Physics 112: Chapter 21-32 PDF

? Water makes life possible: The cells of your body could not function wit...

? Water makes life possible: The cells of your body could not function without water in which to dissolve essential biological molecules. Water is such a good solvent because its molecules (i) have zero net charge; (ii) have zero net charge, but the positive and negative charges are sepa- rated; (iii) have nonzero net charge; (iv) do not respond to electric forces; (v) exert repulsive electric forces on each other. 21 ElEctric chargE and ElEctric FiEld I Learning goaLs n Chapter 5 we mentioned the four kinds of fundamental forces. To this point the only one of these forces that we have examined in any detail is gravity. Looking forward at … Now we are ready to examine the force of electromagnetism, which encom- 21.1 The nature of electric charge, and how we passes both electricity and magnetism. know that electric charge is conserved. Electromagnetic interactions involve particles that have electric charge, an 21.2 How objects become electrically charged. attribute that is as fundamental as mass. Just as objects with mass are acceler- 21.3 How to use Coulomb’s law to calculate the ated by gravitational forces, so electrically charged objects are accelerated by electric force between charges. electric forces. The shock you feel when you scuff your shoes across a carpet 21.4 The distinction between electric force and electric field. and then reach for a metal doorknob is due to charged particles leaping between 21.5 How to calculate the electric field due to a your finger and the doorknob. Electric currents are simply streams of charged collection of charges. particles flowing within wires in response to electric forces. Even the forces that 21.6 How to use the idea of electric field lines to hold atoms together to form solid matter, and that keep the atoms of solid objects visualize and interpret electric fields. from passing through each other, are fundamentally due to electric interactions 21.7 How to calculate the properties of electric between the charged particles within atoms. dipoles. We begin our study of electromagnetism in this chapter by examining the nature of electric charge. We’ll find that charge is quantized and obeys a conser- Looking back at … vation principle. When charges are at rest in our frame of reference, they exert 1.7–1.10 Vector algebra, including the scalar (dot) product and the vector (cross) electrostatic forces on each other. These forces are of tremendous importance in product. chemistry and biology and have many technological applications. Electrostatic 4.3 Newton’s second law. forces are governed by a simple relationship known as Coulomb’s law and are 7.5 Stable and unstable equilibria. most conveniently described by using the concept of electric field. In later 12.5 Streamlines in fluid flow. chapters we’ll expand our discussion to include electric charges in motion. This will lead us to an understanding of magnetism and, remarkably, of the nature of light. While the key ideas of electromagnetism are conceptually simple, applying them to practical problems will make use of many of your mathematical skills, especially your knowledge of geometry and integral calculus. For this reason you may find this chapter and those that follow to be more mathematically demanding than earlier chapters. The reward for your extra effort will be a deeper under- standing of principles that are at the heart of modern physics and technology. 683 684 Chapter 21 electric Charge and electric Field 21.1 eLectric charge The ancient Greeks discovered as early as 600 b.c. that after they rubbed amber with wool, the amber could attract other objects. Today we say that the amber has acquired a net electric charge, or has become charged. The word “electric” is derived from the Greek word elektron, meaning amber. When you scuff your shoes across a nylon carpet, you become electrically charged, and you can charge a comb by passing it through dry hair. Plastic rods and fur (real or fake) are particularly good for demonstrating electrostatics, the interactions between electric charges that are at rest (or nearly so). After we charge both plastic rods in Fig. 21.1a by rubbing them with the piece of fur, we find that the rods repel each other. When we rub glass rods with silk, the glass rods also become charged and repel each other (Fig. 21.1b). But a charged plastic rod attracts a charged glass rod; furthermore, the plastic rod and the fur attract each other, and the glass rod and the silk attract each other (Fig. 21.1c). These experiments and many others like them have shown that there are exactly two kinds of electric charge: the kind on the plastic rod rubbed with fur and the kind on the glass rod rubbed with silk. Benjamin Franklin (1706–1790) suggested calling these two kinds of charge negative and positive, respectively, and these names are still used. The plastic rod and the silk have negative charge; the glass rod and the fur have positive charge. Two positive charges or two negative charges repel each other. A positive charge and a negative charge attract each other. caution Electric attraction and repulsion The attraction and repulsion of two charged objects are sometimes summarized as “Like charges repel, and opposite charges attract.” But “like charges” does not mean that the two charges are exactly