrons is zero. When an atom, or any object, carries no net charge, the object is said to be electrically neutral. The neutrons in the nucleus are electrically neutral particles. The charge on an electron or a proton is the smallest amount of free charge that has been discovered. Charges of larger magnitude are built up on an object by adding or removing , where N is electrons. Thus, any charge of magnitude q is an integer multiple of e; that is, an integer. Because any electric charge q occurs in integer multiples of elementary, indivisible charges of magnitude e, electric charge is said to be quantized. Example 1 emphasizes the quantized nature of electric charge. Example 1 | A Lot of Electrons How many electrons are there in one coulomb of negative charge? Reasoning The negative charge is due to the presence of excess electrons, since they carry negative charge. Because an electron has a charge whose magnitude is , the number of electrons is equal to the charge q divided by the charge e on each electron. Solution The number N of electrons is Copyright © 2007 John Wiley & Sons, Inc. All rights reserved. Charged Objects and the Electric Force It is easy to demonstrate that two electrically charged objects exert a force on one another. Consider Figure 183a, which shows two small balls that have been oppositely charged and are light and free to move. The balls attract each other. On the other hand, balls with the same type of charge, either both positive or both negative, During any process, the net electric charge of an isolated system remains constant (is conserved). Law of Conservation of Electric Charge Electric charges play a role in many situations other than rubbing two surfaces together. They are involved, for instance, in chemical reactions, electric circuits, and radioactive decay. A great number of experiments have verified that in any situation, the law of conservation of electric charge is obeyed. 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. 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. Electricity has many useful applications, and they are related to the fact that it is possible to 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 transferred to 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. 18.2 CONCEPTS AT A GLANCE Like other forces that we have encountered, the electric force (also called the electrostatic force) can alter the motion of an object. It can do so by contributing to the net external force that acts on the object. Newton’s second law, , specifies the acceleration that arises because of the net external force. The Concepts-at-a-Glance chart in Figure 18-4 is an expanded version of the 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. repel each other, as parts b and c of the drawing indicate. The behavior depicted in Figure 18-3 illustrates the following fundamental characteristic of electric charges: The physics of electronic ink. A new technology based on the electric force may revolutionize the way books and other printed matter are made. This technology, called electronic ink, allows letters and graphics on a page to be changed instantly, much like the symbols displayed on a computer monitor. Figure 18-5a illustrates the essential features of electronic ink. 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 dozen extremely tiny white beads that carry a slightly negative charge. The microcapsules are sandwiched between two sheets, an opaque base layer and a transparent top layer, at which the reader looks. When a positive charge is applied to a small region of the base layer, as shown in part b of the drawing, the negatively charged white beads are drawn to it, leaving dark ink at the top layer. Thus, a viewer sees only the dark liquid. When a negative charge is applied to a region of the base layer, the negatively charged white beads are repelled from it and are forced to the top of the microcapsules; now a viewer sees a white area due to the beads. Thus, electronic ink is based on the principle that like charges repel and unlike charges attract each other; a positive charge causes one color to appear, and a Figure 18-4 CONCEPTS AT A GLANCE External electrostatic forces that act on an object must be 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 DNA fragments to different terminal speeds in a gel. The different fragments are represented by the glowing bands. (Eurelios/Phototake) charts in Figures 4-9, 10-4, and 11-5, and emphasizes that any external electric force that acts on an object must be included when determining the net external force to be used in the second law. negative charge causes another color to appear. Each small region, whether dark or light, is known as a pixel (short for “picture element”). Computer chips provide 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. Copyright © 2007 John Wiley & Sons, Inc. All rights reserved. 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. 18.3 Conductors and Insulators Electric charge can not only exist on an object, but it can also move through an 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 readily and, therefore, are known as thermal conductors. On the other hand, substances that conduct heat poorly are referred to as thermal insulators. Figure 18-6 (a) Heat is conducted from the hotter end of the metal bar to the cooler end. (b) Electrons are conducted from the negatively charged end of the metal bar to the positively charged end. A situation analogous to the conduction of heat arises when a metal bar is placed between two charged objects, as in Figure 18-6b. Electrons are conducted through the bar from the negatively charged object toward the positively charged object. Substances that readily conduct electric charge are called electrical conductors. Although there are exceptions, good thermal conductors are generally good electrical conductors. Metals such as copper, aluminum, silver, and gold are excellent electrical conductors and, therefore, are used in electrical wiring. Materials that conduct electric charge poorly are known as electrical insulators. In many cases, thermal insulators are also electrical insulators. Common electrical insulators are rubber, many plastics, and wood. Insulators, such as the rubber or plastic that coats electrical wiring, prevent electric charge from going where it is not wanted. The difference between electrical conductors and insulators is related to atomic structure. As electrons orbit the nucleus, those in the outer orbits experience a weaker force of attraction to the nucleus than do those in the inner orbits. Consequently, the outermost electrons (also called the valence electrons) can be dislodged more easily than the inner ones. In a good conductor, some valence electrons become detached from a parent atom and wander more or less freely throughout the material, belonging to no one atom in particular. The exact number of electrons detached from each atom depends on the nature of the material, but is usually between one and three. When one end of a conducting bar is placed in contact with a negatively charged object and the other end in contact with a positively charged object, as in Figure 18-6b, the “free” electrons are able to move readily away from the negative end and toward the positive end. The ready movement of electrons is the hallmark of a good conductor. In an insulator the situation is different, for there are very few electrons free to move throughout the material. Virtually every electron remains bound to its parent atom. Without the “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. Copyright © 2007 John Wiley & Sons, Inc. All rights reserved. 18.4 Charging by Contact and by Induction When a 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. Once the electrons are on the metal sphere, where they can move readily, they repel one another and spread out over the sphere’s surface. The insulated stand prevents them from flowing 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 over its surface. In a similar manner, the sphere would be left with a positive charge after being rubbed with a positively charged rod. In this case, electrons from the sphere would be transferred to the rod. The process of giving one object a net electric charge by placing it in contact with another object that is already charged is known as charging by contact. 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 distribute themselves over the surface of the sphere. 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 sphere. In the sphere, the free electrons closest to the rod move to the other side, as part a of the drawing indicates. As a result, the part of the sphere nearest the rod becomes positively 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. If the rod were removed, the free electrons would return to their original places, and the charged regions would disappear. 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. Under most conditions the earth is a good electrical conductor. So when a metal wire is attached between the sphere and the ground, as in Figure 18-8b, some of the free electrons leave the sphere and distribute themselves over the much larger earth. If the grounding 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 electric charge without touching the object to a second charged object is called charging by induction. The process could also be used to give the sphere a negative net charge, if a positively 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, instead of metal, the method of producing a net charge by induction would not work, because very little charge would flow through the insulating material and down the grounding wire. However, the electric force of the charged rod would have some effect, as Figure 18-9 illustrates. The electric force would cause the positive and negative charges in the molecules of the insulating material to separate slightly, with the negative charges being “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. Figure 18-9 The negatively charged rod induces a slight positive surface charge on the plastic. 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, and the rod is then and of the spheres after they removed. The following statements refer to the masses are separated and the rod is removed. Which one or more of the statements is true? (a) , (b) if the rod is positive, (c) if the rod is positive, (d) if the rod is negative, (e) 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. In a 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. Copyright © 2007 John Wiley & Sons, Inc. All rights reserved. 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 depends on the amount of charge on the objects and the distance between them. Experiments reveal that the greater the charge and the closer together they are, the greater is the force. To set the stage for explaining these features in more detail, Figure 18-10 shows two charged bodies. These objects are so small, compared to the distance r between them, that they can be regarded as mathematical points. The “point charges” have and . If the charges have unlike signs, as in part a of the picture, each object is magnitudes* attracted to the other by a force that is directed along the line between them; is the electric force is the electric force exerted on object 2 by object 1. If, as in exerted on object 1 by object 2 and part b, the charges have the same sign (both positive or both negative), each object is repelled from the other. The repulsive forces, like the attractive forces, act along the line between the charges. Whether attractive or repulsive, the two forces are equal in magnitude but opposite in direction. These forces always exist as a pair, each one acting on a different object, in accord with Newton’s action–reaction law. Figure 18-10 Each point charge exerts a force on the other. Regardless of whether the forces are (a) attractive or (b) repulsive, they are directed along the line between the charges and have equal magnitudes. The French physicist Charles Augustin de Coulomb (1736–1806) carried out a number of experiments to determine how the electric force that one point charge applies to another depends on the amount of each charge and the separation between them. His result, now known as Coulomb’s law, is stated below. Coulomb’s Law The magnitude F of the electrostatic force exerted by one point charge q1 on another point charge q2 is directly proportional to the magnitudes and of the charges and inversely proportional to the square of the distance r between them: (18.1) in SI units. The electrostatic where k is a proportionality constant: force is directed along the line joining the charges, and it is attractive if the charges have unlike signs and repulsive if the charges have like signs. It is common practice to express k in terms of another constant called the permittivity of free space and has a value of by writing is . Equation 18.1 gives the magnitude of the electrostatic force that each point charge exerts on the other. When using this equation, then, it is important to remember to substitute only the charge magnitudes and , as Example 2 illustrates. (without algebraic signs) for Example 2 | A Large Attractive Force Two objects, whose charges are + 1.0 and −1.0 C, are separated by 1.0 km. Compared to 1.0 km, the sizes of the objects are small. Find the magnitude of the attractive force that either charge exerts on the other. Reasoning Considering 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 magnitudes of the charges are used for the symbols and that appear in the law. Solution The magnitude of the force is (18.1) Need more practice? Interactive LearningWare18.1 An object of mass 215 kg is located just above the surface of the earth ( ). Suppose that the object and the earth have an identical charge q. Therefore, two forces–the gravitational force and the electrostatic force— act on the object. Assuming that the earth’s charge is located at the center of the earth, determine q (magnitude and sign) when the object is in equilibrium. Related Homework: Problem 61 The force calculated in Example 2 corresponds to about 2000 pounds and is so large because charges of ± 1.0 C are enormous. Such large charges are encountered only in the most severe conditions, as in a lightning bolt, where as much as 25 C can be transferred between the cloud and the ground. The typical charges produced in the laboratory are much smaller and are measured conveniently in microcoulombs . Coulomb’s law has a form that is remarkably similar to Newton’s law of gravitation . 