identical, only that both charges have the same algebraic sign (both positive or both negative). “Opposite charges” means that both objects have an electric charge, and those charges have different signs (one positive and the other negative). ❙ 21.1 Experiments in electrostatics. (a) Negatively charged objects repel each other. (b) Positively charged objects repel each other. (c) Positively charged objects and negatively charged objects attract each other. (a) Interaction between plastic rods rubbed (b) Interaction between glass rods rubbed (c) Interaction between objects with opposite on fur on silk charges Plain plastic rods neither Plain glass rods neither The fur-rubbed plastic attract nor repel each attract nor repel each rod and the silk- other... other... rubbed glass rod attract each other... – – – – – + + + ++ Fur Plastic Silk Glass... but after being... but after being... and the fur and silk rubbed with fur, rubbed with silk, each attracts the rod it the rods repel the rods repel rubbed. each other. each other. + + + ++ – – – – – + + + ++ – + – – + + – + ++++ – + + ++ ++  21.1 electric Charge 685 21.2 Schematic diagram of the operation of a laser printer. 2 Laser beam “writes” on the drum, leaving negatively charged areas where the image will be. + + + Toner (positively charged) + + + 1 Wire sprays ions onto drum, giving the drum + + a positive charge. + – – + + + – + 6 Lamp discharges the drum, readying – + + + – 3 Roller applies positively charged toner to drum. it to start the process over. Rotating + + Toner adheres only to negatively charged areas imaging – – + of the drum “written” by the laser. drum – + 5 Fuser rollers heat paper so toner + remains permanently attached. + – + – + + – – – + – – ++ – – – – – – – Paper (feeding to left) – – 4 Wires spray a stronger negative charge on paper so toner will adhere to it. A laser printer (Fig. 21.2) utilizes the forces between charged bodies. The printer’s light-sensitive imaging drum is given a positive charge. As the drum rotates, a laser beam shines on selected areas of the drum, leaving those areas with a negative charge. Positively charged particles of toner adhere only to the areas of the drum “written” by the laser. When a piece of paper is placed in contact with the drum, the toner particles stick to the paper and form an image. electric charge and the structure of Matter When you charge a rod by rubbing it with fur or silk as in Fig. 21.1, there is no visible change in the appearance of the rod. What, then, actually happens to the rod when you charge it? To answer this question, we must look more closely at the structure of atoms, the building blocks of ordinary matter. The structure of atoms can be described in terms of three particles: the neg- 21.3 The structure of an atom. The atively charged electron, the positively charged proton, and the uncharged particular atom depicted here is lithium (see Fig. 21.4a). neutron (Fig. 21.3). The proton and neutron are combinations of other entities called quarks, which have charges of ± 13 and ± 23 times the electron charge. Atom Isolated quarks have not been observed, and there are theoretical reasons to believe that it is impossible in principle to observe a quark in isolation. – Most of the The protons and neutrons in an atom make up a small, very dense core called – atom’s volume the nucleus, with dimensions of the order of 10-15 m. Surrounding the nucleus ~10-10 m is occupied – sparsely by are the electrons, extending out to distances of the order of 10-10 m from the electrons. nucleus. If an atom were a few kilometers across, its nucleus would be the size of a tennis ball. The negatively charged electrons are held within the atom by the attractive electric forces exerted on them by the positively charged nucleus. (The protons and neutrons are held within stable atomic nuclei by an attractive interac- + Tiny compared with the Nucleus rest of the atom, the tion, called the strong nuclear force, that overcomes the electric repulsion of the + + nucleus contains over protons. The strong nuclear force has a short range, and its effects do not extend 99.9% of the atom’s mass. far beyond the nucleus.) ~10-15 m The masses of the individual particles, to the precision that they are presently + Proton: Positive charge known, are Mass = 1.673 * 10-27 kg Mass of electron = m e = 9.109382911402 * 10-31 kg Neutron: No charge Mass of proton = m p = 1.6726217771742 * 10-27 kg Mass = 1.675 * 10-27 kg Mass of neutron = m n = 1.6749273511742 * 10-27 kg – Electron: Negative charge Mass = 9.109 * 10-31 kg The numbers in parentheses are the uncertainties in the last two digits. Note that the masses of the proton and neutron are nearly equal and are roughly 2000 times The charges of the electron and proton are equal in magnitude. the mass of the electron. Over 99.9% of the mass of any atom is concentrated in its nucleus. 686 Chapter 21 electric Charge and electric Field 21.4 (a) A neutral atom has as many Protons (+) Neutrons electrons as it does protons. (b) A positive Electrons (-) ion has a deficit of electrons. (c) A nega- tive ion has an excess of electrons. (The electron “shells” are a schematic represen- tation of the actual electron distribution, a diffuse cloud many times larger than the nucleus.) (a) Neutral lithium atom (Li): (b) Positive lithium ion (Li+): (c) Negative lithium ion (Li−): 3 protons (3+) 3 protons (3+) 3 protons (3+) 4 neutrons 4 neutrons 4 neutrons 3 electrons (3-) 2 electrons (2-) 4 electrons (4-) Electrons equal protons: Fewer electrons than protons: More electrons than protons: Zero net charge Positive net charge Negative net charge The negative charge of the electron has (within experimental error) exactly the same magnitude as the positive charge of the proton. In a neutral atom the number of electrons equals the number of protons in the nucleus, and the net electric charge (the algebraic sum of all the charges) is exactly zero (Fig. 21.4a). The number of protons or electrons in a neutral atom of an element is called the atomic number of the element. If one or more electrons are removed from an atom, what remains is called a positive ion (Fig. 21.4b). A negative ion is an atom that has gained one or more electrons (Fig. 21.4c). This gain or loss of electrons is called ionization. When the total number of protons in a macroscopic body equals the total num- ber of electrons, the total charge is zero and the body as a whole is electrically neutral. To give a body an excess negative charge, we may either add negative charges to a neutral body or remove positive charges from that body. Similarly, we can create an excess positive charge by either adding positive charge or removing negative charge. In most cases, negatively charged (and highly mobile) electrons are added or removed, and a “positively charged body” is one that has lost some of its normal complement of electrons. When we speak of the charge of a body, we always mean its net charge. The net charge is always a very small fraction (typically no more than 10-12) of the total positive charge or negative charge in the body. electric charge is conserved Implicit in the foregoing discussion are two very important principles. First is the principle of conservation of charge: The algebraic sum of all the electric charges in any closed system is constant. If we rub together a plastic rod and a piece of fur, both initially uncharged, the rod acquires a negative charge (since it takes electrons from the fur) and the fur acquires a positive charge of the same magnitude (since it has lost as many elec- trons as the rod has gained). Hence the total electric charge on the two bodies together does not change. In any charging process, charge is not created or destroyed; it is merely transferred from one body to another. Conservation of charge is thought to be a universal conservation law. No experimental evidence for any violation of this principle has ever been observed. Even in high-energy interactions in which particles are created and destroyed, such as the creation of electron–positron pairs, the total charge of any closed system is exactly constant.  21.2 Conductors, Insulators, and Induced Charges 687 The second important principle is: 21.5 Most of the forces on this water skier are electric. Electric interactions between adjacent molecules give rise to the force of The magnitude of charge of the electron or proton is a natural unit of charge. the water on the ski, the tension in the tow rope, and the resistance of the air on the Every observable amount of electric charge is always an integer multiple of this skier’s body. Electric interactions also hold the atoms of the skier’s body together. basic unit. We say that charge is quantized. A familiar example of quantization Only one wholly nonelectric force acts on is money. When you pay cash for an item in a store, you have to do it in one-cent the skier: the force of gravity. increments. Cash can’t be divided into amounts smaller than one cent, and elec- tric charge can’t be divided into amounts smaller than the charge of one electron or proton. (The quark charges, ± 13 and ± 23 of the electron charge, are probably not observable as isolated charges.) Thus the charge on any macroscopic body is always zero or an integer multiple (negative or positive) of the electron charge. Understanding the electric nature of matter gives us insight into many aspects of the physical world (Fig. 21.5). The chemical bonds that hold atoms together to form molecules are due to electric interactions between the atoms. They include the strong ionic bonds that hold sodium and chlorine atoms together to make table salt and the relatively weak bonds between the strands of DNA that record your body’s genetic code. When you stand, the normal force exerted on you by the floor arises from electric forces between charged particles in the atoms of your shoes and the atoms of the floor. The tension force in a stretched string and the adhesive force of glue are likewise due to electric interactions of atoms. test Your understanding of section 21.1 Two charged objects repel each other through the electric force. The charges on the objects are (i) one positive and one negative; (ii) both positive; (iii) both negative; (iv) either (ii) or (iii); (v) any of (i), (ii), or (iii). ❙ 21.2 conductors, insuLators, and induced charges PhET: Balloons and Static Electricity PhET: John Travoltage Some materials permit electric charge to move easily from one region of the ma- terial to another, while others do not. For example, Fig. 21.6a (next page) shows a copper wire supported by a nylon thread. Suppose you touch one end of the wire to a charged plastic rod and attach the other end to a metal ball that is initially uncharged; you then remove the charged rod and the