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 between them. In addition, the force is proportional to the product of an intrinsic property of each of the objects, the magnitudes of the and in Coulomb’s law and the masses m1 and m2 in the gravitation law. But there is a charges major difference between the two laws. The electrostatic force can be either repulsive or attractive, depending on whether or not the charges have the same sign; in contrast, the gravitational force is always an attractive force. Section 5.5 discusses how the gravitational attraction between the earth and a satellite provides the centripetal force that keeps a satellite in orbit. Example 3 illustrates that the electrostatic force of attraction plays a similar role in a famous model of the atom created by the Danish physicist Niels Bohr (1885 – 1962). ANALYZING PROBLEMS MULTIPLE-CONCEPT 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 , as Figure 18-11 shows. The mass of the electron is . Determine the speed of the electron. Figure 18-11 In the Bohr model of the hydrogen atom, the electron (− e) orbits the proton (+ e) at a distance of r = 5.29 × 10−11 m. The velocity of the electron is . 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 , where m and v are, respectively, the mass and speed of the electron and r is the radius of the orbit. This equation can be solved for the speed of the electron. Since the mass and orbital radius are known, we can calculate the electron’s speed provided that a value for the centripetal force can be found. For the electron in the hydrogen atom, the centripetal force is provided almost exclusively by the electrostatic force that 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 gravitational force. However, the gravitational force is negligible in comparison to the electrostatic force. Knowns and Unknowns The data for this problem are: Description Symbol Electron charge −e Electron mass m Proton charge +e Radius of orbit r Value Unknown Variable Orbital speed of electron v ? 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 (Equation 5.3). By solving this equation for the speed, we obtain Equation 1 at the right. The mass and radius in this expression are known. However, the magnitude of the centripetal force is not known, so we will evaluate it in Step 2. (1) Step 2 Coulomb’s Law As the electron orbits the proton in the hydrogen atom, it is attracted to the proton by the electrostatic force. The magnitude F of the electrostatic force is given by Coulomb’s law as (Equation 18.1), where and are the magnitudes of the charges, r is the orbital radius, and . Since the centripetal force is provided almost entirely by the electrostatic force, it follows that . Furthermore, . With these substitutions, Equation 18.1 becomes and All the variables on the right side of this expression are known, so we substitute it into Equation 1, as indicated in the right column. Solution Algebraically combining the results of the modeling steps, we have The speed of the orbiting electron is Related Homework: Problems 16 and 20 Since the electrostatic force depends on the inverse square of the distance between the charges, it becomes larger for smaller distances, such as those involved when a strip of adhesive tape is stuck to a smooth surface. Electrons shift over the small distances between the tape and the surface. As a result, the materials become oppositely charged. Since the distance between the charges is relatively small, the electrostatic force of attraction is large enough to contribute to the adhesive bond. Figure 18-12 shows an image of the sticky surface of a piece of tape after it has been pulled off a metal surface. The image was obtained using an atomic-force microscope and reveals the tiny pits left behind when microscopic portions of the adhesive remain stuck to the metal because of the strong adhesive bonding forces. 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. (Courtesy Louis Scudiero and J. Thomas Dickinson, Washington State University.) The physics of adhesion. The Force on a Point Charge Due to Two or More Other Point Charges Up to now, we have been discussing the electrostatic force on a point charge (magnitude ) due to another point charge (magnitude ). Suppose that a third point charge (magnitude ) is also present. What would be the net force on q1 due to both q2 and q3?. It is convenient to deal with such a problem in parts. First, find the magnitude and direction of the force exerted on q1 by q2 (ignoring q3). Then, determine the force exerted on q1 by q3 (ignoring q2). The net force on q1 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. 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 q1. Figure 18-13 (a) Three charges lying along the x axis. (b) The force exerted on q1 by q2 is , while the force exerted on q1 by q3 is . Reasoning Part b of the drawing shows a free-body diagram of the forces that act on q1. Since q1 and q2 have opposite signs, they attract one another. Thus, the force exerted on q1 by q2 is points to the left. Similarly, the force exerted on q1 by q3 is , and it and is also an attractive force. It points to the right in Figure 18-13b. The magnitudes of these forces can be obtained from Coulomb’s law. The net force is the vector sum of and . Solution The magnitudes of the forces are Since force points in the negative x direction, and points in the positive x direction, the net is The plus sign in the answer indicates that the net force points to the right in the drawing. Problem solving insight When using Coulomb’s law , remember and that the symbols stand for the charge magnitudes. Do not substitute negative numbers for these symbols. 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 q1. Figure 18-14 (a) Three charges lying in a plane. (b) The net force acting on q1 is = + . The angle that Reasoning The force exerted on q1 by q2 is makes with the + x axis is . and is an attractive force because the two charges have opposite signs. It points along the line between the charges. The force exerted on q1 by q3 is and is also an attractive force. It points along the line between q1 and q3. Coulomb’s law specifies the magnitudes of these forces. Since the forces point in different directions (see Figure 1814b), we will use vector components to find the net force. Need more practice? Interactive LearningWare18.2 A positive charge is placed on the x , axis and another positive charge is placed on the y axis .A third charge 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 Solution The magnitudes of the forces are The net force components of is the vector sum of and that lie in the x and y directions are , as part b of the drawing shows. The and , respectively. Our approach to is the same as that used in Chapters 1 and 4. The forces finding x and y components. Then, the x components are combined to give . Once combined to give determined using trigonometry. Force and and are resolved into , and the y components are are known, the magnitude and direction of x component y component + 18 N 0N can be The magnitude F and the angle of the net force are Problem solving insight 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. Check Your Understanding 2 The drawing shows three point charges arranged in different ways. The charges are + q, − q, and − q each has the same magnitude, one being positive and the other two negative. In each part of the drawing the distance d is the same. Rank the arrangements in descending order (largest first) according to the magnitude of the net electrostatic force that acts on the positive charge. (The answer is given at the end of the book.) Background: The electrostatic force that one point charge exerts on another is described by Coulomb’s law. Like any force, this force is a vector quantity, and the rules of vector addition apply. For similar questions (including calculational counterparts), consult Self-Assessment Test 18.1, which is described next. Test your understanding of the material in Sections 18.1, 18.2, 18.3, 18.4 and 18.5 : · The Origin of Electricity · Charged Objects and the Electric Force · Conductors and Insulators · Charging by Contact and by Induction · Coulomb’s Law Copyright © 2007 John Wiley & Sons, Inc. All rights reserved. 18.6 The Electric Field Definition As we know, a charge can experience an electrostatic force due to the presence of other charges. For , which is the vector sum of instance, the positive charge q0 in Figure 18-15 experiences a force the forces exerted by the charges on the rod and the two spheres. It is useful to think of q0 as a test charge for determining the extent to which the surrounding charges generate a force. However, in using a test charge, we must be careful to select one with a very small magnitude, so that it does not alter the locations of the other charges. The next example illustrates how the concept of a test charge is applied. Figure 18-15 A positive charge q0 experiences an electrostatic force the surrounding charges on the ebonite rod and the two spheres. Example 6 | A Test Charge due to The positive test charge shown in Figure 18-15 is and experiences a force in the direction shown in the drawing. (a) Find the force per coulomb that the test charge experiences. (b) Using the result of part (a), predict the force that a charge of would experience if it replaced q0. Reasoning The charges in the environment apply a force coulomb experienced by the test charge is to the test charge q0. The force per . If q0 is replaced by a new charge q, then the force on this new charge is the force per coulomb times q. Solution a. The force per coulomb of charge is b. The result from part (a) indicates that the surrounding charges can exert 2.0 newtons of force would experience a force whose per coulomb of charge. Thus, a charge of magnitude is The direction of this force would be the same as that experienced by the test charge, since both have the same positive sign. The electric force per coulomb, , calculated in Example 6(a) is one illustration of an idea that is very important in the study of electricity. The idea is called the electric field. Equation 18.2 presents the definition of the electric field. Definition of the Electric Field The electric field that exists at a point is the electrostatic force test charge* q0 placed at that point divided by the charge itself: experienced by a small (18.2) The electric field is a vector, and its direction is the same as the direction of the force positive test charge. on a SI Unit of Electric Field: newton per coulomb (N/C) Equation 18.2 indicates that the unit for the electric field is that of force divided by charge, which is a newton/coulomb (N/C) in SI units. It is the surrounding charges that create an electric field at a given point. Any positive or negative charge placed at the point interacts with the field and, as a result, experiences a force, as the next example indicates. Example 7 | An Electric Field Leads to a Force In Figure 18-16a the charges on the two metal spheres and the ebonite rod create an electric field at the spot indicated. This field has a magnitude of 2.0 N/C and is directed as in the drawing. Determine the force on a charge placed at that spot, if the charge has a value of (a) and (b) . Figure 18-16 The electric field 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 while (b) the force on a negative charge points opposite to . Reasoning The electric field at a given spot can exert a variety of forces, depending on the magnitude and sign of the charge placed there. The charge is assumed to be small enough that 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 magnitudes of q0 and : (18.2) Since q0 is positive, the force points in the same direction as the electric field, as part a of the drawing indicates. b. In this case, the magnitude of the force is (18.2) The force on the negative charge points in the direction opposite to the force on the positive charge—that is, opposite to the electric field (see part b of the drawing). At a particular point in space, each of the surrounding charges contributes to the net electric field that exists there. To determine the net field, it is necessary to obtain the various contributions separately and then find the vector sum of them all. Such an approach is an illustration of the principle of linear superposition, as applied to electric fields. (This principle is introduced in Section 17.1, in connection with waves.) Example 8 emphasizes the vector nature of the electric field, and Example 9 illustrates that a charged particle accelerates in an electric field. Example 8 | Electric Fields Add as Vectors Do Figure 18-17 shows two charged objects, A and B. Each contributes as follows to the net electric field at point P: directed to the right, and directed downward. Thus, and are perpendicular. What is the net electric field at P? Figure 18-17 The electric field contributions and , which come from at the two charge distributions, are added vectorially to obtain the net field point P. Reasoning The net electric field is the vector sum of and : . As illustrated in Figure 18-17, and are perpendicular, so is the diagonal of the rectangle and shown in the drawing. Thus, we can use the Pythagorean theorem to find the magnitude of trigonometry to find the directional angle . Solution The magnitude of the net electric field is The direction of is given by the angle in the drawing: ANALYZING PROBLEMS MULTIPLE-CONCEPT Example 9 | A Proton Accelerating