wire. When you bring another charged body up close to the ball (Figs. 21.6b and 21.6c), the ball is attracted or repelled, showing that the ball has become electrically charged. Electric charge has been transferred through the copper wire between the ball and the surface of the plastic rod. The copper wire is called a conductor of electricity. If you repeat the experi- ment using a rubber band or nylon thread in place of the wire, you find that no charge is transferred to the ball. These materials are called insulators. Conductors permit the easy movement of charge through them, while insulators do not. (The supporting nylon threads shown in Fig. 21.6 are insulators, which prevents charge from leaving the metal ball and copper wire.) As an example, carpet fibers on a dry day are good insulators. As you walk across a carpet, the rubbing of your shoes against the fibers causes charge to build up on you, and this charge remains on you because it can’t flow through the insulating fibers. If you then touch a conducting object such as a doorknob, a rapid charge transfer takes place between your finger and the doorknob, and you feel a shock. One way to prevent this is to wind some of the carpet fibers around conducting cores so that any charge that builds up on you can be transferred harmlessly to the carpet. Another solution is to coat the carpet fibers with an antistatic layer that does not easily transfer electrons to or from your shoes; this prevents any charge from building up on you in the first place. 688 Chapter 21 electric Charge and electric Field 21.6 Copper is a good conductor of elec- Most metals are good conductors, while most nonmetals are insulators. Within tricity; nylon is a good insulator. (a) The a solid metal such as copper, one or more outer electrons in each atom become copper wire conducts charge between the detached and can move freely throughout the material, just as the molecules of metal ball and the charged plastic rod to charge the ball negatively. Afterward, the a gas can move through the spaces between the grains in a bucket of sand. The metal ball is (b) repelled by a negatively other electrons remain bound to the positively charged nuclei, which themselves charged plastic rod and (c) attracted to a are bound in nearly fixed positions within the material. In an insulator there are positively charged glass rod. no, or very few, free electrons, and electric charge cannot move freely through (a) the material. Some materials called semiconductors are intermediate in their properties between good conductors and good insulators. Insulating nylon threads – Charged charging by induction – plastic rod – We can charge a metal ball by using a copper wire and an electrically charged – plastic rod, as in Fig. 21.6a. In this process, some of the excess electrons on the Metal Copper – ball wire rod are transferred from it to the ball, leaving the rod with a smaller negative charge. But there is a different technique in which the plastic rod can give another body a charge of opposite sign without losing any of its own charge. This process The wire conducts charge from the negatively is called charging by induction. charged plastic rod to the metal ball. Figure 21.7 shows an example of charging by induction. An uncharged metal (b) ball is supported on an insulating stand (Fig. 21.7a). When you bring a negatively charged rod near it, without actually touching it (Fig. 21.7b), the free electrons in the metal ball are repelled by the excess electrons on the rod, and they shift A negatively charged toward the right, away from the rod. They cannot escape from the ball because plastic rod now repels the supporting stand and the surrounding air are insulators. So we get excess the ball... – negative charge at the right surface of the ball and a deficiency of negative charge – – – – – (that is, a net positive charge) at the left surface. These excess charges are called Charged – induced charges. plastic rod Not all of the free electrons move to the right surface of the ball. As soon as any induced charge develops, it exerts forces toward the left on the other free electrons. These electrons are repelled by the negative induced charge on the right (c) and attracted toward the positive induced charge on the left. The system reaches an equilibrium state in which the force toward the right on an electron, due to... and a positively the charged rod, is just balanced by the force toward the left due to the induced charged glass rod charge. If we remove the charged rod, the free electrons shift back to the left, and attracts the ball. the original neutral condition is restored. – + + What happens if, while the plastic rod is nearby, you touch one end of a – + + conducting wire to the right surface of the ball and the other end to the earth Charged + (Fig. 21.7c)? The earth is a conductor, and it is so large that it can act as a prac- glass rod tically infinite source of extra electrons or sink of unwanted electrons. Some of the negative charge flows through the wire to the earth. Now suppose you disconnect the wire (Fig. 21.7d) and then remove the rod (Fig. 21.7e); a net posi- tive charge is left on the ball. The charge on the negatively charged rod has not changed during this process. The earth acquires a negative charge that is equal in magnitude to the induced positive charge remaining on the ball. 21.7 Charging a metal ball by induction. Electron Electron buildup Metal deficiency ++ –– ++ + ++ ball Negatively – + Wire charged – + – – – – ++ –– – ++ Negative ++ +– – Insulating rod – – –– –– – charge in stand ground – – – – – – – – Ground (a) Uncharged metal ball (b) Negative charge on rod (c) Wire lets electron build- (d) Wire removed; ball now (e) Rod removed; repels electrons, creating up (induced negative has only an electron- electrons rearrange zones of negative and charge) flow into deficient region of themselves, ball has positive induced charge. ground. positive charge. overall electron deficiency (net positive charge).  21.2 Conductors, Insulators, and Induced Charges 689 21.8 The charges within the molecules of an insulating material can shift slightly. As a result, a comb with either sign of charge attracts a neutral insulator. By Newton’s third law the neutral insulator exerts an equal-magnitude attractive force on the comb. (a) A charged comb picking up uncharged (b) How a negatively charged comb attracts an (c) How a positively charged comb attracts an pieces of plastic insulator insulator Electrons in each – + molecule of the neutral – – This time, electrons in + + + insulator shift away – –– –– the molecules shift + + ++ from the comb. – –– –– toward the comb... + ++++ –– – ++ + Positively – Negatively + S charged comb S charged comb F F +– + + – – + + – – +– + + S S −F As a result, the −F... so the – – + + + – – +– +– + + (-) charges in each – +– +– (+) charges in each +– + + – +– + +– + molecule are closer to molecule are closer to – – +– – – the comb than are the (-) the comb, and feel a – + – charges and so feel a stronger stronger force from it, than force from the comb. Therefore the (+) charges. Again, the net the net force is attractive. force is attractive. electric forces on uncharged objects Finally, we note that a charged body can exert forces even on objects that are not charged themselves. If you rub a balloon on the rug and then hold the balloon against the ceiling, it sticks, even though the ceiling has no net electric charge. After you electrify a comb by running it through your hair, you can pick up Demo uncharged bits of paper or plastic with it (Fig. 21.8a). How is this possible? This interaction is an induced-charge effect. Even in an insulator, electric charges can shift back and forth a little when there is charge nearby. This is shown in Fig. 21.8b; the negatively charged plastic comb causes a slight shift- ing of charge within the molecules of the neutral insulator, an effect called polarization. The positive and negative charges in the material are present in equal amounts, but the positive charges are closer to the plastic comb and so feel an attraction that is stronger than the repulsion felt by the negative charges, giving a net attractive force. (In Section 21.3 we will study how electric forces depend on distance.) Note that a neutral insulator is also attracted to a positively charged comb (Fig. 21.8c). Now the charges in the insulator shift in the opposite direction; the negative charges in the insulator are closer to the comb and feel an attractive force that is stronger than the repulsion felt by the positive charges in the insulator. Hence a charged object of either sign exerts an attractive force on an uncharged insulator. Figure 21.9 shows an industrial application of this effect. test Your understanding of section 21.2 You have two lightweight metal spheres, each hanging from an insulating nylon thread. One of the spheres has a net negative charge, while the other sphere has no net charge. (a) If the spheres are close to- gether but do not touch, will they (i) attract each other, (ii) repel each other, or (iii) exert no force on each other? (b) You now allow the two spheres to touch. Once they have touched, will the two spheres (i) attract each other, (ii) repel each other, or (iii) exert no force on each other? ❙ 21.9 The electrostatic painting process (compare Figs. 21.7b and 21.7c). A metal object to be painted is connected to the Metal object earth (“ground”), and the paint droplets are given an electric Spray of charge as they exit the sprayer nozzle. Induced charges of the to be painted negatively opposite sign appear in the object as the droplets approach, charged + + just as in Fig. 21.7b, and they attract the droplets to the sur- paint droplets – face. This process minimizes overspray from clouds of stray + Positive charge – – paint particles and gives a particularly smooth finish. – + is induced on – – – – surface of metal. – – + + Paint sprayer Ground 690 Chapter 21 electric Charge and electric Field BIO application electric forces, 21.3 couLoMb’s Law sweat, and cystic fibrosis One way to test for the genetic disease cystic fibrosis Charles Augustin de Coulomb (1736–1806) studied the interaction forces of (CF) is to measure the salt content of a person’s sweat. Sweat is a mixture of water and ions, charged particles in detail in 1784. He used a torsion balance (Fig. 21.10a) simi- including the sodium (Na+ ) and chloride (Cl- ) lar to the one used 13 years later by Cavendish to study the much weaker gravi- ions that make up ordinary salt (NaCl). When tational interaction, as we discussed in Section 13.1. For point charges, charged sweat is secreted by epithelial cells, some bodies that are very small in comparison with the distance r between them, of the Cl- ions flow from the