in an Electric Field In a vacuum, a proton is moving parallel to a uniform electric field that is directed along the + x axis (see Figure 18-18). The proton starts with a velocity of and accelerates in the same direction as the electric field, which has . Find the velocity of the proton when its displacement is + 2.0 mm a value of from the starting point. Figure 18-18 A proton, moving to the right, accelerates in the presence of the electric field . Reasoning Since we know the initial velocity and displacement of the proton, we can determine its final velocity from an equation of kinematics, provided the proton’s acceleration can be found. The acceleration is given by Newton’s second law as the net 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 Symbol Value Proton charge +e Proton mass m Initial velocity of proton v0 Electric field E Displacement of proton x + 2.0 mm v ? x x Unknown Variable Final velocity of proton x Comment Modeling the Problem Step 1 Kinematics To obtain the final velocity v of the proton we employ Equation 3.6a from the equations of x kinematics: . We have chosen this equation because two of the variables, the initial velocity v0 and the displacement x, are known. Taking the square root of each side of x 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 a is x not known, we will obtain an expression for it in Step 2. (1) Step 2 Newton’s Second Law Newton’s second law, as given in Equation 4.2a, states that the acceleration a of the proton x acting on it divided by the proton’s mass is equal to the net force the electrostatic force F acts on the proton, so it is the net force. Setting x . Only in Newton’s second law gives 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 x F given by x (Equation 18.2), where q0 is the charge. Setting q = e for the proton, 0 we have 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 Algebraically combining the results of the three steps, we have The final velocity of the proton is where the + sign denotes that the final velocity points along the + x axis. Related Homework: Problems 38, 40 and 66 Point Charges A more complete understanding of the electric field concept can be gained by considering the field created by a point charge, as in the following example. Example 10 | The Electric Field of a Point Charge There is an isolated point charge of in a vacuum at the left in Figure 18-19a. Using a , determine the electric field at point P, which is 0.20 m away. test charge of Figure 18-19 (a) At location P, a positive test charge q0 experiences a repulsive force due to the positive point charge q. (b) At P the electric field 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. Reasoning Following the definition of the electric field, we place the test charge q0 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: Equation 18.2 gives the magnitude of the electric field: The electric field points in the same direction as the force on the positive test charge. Since the test charge experiences a force of repulsion directed to the right, the electric field vector also points to the right, as Figure 18-19b shows. The electric field produced by a point charge q can be obtained in general terms from Coulomb’s law. First, note that the magnitude of the force exerted by the charge q on a test charge q0 is . Then, divide this value by to obtain the magnitude of the field. Since eliminated algebraically from the result, the electric field does not depend on the test charge: is (18.3) As in Coulomb’s law, the symbol |q| denotes the magnitude of q in Equation 18.3, without regard to whether q is positive or negative. If q is positive, then is directed away from q, as in Figure 1819b. On the other hand, if q is negative, then is directed toward q, since a negative charge attracts a positive test charge. For instance, Figure 18-19c shows the electric field that would exist at P if there were a charge of − q instead of + q at the left of the drawing. Example 11 reemphasizes the fact that all the surrounding charges make a contribution to the electric field that exists at a given place. Example 11 | The Electric Fields from Separate Charges May Cancel Two positive point charges, and , are separated in a vacuum by a distance of 3.0 m, as Figure 18-20 illustrates. Find the spot on the line between the charges where the net electric field is zero. Figure 18-20 The two point charges q1 and q2 create electric fields and that cancel at a location P on the line between the charges. 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 equals that of . However, since q2 is smaller than q this location must be closer to q2, in order that the field of the smaller charge 1 can balance the field of the larger charge. In the drawing, the cancellation spot is labeled P, and its distance from q1 is d. Solution At P, , and using the expression Rearranging this expression shows that of this equation reveals that , we have . Taking the square root of each side The plus and minus signs on the right occur because either the positive or negative root can be taken. Therefore, there are two possible values for d: + 2.0 m and + 6.0 m. The value + 6.0 m and corresponds to a location off to the right of both charges, where the magnitudes of are equal, but where the directions are the same. Thus, and do not cancel at this spot. The other value for d corresponds to the location shown in the drawing and is the zero-field location: . Problem solving insight Equation 18.3 gives only the magnitude of the electric field produced by a point charge. Therefore, do not use negative numbers for the symbol |q| in this equation. 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. 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? 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. Reasoning and Solution In Figure 18-21a, the charges at corners 1 and 3 are both + q. The positive charge at corner 1 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 1. Thus, the two fields have opposite directions. The magnitudes of the fields are identical because 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 18-21a. 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 of these fields are equal because each charge has the same magnitude and is located at the same distance from the shown center of the rectangle. As a result, the two fields combine to give the net electric field in the drawing. In Figure 18-21b, the charges on corners 2 and 4 are identical to those in part a of the drawing, so . The charges on corners 1 and 3 are identical to this pair gives rise to the electric field labeled those on corners 2 and 4, so they give rise to an electric field labeled , which has the same . The net electric field is the vector sum of and and is also magnitude as shown in the drawing. Clearly, this sum is greater than alone. Therefore, the net field in part b is larger than that in part a. Related Homework: Conceptual Question 12, Problem 34 Need more practice? Interactive LearningWare18.3 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 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 charge were placed at this spot, what would be the electrostatic force (magnitude and direction) exerted on it? Related Homework: Problem 59 The Parallel Plate Capacitor Equation 18.3, which gives the electric field of a point charge, is a very useful result. With the aid of integral calculus, this equation can be applied in a variety of situations where point charges are distributed over one or more surfaces. One such example that has considerable practical importance is the parallel plate capacitor. As Figure 18-22 shows, this device consists of two parallel metal plates, each with area A. A charge + q is spread uniformly over one plate, while a charge − q is spread uniformly over the other plate. In the region between the plates and away from the edges, the electric field points from the positive plate toward the negative plate and is perpendicular to both. It can be shown (see Example 16 in Section 18.9) that the electric field has a magnitude of (18.4) where is the permittivity of free space. In this expression the Greek symbol sigma ( ) denotes the charge per unit area ( ) and is sometimes called the charge density. Except in the region near the edges, the field has the same value at all places between the plates. The field does not depend on the distance from the charges, in distinct contrast to the field created by an isolated point charge. Figure 18-22 A parallel plate capacitor. Check Your Understanding 3 A positive point 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 electric field at corner A is zero. (a) At which corner is the second charge located? (b) Is the second charge positive or negative? (c) Does the second charge have a greater, a smaller, or the same magnitude 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 magnitude and direction depend on the magnitudes and algebraic signs of the charges that create it. For similar questions (including calculational counterparts), consult Self-Assessment Test 18.2, which is described at the end of Section 18.9. Copyright © 2007 John Wiley & Sons, Inc. All rights reserved. 18.7 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 strength of 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. Since the electric field is the electric force per unit charge, the electric field lines are sometimes called lines of force. To introduce 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 repulsive force, as the arrows in the drawing indicate. Therefore, the electric field created by the charge + q is directed radially outward. The electric field lines are lines drawn to show this direction, as part b of the drawing illustrates. They begin on the charge + q and point radially outward. Figure 18-24 shows the field lines in the vicinity of 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 toward negative charges. 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. Figure 18-24 The electric field lines are directed radially inward toward a negative point charge − q. 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 number is 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 charge as from a + q charge. The pattern of electric field lines also provides information about the magnitude or 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 the charges, 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 fact, no matter how many charges are present, the number of lines per unit area passing perpendicularly 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 same number of lines per unit area everywhere, and the electric field has the same strength at all points. For example, Figure 18-25 shows that the field lines between the plates of a parallel plate capacitor are parallel and equally spaced, except near the edges where they bulge outward. The equally spaced, parallel lines indicate that the electric field has the same magnitude and direction at all points in the central region of the capacitor. 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 magnitude and direction at all points. Often, electric field lines are curved, as in the case of an electric dipole. An electric dipole consists of two separated point charges that have the same magnitude but opposite signs. The electric field of a dipole is proportional to the product of the magnitude of one of the charges and the distance between the charges. This product is called the dipole moment. Many molecules, such as H2O and HCl, have dipole moments. Figure 18-26 depicts the field lines in the vicinity of a dipole. For a curved field line, the electric field vector at a point is tangent to the line at that point (see points 1, 2, and 3 in the drawing). The pattern of the lines for the dipole indicates that the electric field is greatest in the region between and immediately surrounding the two charges, since the lines are closest together there. 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 1, 2, or 3, the field created by the dipole is tangent to the line through the point. Notice in Figure 18-26 that any given field line starts on the positive charge and ends on the negative charge. In general, electric field lines always begin on a positive charge and end on a negative charge and do not start or stop in midspace. Furthermore, the number of lines leaving a positive charge or entering a negative charge is proportional to the magnitude of the charge. This means, for example, that if 100 lines are drawn leaving a charge, then 75 lines would have to end on a charge and 25 lines on a charge. Thus, 100 lines leave the charge of and end , So the lines begin and end on equal amounts of total charge. on a total charge of The electric field lines are also curved in the vicinity of two identical charges. Figure 18-27 shows 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 that the electric field is relatively weak between the charges. 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. Some of the important properties of electric field lines are considered in Conceptual Example 13. Conceptual Example 13 | Drawing Electric Field Lines Figure 18-28a 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 things wrong with this drawing. What are they? Figure 18-28 (a) Incorrectly and (b) correctly drawn electric field lines. Reasoning and Solution One aspect of Figure 18-28a that is incorrect is that electric field lines cross at point P. Field lines can never cross, and here’s why. An electric charge placed at P experiences a 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 that point. 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 any point, 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 negative charges. Remember that the number of field lines leaving a positive charge or entering a negative charge is proportional to the magnitude of the charge. The − 2q charge has half the magnitude of 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 − q charges, 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 evenly spaced. This would indicate that the electric field everywhere in this region has a constant magnitude and direction, as is the case in the central region of a parallel plate capacitor. But the electric field between the + 4q and − q charges is not constant everywhere. It certainly is stronger in places close to the + 4q or − q charge than it is midway between them. The field lines, therefore, should be drawn with a curved nature, similar (but not identical) to those that surround a dipole. Figure 18-28b shows more nearly correct representations of the field lines for the four charges. Related Homework: Problems 25 and 26 Copyright © 2007 John Wiley & Sons, Inc. All rights reserved. 18.8 The Electric Field Inside a Conductor: Shielding In conducting materials such as copper, electric charges move readily in response to the forces that electric fields exert. This property of conducting materials has a major effect on the electric field that can exist within and around them. Suppose that a piece of copper carries a number of excess electrons somewhere within it, as in Figure 18-29a. Each electron would experience a force of repulsion because of the electric field of its neighbors. And, since copper is a conductor, the excess electrons move readily in response to that force. In fact, as a consequence of the 1/r2 dependence on distance in Coulomb’s law, they rush to the surface of the copper. Once static equilibrium is established with all of the excess charge on the surface, no further movement of charge occurs, as part b of the drawing indicates. Similarly, excess positive charge also moves to the surface of a conductor. In general, at equilibrium under electrostatic conditions, any excess charge resides on the surface of a conductor. Figure 18-29 (a) Excess charge within a conductor (copper) moves quickly (b) to the surface. Now consider the interior of the copper in Figure 18-29b. The interior is electrically neutral, although there are still free electrons that can move under the influence of an electric field. The absence of a net movement of these free electrons indicates that there is no net electric field present within the conductor. In fact, the excess charges arrange themselves on the conductor surface precisely in the manner needed to make the electric field zero within the material. Thus, at equilibrium under electrostatic conditions, the electric field is zero at any point within a conducting material. This fact has some fascinating implications. Figure 18-30a shows an uncharged, solid, cylindrical conductor at equilibrium in the central region of a parallel plate capacitor. Induced charges on the surface of the cylinder alter the electric field lines of the capacitor. Since an electric field cannot exist within the conductor under these conditions, the electric field lines do not penetrate the cylinder. Instead, they end or begin on the induced charges. Consequently, a test charge placed inside the conductor would feel no force due to the presence of the charges on the capacitor. In other words, the conductor shields any charge within it from electric fields created outside the conductor. The shielding results from the induced charges on the conductor surface. Figure 18-30 (a) A cylindrical conductor (shown as an end view) is placed between the oppositely charged plates of a capacitor. The electric field lines do not penetrate the conductor. The blowup shows that, just outside the conductor, the electric field lines are perpendicular to its surface. (b) The electric field is zero in a cavity within the conductor. Since the electric field is zero inside the conductor, nothing is disturbed if a cavity is cut from the interior of the material, as in part b of the drawing. Thus, the interior of the cavity is also shielded from external electric fields, a fact that has important applications, particularly for shielding electronic circuits. “Stray” electric fields are produced by various electrical appliances (e.g., hair dryers, blenders, and vacuum cleaners), and these fields can interfere with the operation of sensitive electronic circuits, such as those in stereo amplifiers, televisions, and computers. To eliminate such interference, circuits are often enclosed within metal boxes that provide shielding from external fields. The physics of shielding electronic circuits. The blowup in Figure 18-30a shows another aspect of how conductors alter the electric field lines created by external charges. The lines are altered because the electric field just outside the surface of a conductor is perpendicular to the surface at equilibrium under electrostatic conditions. If the field were not perpendicular, there would be a component of the field parallel to the surface. Since the free electrons on the surface of the conductor can move, they would do so under the force exerted by that parallel component. In reality, however, no electron flow occurs at equilibrium. Therefore, there can be no parallel component, and the electric field is perpendicular to the surface. The preceding discussion deals with features of the electric field within and around a conductor at equilibrium under electrostatic conditions. These features are related to the fact that conductors contain mobile free electrons and do not apply to insulators, which contain very few free electrons. Conceptual Example 14 further explores the behavior of a conducting material in the presence of an electric field. Conceptual Example 14 | A Conductor in an Electric Field A charge + q is suspended at the center of a hollow, electrically neutral, spherical conductor, as Figure 18-31 illustrates. Show that this charge induces (a) a charge of − q on the interior surface and (b) a charge of + q on the exterior surface of the conductor. Figure 18-31 A positive charge + q is suspended at the center of a hollow spherical conductor that is electrically neutral. Induced charges appear on the inner and outer surfaces of the conductor. The electric field within the conductor itself is zero. Reasoning and Solution a. Electric field lines emanate from the positive charge + q. Since the electric field inside the metal conductor must be zero at equilibrium under electrostatic conditions, each field line ends when it reaches the conductor, as the picture shows. Since field lines terminate only on negative charges, there must be an induced negative charge on the interior surface of the conductor. Furthermore, the lines begin and end on equal amounts of charge, so the magnitude of the total induced charge is the same as the magnitude of the charge at the center. Thus, the total induced charge on the interior surface is − q. b. Before the charge + q is introduced, the conductor is electrically neutral. Therefore, it carries no net charge. We have also seen that there can be no excess charge within the metal. Thus, since an induced charge of − q appears on the interior surface, a charge of + q must be induced on the outer surface. The positive charge on the outer surface generates field lines that radiate outward (see the drawing) as if they originated from the central charge and the conductor were absent. The conductor does not shield the outside from the field produced by the charge on the inside. Related Homework: Problem 57 Copyright © 2007 John Wiley & Sons, Inc. All rights reserved. Gauss’ Law We begin by developing a version of Gauss’ law that applies only to a point charge, which we assume to be positive. The electric field lines for a positive point charge radiate outward in all directions from the charge, as , Figure 18-23b indicates. The magnitude E of the electric field at a distance r from the charge is according to Equation 18.3, in which we have replaced the symbol |q| with the symbol q since we are assuming that Figure 18-32 CONCEPTS AT A GLANCE The electric field and the surface through which it passes are brought together to define the concept of electric flux. It is in terms of electric flux that Gauss’ law is formulated. When the electric field is produced by a distribution of charges, such as that on the silvery metal sphere in the photograph, Gauss’ law relates the electric flux to the charge. (© SSPL/ The Image Works) CONCEPTS AT A GLANCE In presenting Gauss’ law, it will be necessary to introduce a new idea called electric flux, and the Concepts-at-a-Glance chart in Figure 18-32 outlines the approach we will follow. The idea of flux involves both the electric field and the surface through which it passes. By bringing together the electric field and the surface through which it passes, we will be able to define electric flux and then present Gauss’ law. Section 18.6 discusses how a point charge creates an electric field in the space around the charge. There are also many situations in which an electric field is produced by charges that are spread out over a region, rather than by a single point charge. Such an extended collection of charges is called a “charge distribution.” For example, the electric field within the parallel plate capacitor in Figure 18-22 is produced by positive charges spread uniformly over one plate and an equal number of negative charges spread over the other plate. As we will see, Gauss’ law describes the relationship between a charge distribution and the electric field it produces. This law was formulated by the German mathematician and physicist Carl Friedrich Gauss (1777–1855). 18.9 , where . is Figure 18-34 shows a charge distribution whose net charge is labeled Q. The charge distribution is surrounded by a Gaussian surface—that is, an imaginary closed surface. The surface can have any arbitrary shape (it need not be spherical), but it must be closed (an open surface would be like that of half an eggshell). The direction of the electric field is not necessarily perpendicular to the Gaussian surface. Furthermore, the magnitude of the electric field need not be constant on the surface, but can vary from point to point. Equation 18.5 is the result we have been seeking, for it is the form of Gauss’ law that applies to a point charge. depends only on the charge q within This result indicates that, aside from the constant , the electric flux the Gaussian surface and is independent of the radius r of the surface. We will now see how to generalize Equation 18.5 to account for distributions of charges and Gaussian surfaces with arbitrary shapes. Figure 18-33 A positive point charge is located at the center of an imaginary spherical surface of radius r. Such a surface is one example of a Gaussian surface. Here the electric field is perpendicular to the surface and has the same magnitude everywhere on it. The left side of Equation 18.5 is the product of the magnitude E of the electric field at any point on the Gaussian surface and the area A of the surface. In Gauss’ law this product is especially important and is called the electric flux, : . (It will be necessary to modify this definition of flux when we consider the general case of a Gaussian surface with an arbitrary shape.) We now place this point charge at the center of an imaginary spherical surface of radius r, as Figure 18-33 shows. Such a hypothetical closed surface is called a Gaussian surface, although in general it need not be spherical. The surface area A of a sphere is , and the magnitude of the electric field can be written in terms of this area as , or the permittivity of free space. With this substitution, the magnitude of the electric field becomes the charge is positive. As mentioned in Section 18.5, the constant k can be expressed as (18.5) Gauss’ Law where, as usual, the symbol means “the sum of.” Gauss’ law relates the electric flux Q enclosed by the arbitrarily shaped Gaussian surface. to the net charge To determine the electric flux through such a surface, we divide the surface into many tiny sections with areas , and so on. Each section is so small that it is essentially flat and the electric field is a constant (both in magnitude and direction) over it. For reference, a dashed line called the “normal” is drawn perpendicular to each section on the outside of the surface. To determine the electric flux for each of the sections, we use only that is perpendicular to the surface—that is, the component of the electric field that the component of passes through the surface. From the drawing it can be seen that this component has a magnitude of E , where ) is the angle between the electric field and the normal. The electric flux through any one section is then (E . The electric flux that passes through the entire Gaussian surface is the sum of all of these individual , or fluxes: Figure 18-34 The charge distribution Q is surrounded by an arbitrarily shaped Gaussian surface. The electric flux through any tiny segment of the surface is the product of and the area of the segment: . The angle is the angle between the electric field and the normal to the surface. (18.6) , Reasoning Because the charge is distributed uniformly over the spherical shell, the electric field is symmetrical. This means that the electric field is directed radially outward in all directions, and its magnitude is the same at all points that are equidistant from the shell. All such points lie on a sphere, so the symmetry is called spherical symmetry. With this symmetry in mind, we will use a spherical Gaussian surface to evaluate the electric flux . We will then use Gauss’ law to determine the magnitude of the electric field. (b) inside the shell, respectively. Figure 18-35 A uniform distribution of positive charge resides on a thin spherical shell of radius R. The spherical Gaussian surfaces (a) S and (b) S1 are used in Example 15 to evaluate the electric flux (a) outside and Figures 18-35a and b show a thin spherical shell of radius R. A positive charge q is spread uniformly over the shell. Find the magnitude of the electric field at any point (a) outside the shell and (b) inside the shell. Example 15 | The Electric Field of a Charged Thin Spherical Shell Although we arrived at Gauss’ law by assuming the net