sweat back into Coulomb found that the electric force is proportional to 1>r 2. That is, when the these cells (a process called reabsorption). The electric attraction between negative and distance r doubles, the force decreases to one-quarter of its initial value; when positive charges pulls Na+ ions along with the distance is halved, the force increases to four times its initial value. the Cl-. Water molecules cannot flow back The electric force between two point charges also depends on the quantity of into the epithelial cells, so sweat on the skin charge on each body, which we will denote by q or Q. To explore this depen- has a low salt content. However, in persons dence, Coulomb divided a charge into two equal parts by placing a small charged with CF the reabsorption of Cl- ions is blocked. Hence the sweat of persons with CF is unusu- spherical conductor into contact with an identical but uncharged sphere; by sym- ally salty, with up to four times the normal metry, the charge is shared equally between the two spheres. (Note the essential concentration of Cl- and Na+. role of the principle of conservation of charge in this procedure.) Thus he could obtain one-half, one-quarter, and so on, of any initial charge. He found that the forces that two point charges q1 and q2 exert on each other are proportional to each charge and therefore are proportional to the product q1 q2 of the two charges. Thus Coulomb established what we now call Coulomb’s law: The magnitude of the electric force between two point charges is directly pro- portional to the product of the charges and inversely proportional to the square of the distance between them. In mathematical terms, the magnitude F of the force that each of two point charges q1 and q2 a distance r apart exerts on the other can be expressed as 0 q1 q2 0 F = k (21.1) r2 where k is a proportionality constant whose numerical value depends on the sys- tem of units used. The absolute value bars are used in Eq. (21.1) because the charges q1 and q2 can be either positive or negative, while the force magnitude F is always positive. The directions of the forces the two charges exert on each other are always along the line joining them. When the charges q1 and q2 have the same sign, either both positive or both negative, the forces are repulsive; when the charges 21.10 (a) Measuring the electric force (a) A torsion balance of the type used by (b) Interactions between point charges between point charges. (b) The electric Coulomb to measure the electric force S forces between pointS charges obey S F2 on 1 Newton’s third law: F1 on 2 = −F2 on 1. Charges of the + r same sign repel. q1 The negatively charged ball attracts + S F1 on 2 the positively charged S S one; the positive ball F1 on 2 = −F2 on 1 q2 Torsion fiber moves until the elastic 0 q1q2 0 F1 on 2 = F2 on 1 = k forces in the torsion r2 fiber balance the Charges electrostatic attraction. of opposite – r sign attract. q1 S F2 on 1 Charged S F1 on 2 + pith balls q2 –+ + Scale  21.3 Coulomb’s Law 691 have opposite signs, the forces are attractive (Fig. 21.10b). The two forces obey Newton’s third law; they are always equal in magnitude and opposite in direc- tion, even when the charges are not equal in magnitude. The proportionality of the electric force to 1>r 2 has been verified with great precision. There is no reason to suspect that the exponent is different from pre- cisely 2. Thus the form of Eq. (21.1) is the same as that of the law of gravitation. But electric and gravitational interactions are two distinct classes of phenomena. Electric interactions depend on electric charges and can be either attractive or repulsive, while gravitational interactions depend on mass and are always attrac- tive (because there is no such thing as negative mass). fundamental electric constants The value of the proportionality constant k in Coulomb’s law depends on the system of units used. In our study of electricity and magnetism we will use SI units exclusively. The SI electric units include most of the familiar units such as the volt, the ampere, the ohm, and the watt. (There is no British system of electric units.) The SI unit of electric charge is called one coulomb (1 C). In SI units the constant k in Eq. (21.1) is k = 8.987551787 * 109 N # m2>C2 ≅ 8.988 * 109 N # m2>C2 The value of k is known to such a large number of significant figures because this value is closely related to the speed of light in vacuum. (We will show this in Chapter 32 when we study electromagnetic radiation.) As we discussed in Section 1.3, this speed is defined to be exactly c = 2.99792458 * 108 m>s. The numerical value of k is defined in terms of c to be precisely k = 110-7 N # s2>C22c2 You should check this expression to confirm that k has the right units. In principle we can measure the electric force F between two equal charges q at a measured distance r and use Coulomb’s law to determine the charge. Thus we could regard the value of k as an operational definition of the coulomb. For reasons of experimental precision it is better to define the coulomb instead in terms of a unit of electric current (charge per unit time), the ampere, equal to 1 coulomb per second. We will return to this definition in Chapter 28. In SI units we usually write the constant k in Eq. (21.1) as 1>4pP0 , where P0 (“epsilon-nought” or “epsilon-zero”) is called the electric constant. This shorthand