charge Q was positive, Equation 18.7 also applies when Q is also negative. Gauss’ law is often used to find the magnitude of is negative. In this case, the electric flux the electric field produced by a distribution of charges. The law is most useful when the distribution is uniform and symmetrical. In the next two examples we will see how to apply Gauss’ law in such situations. SI Unit of Electric Flux: N·m2/C The electric flux through a Gaussian surface is equal to the net charge Q enclosed by the surface divided by the permittivity of free space: (18.7) . In accord with Gauss’ law, Example 16 | The Electric Field Inside a Parallel Plate Capacitor Gauss’ law allows us to deduce that there is no electric field inside a uniform spherical shell of charge. An electric field exists only on the outside. , where Q is the net charge inside the Gaussian surface. But now Q = 0 C, the electric flux must be equal to or since all the charge lies on the shell that is outside the surface S1. Consequently, we have the electric flux through the Gaussian surface is the electric field (if it exists) must also have spherical symmetry. Therefore, using reasoning like that in part (a), b. To find the magnitude of the electric field inside the charged shell, we select a spherical Gaussian surface that lies inside the shell and is concentric with it. See the surface labeled S1 in Figure 18-35b. Inside the charged shell, This is a surprising result, for it is the same as that for a point charge (see Equation 18.3 with ). Thus, the electric field outside a uniformly charged spherical shell is the same as if all the charge q were concentrated as a point charge at the center of the shell. . Thus, we can solve for E and find that . Since the only charge within the Gaussian surface is the charge q on the shell, it follows is just the sum of the tiny areas that make up the Gaussian surface. Since the area of a spherical , we have . Setting the electric flux equal to , as specified by Gauss’ law, that the net charge within the Gaussian surface is yields The term surface is a. To find the magnitude of the electric field outside the charged shell, we evaluate the electric flux by using a spherical Gaussian surface of radius r (r > R) that is concentric with the shell. See is everywhere perpendicular to the the surface labeled S in Figure 18-35a. Since the electric field Gaussian surface, and . In addition, E has the same value at all points on the surface, since they are equidistant from the charged shell. Being constant over the surface, E can be factored outside the summation, with the result that Solution Surface 1 —the flat left end of the cylinder—is embedded inside the positive metal plate. As discussed in Section 18.8, the electric field is zero everywhere inside a conductor that is in equilibrium under electrostatic conditions. Since , the electric flux through this surface is also zero: Solution Figure 18-36b shows that we have placed our Gaussian cylinder so that its left end is inside the positive metal plate, and the right end is in the space between the plates. To determine the electric flux through this Gaussian surface, we evaluate the flux through each of the three parts—labeled 1, 2, and 3 in the drawing—that make up the total surface of the cylinder and then add up the fluxes. Figure 18-36 (a) A side view of a parallel plate capacitor, showing some of the electric field lines. (b) The Gaussian surface is a cylinder oriented so its axis is perpendicular to the positive plate and its left end is inside the plate. Reasoning Figure 18-36a shows the electric field inside a parallel plate capacitor. Because the positive and negative charges are distributed uniformly over the surfaces of the plates, symmetry requires that the electric field be perpendicular to the plates. We will take advantage of this symmetry by choosing our Gaussian surface to be a small cylinder whose axis is perpendicular to the plates (see part b of the figure). With this choice, we will be able to evaluate the electric flux and then, with the aid of Gauss’ law, determine E. According to Equation 18.4, the electric field inside a parallel plate capacitor, and away from the edges, is constant , where is the charge density (the charge per unit area) on a plate. Use Gauss’ and has a magnitude of law to prove this result. For similar questions (including calculational counterparts), consult Self-Assessment Test 18.2, which is described next. Background: Electric flux is defined in terms of the electric field and the surface area through which the field passes. Gauss’ law is formulated in terms of electric flux and relates the flux to the net electric charge. The drawing shows an arrangement of three charges. In parts (a) and (b) different Gaussian surfaces are shown. Through which surface, if either, does the greater electric flux pass? (The answer is given at the end of the book.) Check Your Understanding 4 , where Q is the net charge inside the According to Gauss’ law, we set the electric flux equal to . But Q/A is the charge per unit area, , on the plate. Therefore, we arrive at the Gaussian cylinder: . Notice that the distance of the right end of value of the electric field inside a parallel plate capacitor: the Gaussian cylinder from the positive plate does not appear in this result, indicating that the electric field has the same value everywhere between the plates. The electric flux through the entire Gaussian cylinder is the sum of the three fluxes determined above: Surface 3—the flat right end of the cylinder—is perpendicular to the electric field between the plates, . The electric field is constant over this surface, so E can be taken outside the summation so is the area of surface 3, we find that the electric flux through this surface is in Equation 18.6. Noting that Surface 2—the curved wall of the cylinder—is everywhere parallel to the electric field between the plates, so . Therefore, the electric flux through this surface is also zero: that Copyright © 2007 John Wiley & Sons, Inc. All rights reserved. · Gauss’ Law · Electric Field Lines · The Electric Field Test your understanding of the material in Sections 18.6, 18.7, 18.8 and 18.9 : 18.10 *Copiers and Computer Printers The electrostatic force that charged particles exert on one another plays the central role in an office copier. The copying process is called xerography, from the Greek xeros view and graphos, meaning “dry writing.” The heart of a copier is the xerographic drum, an aluminum cylinder coated with a layer of selenium (see Figure 1837a). Aluminum is an excellent electrical conductor. Selenium, on the other hand, is a photoconductor: it is an insulator in the dark but becomes a conductor when exposed to light. Consequently, a positive charge deposited on the selenium surface will remain there, provided the selenium is kept in the dark. When the drum is exposed to light, however, electrons from the aluminum pass through the conducting selenium and neutralize the positive charge. Figure 18-37 (a) This cutaway shows the essential elements of a copying machine. (b) The five steps in the xerographic process. The physics of xerography. The photoconductive property of selenium is critical to the xerographic process, as Figure 18-37b illustrates. First, an electrode called a corotron gives the entire selenium surface a positive charge in the dark. Second, a series of lenses and mirrors focuses an image of a document onto the revolving drum. The dark and light areas of the document produce corresponding areas on the drum. The dark areas retain their positive charge, but the light areas become conducting and lose their positive charge, ending up neutralized. Thus, a positive-charge image of the document remains on the selenium surface. In the third step, a special dry black powder, called the toner, is given a negative charge and then spread onto the drum, where it adheres selectively to the positively charged areas. The fourth step involves transferring the toner onto a blank piece of paper. However, the attraction of the positive-charge image holds the toner to the drum. To transfer the toner, the paper is given a greater positive charge than that of the image, with the aid of another corotron. Last, the paper and adhering toner pass through heated pressure rollers, which melt the toner into the fibers of the paper and produce the finished copy. A laser printer is used with computers to provide high-quality copies of text and graphics. It is similar in operation to the xerographic machine, except that the information to be reproduced is not on paper. Instead, the information is transferred from the computer’s memory to the printer, and laser light is used to copy it onto the seleniumaluminum drum. A laser beam, focused to a fine point, is scanned rapidly from side to side across the rotating drum, as Figure 18-38 indicates. While the light remains on, the positive charge on the drum is neutralized. As the laser beam moves, the computer turns the beam off at the right moments during each scan to produce the desired positive-charge image, which is the letter “A” in the picture. Figure 18-38 As the laser beam scans back and forth across the surface of the xerographic drum, a positivecharge image of the letter “A” is created. The physics of a laser printer. An inkjet printer is another type of printer that uses electric charges in its operation. While shuttling back and forth across the paper, the inkjet printhead ejects a thin stream of ink. Figure 18-39 illustrates the elements of one type of printhead. The ink is forced out of a small nozzle and breaks up into extremely small droplets, with diameters that can be as small as 9 × 10−6 m. About 150 000 droplets leave the nozzle each second and travel with a speed of approximately 18 m/s toward the paper. During their flight, the droplets pass through two electrical components, an electrode and the deflection plates (a parallel plate capacitor). When the printhead moves over regions of the paper that are not to be inked, the charging control is turned on and an electric field is established between the printhead and the electrode. As the drops pass through the electric field, they acquire a net charge by the process of induction. The deflection plates divert the charged droplets into a gutter and thus prevent them from reaching the paper. Whenever ink is to be placed on the paper, the charging control, responding to instructions from the computer, turns off the electric field. The uncharged droplets fly straight through the deflection plates and strike the paper. Figure 18-39 An inkjet printhead ejects a steady flow of ink droplets. Charged droplets are deflected into a gutter by the deflection plates, while uncharged droplets fly straight onto the paper. Letters formed by an inkjet printer look normal, except when greatly enlarged and the patterns from the drops become apparent. The physics of an inkjet printer. Copyright © 2007 John Wiley & Sons, Inc. All rights reserved. 18.11 Concepts & Calculations In this chapter we have studied electric forces and electric fields. We conclude now by presenting some examples that review important features of these concepts. The three-part format of the examples stresses the role of conceptual understanding in problem solving. First, the problem statement is given. Then, there is a concept question-and-answer section, followed by the solution section. The purpose of the concept question-and-answer section is to provide help in understanding the solution and to illustrate how a review of the concepts can help in anticipating some of the characteristics of the numerical answers. Concepts & Calculations Example 17 | The Vector Nature of Electric Forces The charges on three identical metal spheres are , , and . The spheres are brought together so they simultaneously touch each other. They are then separated and placed on the x and y axes, as in Figure 18-40a. What is the net force (magnitude and direction) exerted on the sphere at the origin? Treat the spheres as if they were particles. Figure 18-40 (a) Three equal charges lie on the x and y axes. (b) The net force . exerted on q1 by the other two charges is Concept Questions and Answers Is the net charge on the system comprising the three spheres the same before and after touching? Answer Yes. The conservation of electric charge states that during any process the net electric charge of an isolated system remains constant (is conserved). Therefore, the net charge on the three spheres before they touch ( ) is the same as the net charge after they touch. After the spheres touch and are separated, do they have identical charges? Answer Yes. Since the spheres are identical, the net charge distributes itself equally over the three spheres. After they are separated, each has one-third of the net charge: . Do q2 and q3 exert forces of equal magnitude on q1? Answer Yes. The charges q2 and q3 have equal magnitudes and are the same distance from q1. According to Coulomb’s law, then, they exert forces of equal magnitude on q1. Is the magnitude of the net force exerted on q1 equal to 2F, where F is the magnitude of the force that either q2 or q3 exerts on q1? Answer No. Although the two forces that act on q1 have equal magnitudes, they have different directions. The forces are repulsive forces, since all of the charges in part a of the drawing are exerted on q1 by q2 and the force identical. Figure 18-40b shows the force exerted on q1 by q3. To obtain the net force , we must take these directions into account by using vector addition. Solution The magnitude F12 of the force exerted on q1 by q2 is given by Coulomb’s law, Equation 18.1, as Problem solving insight Often charge magnitudes are specified in microcoulombs . When using Coulomb’s law, be sure to convert microcoulombs into coulombs before substituting for the charge and . magnitudes Note that we have used the magnitudes of q1 and q2 in Coulomb’s law. As mentioned previously, the magnitude of the force exerted on q1 by q3 has the same value as F12, and so . Since the forces and are perpendicular to each other (see Figure 18-40b), we may use the Pythagorean theorem to find the magnitude F of the net force: The angle that the net force makes with the −x axis (see part b of the drawing) is Concepts & Calculations Example 18 | Becoming Familiar with Electric Fields Two point charges are lying on the y axis in Figure 18-41a: and . They are equidistant from the point P, which lies on the x axis. (a) What is the and mass m = 1.20 g is net electric field at P? (b) A small object of charge placed at P. When it is released, what is its acceleration? Figure 18-41 (a) Two charges q1 and q2 produce an electric field at the point P. (b) The electric fields and add to give the net electric field . Concept Questions and Answers There is no charge at P in part (a). Is there an electric field at P? Answer Yes. An electric field is produced by the charges q1 and q2, and it exists throughout the entire region that surrounds them. If a test charge were placed at this point, it would experience a force due to the electric field. The force would be the product of the charge and the electric field. The charge q1 produces an electric field at the point P. What is the direction of this field? Answer The electric field created by a charge always points away from a positive charge and points toward it (see toward a negative charge. Since q1 is negative, the electric field Figure 18-41b). What is the direction of the electric field produced by q2 at P? Answer Since q2 is positive, the electric field that it produces points away from q2, as shown in the drawing. Is the magnitude of the net electric field equal to E1 + E2, where E1 and E2 are the magnitudes of the electric fields produced by q1 and q2? Answer No, because the electric fields have different directions. We must add the individual fields as vectors to obtain the net electric field. Only then can we determine its magnitude. Solution a. The magnitude of the electric fields that q1 and q2 produce at P are given by Equation 18.3, where the distances are specified in the drawing: The x and y components of these fields and the total field table: are given in the following Electric field x component = 0 N/C The net electric field y component = + 7.56 × 104 N/C has only a component along the + y axis. So, of an object placed b. According to Newton’s second law, Equation 4.2, the acceleration is at this point is equal to the net force acting on it divided by its mass. The net force , as indicated by Equation the product of the charge and the net electric field, 18.2. Thus, the acceleration is At the end of the problem set for this chapter, you will find homework problems that contain both conceptual and quantitative parts. These problems are grouped under the heading Concepts & Calculations. Copyright © 2007 John Wiley & Sons, Inc. All rights reserved. Concept Summary If you need more help with a concept, use the text examples and media noted next to the discussion or equation. Go to for: Interactive LearningWare (ILW)—examples solved in a five-step interactive format. Concept Simulations (CS)—animated text figures or animations of important concepts. Interactive Solutions (IS)—models for certain types of problems in the chapter homework. The calculations are carried out interactively. Self-Assessment Tests—qualitative and quantitative questions with extensive feedback. Topic Discussion The coulomb (C) 18.1 THE ORIGIN OF ELECTRICITY There are two kinds of electric charge: positive and negative. The SI unit of electric charge is the coulomb (C). The magnitude of the charge on an electron or a proton is Magnitude of charge on electron or proton Since the symbol e denotes a magnitude, it has no algebraic sign. Thus, the electron carries a charge of − e, and the proton carries a charge of + e. Charge is quantized The charge on any object, whether positive or negative, is quantized, in the sense that the charge consists of an integer number of protons or electrons. Law of conservation of electric charge 18.2 CHARGED OBJECTS AND THE ELECTRIC FORCE The law of conservation of electric charge states that the net electric charge of an isolated system remains constant during any process. Electric repulsion and attraction Like charges repel and unlike charges attract each other. Conductor 18.3 CONDUCTORS AND INSULATORS An electrical conductor is a material, such as copper, that conducts electric charge readily. Learning Aids Insulator An electrical insulator is a material, such as rubber, that conducts electric charge poorly. Charging by contact 18.4 CHARGING BY CONTACT AND BY INDUCTION Charging by contact is the process of giving one object a net electric charge by placing it in contact with an object that is already charged. Charging by induction Charging by induction is the process of giving an object a net electric charge without touching it to a charged object. Point charge 18.5 COULOMB’s LAW A point charge is a charge that occupies so little space that it can be regarded as a mathematical point. Coulomb’s law gives the magnitude F of the electric force that two point charges q1 and q2 exert on each other: Coulomb’s law (18.1) where and are the magnitudes of the charges and have no algebraic sign. The term k is a constant and . The force has the value specified by Equation 18.1 acts along the line between the two charges. Permittivity of free space The permittivity of free space relation is defined by the Use Self-Assessment Test 18.1 to evaluate your understanding of Sections 18.1, 18.2, 18.3, 18.4 and 18.5 . 18.6 THE ELECTRIC FIELD The electric field at a given spot is a vector and is the electrostatic force experienced by a very small test charge q0 placed at that spot divided by the charge itself: Electric field (18.2) The direction of the electric field is the same as the direction of the force on a positive test charge. The SI unit for the electric field is the newton per coulomb (N/ C). The source of the electric field at any spot is the charged objects surrounding that spot. The magnitude of the electric field created by a point charge q is Electric field of a point charge (18.3) where |q| is the magnitude of the charge and has no algebraic sign and r is the distance from the charge. The points away from a positive charge electric field and toward a negative charge. For a parallel plate capacitor that has a charge per unit area of on each plate, the magnitude of the electric field between the plates is Electric field of a parallel plate capacitor Electric field lines Direction of electric field Strength of electric field (18.4) 18.7 ELECTRIC FIELD LINES Electric field lines are lines that can be thought of as a “map,” insofar as the lines provide information about the direction and 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 passing perpendicularly through a surface is the greatest—that is, where the lines are packed together most tightly. Excess charge carried by a conductor at equilibrium 18.8 THE ELECTRIC FIELD INSIDE A CONDUCTOR: SHIELDING Excess negative or positive charge resides on the surface of a conductor at equilibrium under electrostatic conditions. In such a situation, the electric field at any point within the conducting material is zero, and the electric field just outside the surface of the conductor is perpendicular to the surface. through a 18.9 Gauss’ LAW The electric flux surface is related to the magnitude E of the 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) Gauss’ law states that the electric flux through a closed surface (a Gaussian surface) is equal to the net charge Q enclosed by the surface divided by , the permittivity of free space: Gauss’ law (18.7) Use Self-Assessment Test 18.2 to evaluate your understanding of Sections 18.6, 18.7, 18.8 and 18.9 . Copyright © 2007 John Wiley & Sons, Inc. All rights reserved. Conceptual Questions ssm Solution is in the Student Solutions Manual 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 left with a charge? Account for your answer. 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 happens to 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 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 natural oils and drying it thoroughly, run a plastic comb through it. Small bits of paper will be attracted to the comb. Explain why. 5 Blow up a balloon and rub it against your shirt a number of times. In so doing you give the balloon a net electric charge. Now touch the balloon to the ceiling. On being released, the balloon will remain stuck to the ceiling. Why? 6 A proton and an electron are held in place on the x axis. The proton is at x = −d, while the electron is at x = + d. They are released simultaneously, and the only force that affects their motions is the electrostatic force of attraction that each applies to the other. Which particle reaches the origin first? Give your reasoning. 7 A particle is attached to a spring and is pushed so that the spring is compressed more and more. As a result, the spring exerts a greater and greater force on the particle. Similarly, a charged particle experiences a greater and greater force when pushed closer and closer to another particle that is fixed in position and has a charge of the same polarity. In spite of the similarity, the charged particle will not exhibit simple harmonic motion on being released, as will the particle on the spring. Explain why not. 8 ssm Identical point charges are fixed to opposite corners of a square. Where does a third point charge experience the greater force, at one of the empty corners or at the center of the square? Account for your answer. 9 On a thin, nonconducting rod, positive charges are spread evenly, so that there is the same amount of charge per unit length at every point. On another identical rod, positive charges are spread evenly over only the left half, and the same amount of negative charges are spread evenly over the right half. For each rod, deduce the direction of the electric field at a point that is located directly above the midpoint of the rod. Give your reasoning. 10 There is an electric field at point P. A very small charge is placed at this point and experiences a force. Another very small charge is then placed at this point and experiences a force that differs in both magnitude and direction from that experienced by the first charge. How can these two different forces result from the single electric field that exists at point P? 11 Three point charges are fixed to the corners of a square, one to a corner, in such a way that the net electric field at the empty corner is zero. Do these charges all have (a) the same sign and (b) the same magnitude (but, possibly, different signs)? Justify your answers. 12 Review Conceptual Example 12 as an aid in answering this question. Suppose in Figure 18-21 that charges + q are placed on corners 1 and 3 of the rectangle, and charges − q are placed on corners 2 and 4. What is 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 where the 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 greater than the magnitude of the positive charge. Is there any place on the line through the charges where the electric field is zero? Justify your answer. 14 ssm Drawings I and II show two examples of electric field lines. Decide which of the following statements are true and which are false, defending your choice in each case. (a) In both I and II the electric field is the same everywhere. (b) As you move from left to right in each case, the electric field becomes stronger. (c) The electric field in I is the same everywhere but becomes stronger in II as you move from left to right. (d) The electric fields in both I and II could be created by negative charges located somewhere on the left and positive charges somewhere on the right. (e) Both I and II 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 between the 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 through the surface is . Does the fact that imply that the electric field at also zero; that is, 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 q1, q2, and q3. A Gaussian surface is drawn around q1 and q2. (a) Which charges determine the electric flux through the Gaussian surface? (b) Which charges produce the electric field at the point P? Justify your answers. 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? Copyright © 2007 John Wiley & Sons, Inc. All rights reserved. . What is the net charge after electrons have been placed on it? of water? (b) What is the net charge of all these electrons? Section 18.5 Coulomb’s Law 7 ssm Two charges attract each other with a force of 1.5 N. What will be the force if the distance between them is reduced to one-ninth of its original value? electrons are there in one liter . The objects are 4 Four identical metallic objects carry the following charges: + 1.6, + 6.2, − 4.8, and brought simultaneously into contact, so that each touches the others. Then they are separated, (a) What is the final charge on each object? (b) How many electrons (or protons) make up the final charge on each object? 