simplifies many formulas that we will encounter in later chapters. From now on, we will usually write Coulomb’s law as Coulomb’s law: Values of the Magnitude of electric 1 0 q1q2 0 two charges F = (21.2) force between two 4pP0 r2 Distance between the point charges Electric constant two charges The constants in Eq. (21.2) are approximately P0 = 8.854 * 10-12 C2>N # m2 = k = 8.988 * 109 N # m2>C2 1 and 4pP0 In examples and problems we will often use the approximate value = 9.0 * 109 N # m2>C2 1 4pP0 As we mentioned in Section 21.1, the most fundamental unit of charge is the magnitude of the charge of an electron or a proton, which is denoted by e. The most precise value available as of the writing of this book is e = 1.6021765651352 * 10-19 C 692 Chapter 21 electric Charge and electric Field One coulomb represents the negative of the total charge of about 6 * 1018 elec- trons. For comparison, a copper cube 1 cm on a side contains about 2.4 * 1024 electrons. About 1019 electrons pass through the glowing filament of a flashlight bulb every second. In electrostatics problems (problems that involve charges at rest), it’s very unusual to encounter charges as large as 1 coulomb. Two 1-C charges sepa- rated by 1 m would exert forces on each other of magnitude 9 * 109 N (about 1 million tons)! The total charge of all the electrons in a copper one-cent coin is even greater, about 1.4 * 105 C, which shows that we can’t disturb electric neu- trality very much without using enormous forces. More typical values of charge range from about a microcoulomb 11 mC = 10-6 C2 to about a nanocoulomb 11 nC = 10-9 C2. Solution ExamplE 21.1 ElEctric forcE vErsus gravitational forcE An a particle (the nucleus of a helium atom) has mass m = 21.11 Our sketch for this problem. 6.64 * 10-27 kg and charge q = + 2e = 3.2 * 10-19 C. Com- pare the magnitude of the electric repulsion between two a (“alpha”) particles with that of the gravitational attraction be- tween them. soLution identifY and set up: This problem involves Newton’s law for Fe q2 1 the gravitational force Fg between particles (see Section 13.1) and = Fg 4pP0G m2 Coulomb’s law for the electric force Fe between point charges. To compare these forces, we make our target variable the ratio Fe >Fg. 9.0 * 109 N # m2>C2 13.2 * 10-19 C22 = = 3.1 * 1035 6.67 * 10-11 N # m2>kg 2 16.64 * 10-27 kg22 We use Eq. (21.2) for Fe and Eq. (13.1) for Fg. execute: Figure 21.11 shows our sketch. From Eqs. (21.2) and (13.1), evaLuate: This astonishingly large number shows that the gravi- tational force in this situation is completely negligible in compar- 1 q2 m2 ison to the electric force. This is always true for interactions of Fe = Fg = G atomic and subnuclear particles. But within objects the size of a 4pP0 r 2 r2 person or a planet, the positive and negative charges are nearly These are both inverse-square forces, so the r 2 factors cancel equal in magnitude, and the net electric force is usually much when we take the ratio: smaller than the gravitational force. superposition of forces Coulomb’s law as we have stated it describes only the interaction of two point charges. Experiments show that when two charges exert forces simultaneously on a third charge, the total force acting on that charge is the vector sum of the forces that the two charges would exert individually. This important property, called the principle of superposition of forces, holds for any number of charges. By using this principle, we can apply Coulomb’s law to any collection of charges. Two of the examples at the end of this section use the superposition principle. Strictly speaking, Coulomb’s law as we have stated it should be used only for point charges in a vacuum. If matter is present in the space between the charges, the net force acting on each charge is altered because charges are induced in the molecules of the intervening material. We will describe this effect later. As a practical matter, though, we can use Coulomb’s law unaltered for point charges in air. At normal atmospheric pressure, the presence of air changes the electric force from its vacuum value by only about one part in 2000.  21.3 Coulomb’s Law 693 problEm-solving stratEgy 21.1 coulomb’s law identifY the relevant concepts: Coulomb’s law describes the 4. Use consistent units; SI units are completely consistent. With electric force between charged particles. 1>4pP0 = 9.0 * 109 N # m2>C2, distances must be in meters, charges in coulombs, and forces in newtons. set up the problem using the following steps: 5. Some examples and problems in this and later chapters involve 1. Sketch the locations of the charged particles and label each continuous distributions of charge along a line, over a surface, particle with its charge. or throughout a volume. In these cases the vector sum in step 3 2. If the charges do not all lie on a single line, set up an xy- becomes a vector integral. We divide the charge distribution coordinate system. into infinitesimal pieces, use Coulomb’s law for each piece, and 3. The problem will ask you to find the electric force on one or integrate to find the vector sum. Sometimes this can be done more particles. Identify which these are. without actual integration. 