5 ssm Consider three identical metal spheres, A, B, and C. Sphere A carries a charge of + 5q. Sphere B carries a charge of − q. Sphere C carries no net charge. Spheres A and B are touched together and then separated. Sphere C is then touched to sphere A and separated from it. Last, sphere C is touched to sphere B and separated from it. (a) How much charge ends up on sphere C? What is the total charge on the three spheres (b) before they are allowed to touch each other and (c) after they have touched? * 6 Water has a mass per mole of 18.0 g/mol, and each water molecule (H2O) has 10 electrons. (a) How many 3 A metal sphere has a charge of 1 ssm How many electrons must be removed from an electrically neutral silver dollar to give it a charge of , while a rod carries a charge of How many electrons must 2 A plate carries a charge of be transferred from the plate to the rod, so that both objects have the same charge? Section 18.4 Charging by Contact and by Induction Section 18.3 Conductors and Insulators, Section 18.2 Charged Objects and the Electric Force, Section 18.1 The Origin of Electricity, This icon represents a biomedical application. ssm Solution is in the Student Solutions Manual. are presented in a guided tutorial format that provides enhanced interactivity. See Preface for additional details. Note to Instructors: Most of the homework problems in this chapter are available for assignment via an online homework management program such as WileyPLUS or WebAssign, and those marked with the icon Problems that are not marked with a star are considered the easiest to solve. Problems that are marked with a single star (*) are more difficult, while those marked with a double star (**) are the most difficult. Problems ? . Find (a) the charge on each particle and (b) the mass of particle 2. and . They are separated by 14 Two tiny conducting spheres are identical and carry charges of a distance of 2.50 cm. (a) What is the magnitude of the force that each sphere experiences, and is the force attractive or repulsive? (b) The spheres are brought into contact and then separated to a distance of 2.50 cm. Determine the magnitude of the force that each sphere now experiences, and state whether the force is attractive or repulsive. 15 Interactive Solution 18.15 at provides a model for solving this type of problem. Two small objects, A and B, are fixed in place and separated by 3.00 cm in a vacuum. Object A has a charge of , and object B has a charge of . How many electrons must be removed from A and put onto B to make the electrostatic force that acts on each object an attractive force whose magnitude is 68.0 N? * 16 Multiple-Concept Example 3 provides some pertinent background for this problem. Suppose a single electron orbits about a nucleus containing two protons (+ 2e), as would be the case for a helium atom from which one of . Determine the the two naturally occurring electrons is removed. The radius of the orbit is magnitude of the electron’s centripetal acceleration. of 13 ssm www Two particles, with identical positive charges and a separation of , are released from rest. Immediately after the release, particle 1 has an acceleration whose magnitude is , while particle 2 has an acceleration whose magnitude is . Particle 1 has a mass 9 ssm www Two very small spheres are initially neutral and separated by a distance of 0.50 m. Suppose that electrons are removed from one sphere and placed on the other. (a) What is the magnitude of the electrostatic force that acts on each sphere? (b) Is the force attractive or repulsive? Why? 10 A charge + q is located at the origin, while an identical charge is located on the x axis at x = + 0.50 m. A third charge of + 2 q is located on the x axis at such a place that the net electrostatic force on the charge at the origin doubles, its direction remaining unchanged. Where should the third charge be located? 11 ssm Consult Concept Simulation 18.1 at for insight into this problem. Three charges are fixed to an x, y coordinate system. A charge of is on the y axis at y = + 3.0 m. A charge of is at the origin. Last, is on the x axis at x = + 3.0 m. Determine the magnitude and direction of the net electrostatic a charge of force on the charge at x = + 3.0 m. Specify the direction relative to the − x axis. is fixed at the center of a compass. Two additional charges are fixed on the circle of 12 A charge of the compass (radius = 0.100 m). The charges on the circle are at the position due north and at the position due east. What is the magnitude and direction of the net electrostatic force acting on the charge at the center? Specify the direction relative to due east. 8 Two spherical objects are separated by a distance of . The objects are initially electrically neutral and are very small compared to the distance between them. Each object acquires the same negative charge due to the addition of electrons. As a result, each object experiences an electrostatic force that has a magnitude of . How many electrons did it take to produce the charge on one of the objects? * 20 Multiple-Concept Example 3 illustrates several of the concepts used in this problem. A single electron orbits a lithium nucleus that contains three protons (+ 3e). The radius of the orbit is . Determine the kinetic energy of the electron. * 21 An electrically neutral model airplane is flying in a horizontal circle on a 3.0-m guideline, which is nearly parallel to the ground. The line breaks when the kinetic energy of the plane is 50.0 J. Reconsider the same situation, except that now there is a point charge of + q on the plane and a point charge of − q at the other end of the guideline. In this case, the line breaks when the kinetic energy of the plane is 51.8 J. Find the magnitude of the charges. * 19 ssm Two spheres are mounted on identical horizontal springs and reston a frictionless table, as in the drawing. When the spheres are uncharged, the spacing between them is 0.0500 m, and the springs are unstrained. When , the spacing doubles. Assuming that the spheres have a negligible each sphere has a charge of diameter, determine the spring constant of the springs. points vertically downward and has a magnitude of 405 N. Determine the magnitudes and algebraic signs of the charges qA and qB. * 18 The drawing shows an equilateral triangle, each side of which has a length of 2.00 cm. Point charges are fixed to charge experiences a net force due to the charges qA and qB. This net force each corner, as shown. The 1.50 g and it were free to move, what would be its acceleration? * 17 Interactive LearningWare 18.2 at provides one approach to solving problems such as this one. The drawing shows three point charges fixed in place. The charge at the coordinate origin has a value of ; the other two charges have identical magnitudes, but opposite signs: and . (a) Determine the net force (magnitude and direction) exerted on q1 by the other two charges. (b) If q1 had a mass of 28 Four point charges have the same magnitude of and are fixed to the corners of a square that is 4.0 cm on a side. Three of the charges are positive and one is negative. Determine the magnitude of the net electric field that exists at the center of the square. , are fixed in place and separated by 3.0 m. (a) At what spot along a line through 29 Two charges, −16 and the charges is the net electric field zero? Locate this spot relative to the positive charge. (Hint: The spot does placed at this spot? not necessarily lie between the two charges.) (b) What would be the force on a charge of Section 18.8 The Electric Field Inside a Conductor: Shielding 25 Concept Simulation 18.2 at provides background concerning the electric field lines that are the focus of this problem. Review the important features of electric field lines discussed in Conceptual Example 13. Three point charges (+ q, + 2q, and −3q) are at the corners of an equilateral triangle. Sketch in six electric field lines between the three charges 26 Review Conceptual Example 13 as an aid in working this problem. Charges of −4q are fixed to diagonally opposite corners of a square. A charge of + 5q is fixed to one of the remaining corners, and a charge of + 3q is fixed to the last corner. Assuming that ten electric field lines emerge from the + 5q charge, sketch the field lines in the vicinity of the four charges. . What electric field (magnitude and direction) 27 ssm www A tiny ball (mass = 0.012 kg) carries a charge of is needed to cause the ball to float above the ground? Section 18.7 Electric Field Lines, Section 18.6 The Electric Field, . Two are positive and two are negative. The charges are ** 24 There are four charges, each with a magnitude of fixed to the corners of a 0.30-m square, one to a corner, in such a way that the net force on any charge is directed toward the center of the square. Find the magnitude of the net electrostatic force experienced by any charge. is held 0.150 m away from the sphere and directly to the right of it, so negligible mass. A charge of the wire makes an angle with the vertical (see the drawing). Find (a) the angle and (b) the tension in the wire. ** 22 Two identical small insulating balls are suspended by separate 0.25-m threads that are attached to a common point on the ceiling. Each ball has a mass of 8.0 × 10−4 kg. Initially the balls are uncharged and hang straight down. They are then given identical positive charges and, as a result, spread apart with an angle of 36° between the threads. Determine (a) the charge on each ball and (b) the tension in the threads. ** 23 ssm A small spherical insulator of mass and charge is hung by a thin wire of The membrane surrounding a living cell consists of an inner and an outer wall that are separated by a is at x1 = + 9.0 cm. Find the net electric field (magnitude and direction) at (a) x = 0 cm and that is perpendicular to the . It is placed at a certain spot where there is an electric field. When in the direction of the + x axis. Determine released, the object experiences an acceleration of the magnitude and direction of the electric field. * 38 Multiple-Concept Example 9 illustrates some of the concepts pertinent to this problem. An electron is released from rest at the negative plate of a parallel plate capacitor. The charge per unit area on each plate , and the plates are separated by a distance of . How fast is the electron is moving just before it reaches the positive plate? and a charge of the plate radius for the first capacitor. * 37 Interactive Solution 18.37 at provides a model for problems of this kind. A small object has a mass of r between the two charges? * 36 Two parallel plate capacitors have circular plates. The magnitude of the charge on these plates is the same. , while the field within However, the electric field between the plates of the first capacitor is . Determine the ratio r2/r1 of the plate radius for the second capacitor to the second capacitor is is . The charge per unit area on each plate has a magnitude of . The force on q1 due to q2 equals the force on q1 due to the electric field of the parallel plate capacitor. What is the distance rod. With respect to the z axis, find the magnitude of the net torque applied to the rod. 33 ssm A small drop of water is suspended motionless in air by a uniform electric field that is directed upward and has a magnitude of 8480 N/C. The mass of the water drop is . (a) Is the excess charge on the water drop positive or negative? Why? (b) How many excess electrons or protons reside on the drop? 34 Review Conceptual Example 12 before attempting to work this problem. The magnitude of each of the charges . The lengths of the sides of the rectangles are 3.00 cm and 5.00 cm. Find in Figure 18-21 is the magnitude of the electric field at the center of the rectangle in Figures 18-21a and b. * 35 Two charges are placed between the plates of a parallel plate capacitor. One charge is + q1 and the other x, y plane there is a constant external electric field (b) x = + 6.0 cm. 32 A long, thin rod (length = 4.0 m) lies along the x axis, with its midpoint at the origin. In a vacuum, a point charge is fixed to one end of the rod, and a point charge is fixed to the other end. Everywhere in the other small space. Assume that the membrane acts like a parallel plate capacitor in which the effective charge density on . (a) What is the magnitude of the electric field the inner and outer walls has a magnitude of within the cell membrane? (b) Find the magnitude of the electric force that would be exerted on a potassium ion (K +; charge = + e) placed inside the membrane. is at x1 = + 3.0 cm and the 31 ssm Two charges are placed on the x axis. One of the charges 30 ? ** 43 A small plastic ball with a mass of and with a charge of is suspended from an insulating thread and hangs between the plates of a capacitor (see the drawing). The ball is in equilibrium, with the thread making an angle of 30.0° with respect to the vertical. The area of each plate is 0.0150 m2. What is the magnitude of the charge on each plate? ** 42 The drawing shows an electron entering the lower left side of a parallel plate capacitor and exiting at the upper . The capacitor is 2.00 cm long, and its plates right side. The initial speed of the electron is are separated by 0.150 cm. Assume that the electric field between the plates is uniform everywhere and find its magnitude. a circle. At the center of the circle they produce a net electric field that is directed upward along the vertical axis. Determine the ratio / of the charge magnitudes. * 41 Interactive Solution 18.41 at for help with this problem. The drawing shows two positive charges q1 and q2 fixed to of + 480 N/C. What is the displacement of the particle after a time of * 40 In Multiple-Concept Example 9 you can see the concepts that are important in this problem. A particle of charge and mass is released from rest in a region where there is a constant electric field at the (empty) upper right corner is zero. Find the magnitudes of q1 and q2. Express your an swers in terms of q. * 39 ssm www A rectangle has a length of 2d and a height of d. Each of the following three charges is located at a corner of the rectangle: + q1 (upper left corner), + q2 (lower right corner), and −q (lower left corner). The net electric field . Find the angle in the drawing. P What is Q? 49 A vertical wall in a house faces due east. A uniform electric field has a magnitude of 150 N/C. This field is parallel to the ground and points 35° north of east. What is the electric flux through the wall? 50 A charge Q is located inside a rectangular box. The electric flux through each of the six surfaces of the box is: 48 A rectangular surface is oriented in a uniform electric field of 580 N/C. What is the maximum possible electric flux through the surface? 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 charge, (b) a single charge, and (c) both of the charges in (a) and (b). 47 ssm The drawing shows an edge-on view of two planar surfaces that intersect and are mutually perpendicular. Surface 1 has an area of 1.7 m2, while surface 2 has an area of 3.2 m2. The electric field in the drawing is uniform and has a magnitude of 250 N/C. Find the electric flux through (a) surface 1 and (b) surface 2. is M ** 44 Two particles are in a uniform electric field whose value is + 2500 N/C. The mass and charge of particle 1 and , while the corresponding values for particle 2 are are and . 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 accelerate, 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 at the midpoint M between the charges has a magnitude E . The field directly above the midpoint at point P has a magnitude E . The ratio of these two field magnitudes . . (Hint: For a Gaussian surface, use a sphere of radius r sphere at a radius r has a magnitude of Copyright © 2007 John Wiley & Sons, Inc. All rights reserved. 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 for a perspective that is useful in solving this problem. Two spherical shells have a common center. A charge is spread uniformly over the inner shell, which has a radius of 0.050 m. A charge is spread uniformly over the outer shell, which has a radius of 0.15 m. Find the magnitude and direction of the electric field at a distance (measured from the common center) 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 an x, 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. The magnitude of the field is 1500 N/C. (a) Find the electric flux 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 the cubical surface is zero, as Gauss’ law predicts, since there is no net charge within the cube. ** 54 A long, thin, straight wire of length L has a positive charge Q distributed uniformly along it. Use Gauss’ law to show that the electric field created by this wire at a radial distance r has a magnitude of , where . (Hint: For a Gaussian surface, use a cylinder aligned with its axis along the wire and note that the cylinder has a flat surface at either end, as well as a curved surface.) . Use Gauss’ law to show that the electric field at a point within the charge per unit volume, therefore, is * 51 Interactive Solution 18.51 at offers help with this problem in an interactive environment. A solid nonconducting sphere has a positive charge q spread uniformly throughout its volume. The charge density or Additional Problems 55 An electric field of 260 000 N/C points due west at a certain spot. What are the at this spot? magnitude and direction of the force that acts on a charge of 56 The force of repulsion that two like charges exert on each other is 3.5 N. What will the force be if the distance between the charges is increased to five times its original value? 57 Conceptual Example 14 deals with the hollow spherical conductor in Figure 18-31. The conductor is initially electrically neutral, and then a charge + q is placed at the center of the hollow space. Suppose the conductor initially has a net charge of + 2q instead of being neutral. What is the total charge on the interior and on the exterior surface when the + q charge is placed at the center? 58 At a distance r1 from a point charge, the magnitude of the electric field created by the charge is 248 N/C. At a distance r2 from the charge, the field has a magnitude of 132 N/ C. Find the ratio r2/r1. 59 ssm Background pertinent to this problem is available in Interactive LearningWare 18.3 at . A point charge is placed in an external uniform electric field of . At what distance from the charge is the net electric field zero? 60 In a vacuum, two particles have charges of q1 and q2, where . They are separated by a distance of 0.26 m, and particle 1 experiences an attractive force of 3.4 N. What is q2 (magnitude and sign)? 61 ssm Interactive LearningWare 18.1 at offers some perspective on this problem. Two tiny spheres have the same mass and carry charges of the same magnitude. The mass of each sphere is . The gravitational force that each sphere exerts on the other is balanced by the electric force. (a) What algebraic signs can the charges have? (b) Determine the charge magnitude. is located in an electric field. The x and y components of the 62 A charge of and , respectively. (a) electric field are What is the magnitude of the force on the charge? (b) Determine the angle that the force makes with the + x axis. at x1 = + 4.0 cm, and * 63 Two charges are located along the x axis: at x2 = − 4.0 cm. Two other charges are located on the y axis: at y3 = + 5.0 cm, and electric field (magnitude and direction) at the origin. at y4 = + 7.0 cm. Find the net * 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. * 65 A proton is moving parallel to a uniform electric field. The electric field accelerates the proton and thereby increases its linear momentum to from in a time of . 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 and mass , is placed in a constant electric field. Starting from rest, the object accelerates to a speed of in a time of 0.96 s. Determine the magnitude of the electric field. * 67 ssm In the rectangle in the drawing, a charge is to be placed at the empty corner to make the net force on the charge at 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. Copyright © 2007 John Wiley & Sons, Inc. All rights reserved. Concepts & Calculations Note: Each of these problems consists of Concept Questions followed by a related quantitative Problem. The Concept Questions involve little or no mathematics. They focus on the concepts with which the problems deal. Recognizing the concepts is the essential initial step in any problem-solving technique. 69 Concept Questions Two identical metal spheres have charges of q1 and q2. 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 sphere the same? Why? Problem Four identical metal spheres have charges of , , , and . (a) Two of the spheres are brought together so they touch and then they are separated. Which spheres are they, if the final charge on each of the two is ? (b) In a similar manner, which three spheres are brought together and then separated, if the final charge on each of the three is (c) How many electrons would have to be added to one of the spheres in part (b) to make it electrically neutral? 70 Concept Questions The drawings show three charges that have the same magnitude, but different signs. In all cases the distance d between charges 1 and 2 and between 2 and 3 is 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 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 |q| = 8.6 μC, and the distance between them is 3.8 mm. Determine the magnitude of the net force on charge 2 for each of the three drawings. Verify that your answers are consistent with your answers to the Concept Questions. 71 Concept Questions Suppose you want to neutralize the V gravitational attraction between the earth and the moon by placing equal amounts of charge on each. (a) Should the charges be both positive, both negative, or one positive and the other negative? Why? (b) Do you need to know the distance between the earth and the moon to find the magnitude of the charge? Why or why not? and , Problem The masses of the earth and moon are respectively. Identical amounts of charge are placed on each body, such that the net force (gravitational plus electrical) on each is zero. What is the magnitude of the charge? 72 Concept Questions Suppose you want to determine the electric field in a certain region of space. You have a small object of known charge and an instrument that measures the magnitude and direction of the force exerted on the object by the electric field. How would you determine the magnitude and direction of the electric field if the object were (a) positively charged and (b) negatively charged? Problem (a) The object has a charge of and the instrument indicates that the electric force exerted on it is , due east. What is the magnitude and direction of the electric field? (b) What is the magnitude and direction of the electric field if the object has a charge of and the instrument indicates that the force is due west? 73 Concept Question The drawing shows two situations in which charges are placed on the x and y axes. They are all located at the same distance from the origin O. Without doing any calculations, does the net electric field at the origin in part (a) have a magnitude that is greater than, less than, or equal to the magnitude of the field at the origin in part (b)? Justify your answer. Problem The distance between each of the charges and the origin is 6.1 cm. For each of the situations shown in the drawing, determine the magnitude of the net electric field at the origin O. Check to see that your results are consistent with your answer to the Concept Question. 74 Concept Questions A proton and an electron are moving due east in a constant electric field that also points due east. (a) Does each experience an electric force of the same magnitude and direction? (b) What is the direction of the proton’s acceleration, and what is the direction of the electron’s acceleration? (c) Is the magnitude of the proton’s acceleration greater than, less than, or the same as that of the electron’s acceleration? Explain your answers. Problem The electric field points due east and has a magnitude of . Determine the magnitude of the acceleration of the proton and the electron. Check that your answers are consistent with part (c) of the Concept Questions. 75 Concept Questions Three point charges have equal magnitudes, two being positive and one negative. These charges are fixed to the corners of an equilateral triangle, as the drawing shows. (a) The charge at any one corner experiences forces from the charges at the other corners. Do the individual forces exerted by the charges have the same or different magnitudes? (b) At which one or more corners does (do) the charge(s) experience a net force that has the greatest magnitude? (c) At which one or more corners does (do) the charge(s) experience a net force that has the smallest magnitude? Problem The magnitude of each of the charges is , and the lengths of the sides of the triangle are 3.0 cm. Calculate the magnitude of the net force that each charge experiences. Be sure that your answers are consistent with your answers to the Concept Questions. 76 Concept Questions The drawing shows a positive point charge + q1, a second point charge q2 that may be positive or negative, and a spot labeled P, all on the same straight line. The distance d between the two charges is the same as the distance between q1 and the point P. With q2 present, the magnitude of the net electric field at P is twice what it is when q1 is present alone. (a) When the second charge is positive, is its magnitude smaller than, equal to, or greater than the magnitude of q1? Explain your reasoning. (b) When the second charge is negative, is its magnitude smaller than, equal to, or greater than that in question (a)? Account for your answer. Problem Given that , determine q2 when it is (a) positive and (b) negative. Verify that your answers are consistent with your answers to the Concept Questions. Copyright © 2007 John Wiley & Sons, Inc. All rights reserved. CHAPTER 18 ELECTRIC FORCES AND ELECTRIC FIELDS CONCEPTUAL QUESTIONS ____________________________________________________________________________________________ 8. 2 REASONING AND SOLUTION Identical point 1 charges are fixed to opposite corners of a square, as shown in the figure at the right. There exists an electric field at each point in space in the vicinity of this configuration. The electric field at a given point is the resultant of the electric field at that C point due to the charge at corner 1 and the charge at corner 3. 3 The magnitude of the electric field due to a single 4 point charge q is given by Equation 18.3: E k q / r 2 , where r is the distance from the charge to the point in question. Since the point charges at corners 1 and 3 are identical, they each have the same value of q. Furthermore, they are equidistant from the center point C. Therefore, at the center of the square, each charge gives rise to an electric field that is equal in magnitude and opposite in direction. The resultant electric field at the center of the square is, therefore, zero. Note that this result is independent of the polarity of the charges. The distance from either empty corner, 2 or 4, to either of the charges is the same. Since the charges are equidistant from either empty corner, each charge gives rise to an electric field that is equal in magnitude. Since the direction of the electric field due to a point charge is radial (radially inward for negative charges and radially outward for positive charges), we see that the electric fields due to each of the two point charges will be mutually perpendicular. Their resultant can be found by using the Pythagorean theorem. The magnitude of the force experienced by a third point charge placed in this system is, from Equation 18.2: F q0 E , where q0 is the magnitude of the charge and E is the magnitude of the electric field at the location of the charge. Since the electric field is zero at the center of the square, a third point charge will experience no force there. Thus, a third point charge will experience the greater force at one of the empty corners of the square. ____________________________________________________________________________________________ 14. REASONING AND SOLUTION In general, electric field lines are always directed away from positive charges and toward negative charges. At a point in space, the direction of the electric field is tangent to the electric field line that passes through that point. The magnitude of the electric field is strongest in regions where the field lines are closest together. Furthermore, the text states that electric field lines created by a positive point charge are directed radially outward, while those created by a negative point charge are directed radially inward. a. In both drawings I and II the electric field is the same everywhere. This statement is false. By inspection, we see that the field lines in II get closer together as we proceed from