6. Exploit any symmetries in the charge distribution to simplify execute the solution as follows: your problem solving. For example, two identical charges q 1. For each particle that exerts an electric force on a given particle exert zero net electric force on a charge Q midway between of interest, use Eq. (21.2) to calculate the magnitude of that them, because the forces on Q have equal magnitude and oppo- force. site direction. 2. Using those magnitudes, sketch a free-body diagram showing the electric-force vectors acting on each particle of interest. The evaLuate your answer: Check whether your numerical results force exerted by particle 1 on particle 2 points from particle 2 are reasonable. Confirm that the direction of the net electric force toward particle 1 if the charges have opposite signs, but points agrees with the principle that charges of the same sign repel and from particle 2 directly away from particle 1 if the charges have charges of opposite sign attract. the same sign. 3. Use the principle of superposition to calculate the total electric force—a vector sum—on each particle of interest. (Review the vector algebra in Sections 1.7 through 1.9. The method of com- ponents is often helpful.) Solution ExamplE 21.2 forcE bEtwEEn two point chargEs Two point charges, q1 = + 25 nC and q2 = - 75 nC, are sepa- 21.12 What force does q1 exert on q2 , and what force does q2 rated by a distance r = 3.0 cm (Fig. 21.12a). Find the magnitude exert on q1? Gravitational forces are negligible. and direction of the electric force (a) that q1 exerts on q2 and (a) The two charges (b) Free-body diagram (c) Free-body diagram (b) that q2 exerts on q1. for charge q2 for charge q1 q1 q2 S F1 on 2 q2 q1 S soLution F2 on 1 identifY and set up: This problem asks for the electric forces r that two charges exert on each other. We use Coulomb’s law, Eq. (21.2), to calculate the magnitudes of the forces. The signs of the charges will determine the directions of the forces. execute: (a) After converting the units of r to meters and the (b) Proceeding as in part (a), we have units of q1 and q2 to coulombs, Eq. (21.2) gives us 1 0 q2 q1 0 1 0 q1 q2 0 F2 on 1 = = F1 on 2 = 0.019 N F1 on 2 = 4pP0 r 2 4pP0 r 2 0 1+ 25 * 10-9 C21-75 * 10-9 C2 0 The attractive force that acts on q1 is to the right, toward q2 = 19.0 * 109 N # m2>C22 (Fig. 21.12c). 10.030 m22 evaLuate: Newton’s third law applies to the electric force. Even = 0.019 N though the charges have different magnitudes, the magnitude of The charges have opposite signs, so the force is attractive (to the the force that q2 exerts on q1 is the same as the magnitude of the left in Fig. 21.12b); that is, the force that acts on q2 is directed force that q1 exerts on q2, and these two forces are in opposite toward q1 along the line joining the two charges. directions. 694 Chapter 21 electric Charge and electric Field Solution ExamplE 21.3 vEctor addition of ElEctric forcEs on a linE Two point charges are located on the x-axis of a coordinate system: 21.13 Our sketches for this problem. q1 = 1.0 nC is at x = + 2.0 cm, and q2 = -3.0 nC is at x = (a) Our diagram of the situation (b) Free-body diagram for q3 + 4.0 cm. What is the total electric force exerted by q1 and q2 on a charge q3 = 5.0 nC at x = 0? soLution identifY and set up: Figure 21.13a shows the situation. To find the total force on q3 , our target variable, we find the vector sum of the two electric forces on it. execute: Figure 21.13b is a free-body diagram for q3 , which In theS same way you can showS that F2 on 3 = 84 mN. We thus is repelled by q1 (which has the same sign) and attracted to q2 have F1 on 3 = 1- 112 mN2dn and F2 on 3 = 184 mN2dn. The net force S (which has the opposite sign): F1 on 3 is in the - x-direction and on q3 is S F2 on 3 is in the + x-direction. After unit conversions, we have from S S S F3 = F1 on 3 + F2 on 3 = 1- 112 mN2dn + 184 mN2dn = 1- 28 mN2dn Eq. (21.2) evaLuate: As a check, note that the magnitude of q2 is three 1 0 q1q3 0 F1 on 3 = times that of q1, but q2 is twice as far from q3 as q1. Equation (21.2) 4pP0 r 132 then says that F2 on 3 must be 3>22 = 3>4 = 0.75 as large as 11.0 * 10-9 C215.0 * 10-9 C2 F1 on 3. This agrees with our calculated values: F2 on 3>F1 on 3 = = 19.0 * 109 N # m2>C22 184 mN2>1112 mN2 = 0.75. Because FS2 on 3 is the weaker force, 10.020 m22 the direction of the net force is that of F1 on 3—that is, in the nega- = 1.12 * 10-4 N = 112 mN tive x-direction. Solution ExamplE 21.4 vEctor addition of ElEctric forcEs in a planE S S Two equal positive charges q1 = q2 = 2.0 mC are located at execute: Figure 21.14 shows the forces F1 on Q and F2 on Q due to x = 0, y = 0.30 m and x = 0, y = - 0.30 m, respectively. What the identical charges q1 and q2 , which are at equal distances are the magnitude and direction of the total electric force that q1 from Q. From Coulomb’s law, both forces have magnitude and q2 exert on a third charge Q = 4.0 mC at x = 0.40 m, y = 0? F1 or 2 on Q = 19.0 * 109 N # m2>C22 soLution 14.0 * 10-6 C212.0 * 10-6 C2 * = 0.29 N identifY and set up: As in Example 21.3, we must compute the 10.50 m22 force that each charge exerts on Q and then find the vector sum of those forces. Figure 21.14 shows the situation. Since the three The x-components of the two forces are equal: charges do not all lie on a line, the best way to calculate the forces 0.40 m is to use components. 1F1 or 2 on Q2x = 1F1 or 2 on Q2cos a = 10.29 N2 = 0.23 N 0.50 m From symmetry we see that the y-components of the two forces are S 21.14 Our sketch for this problem. equal and opposite. Hence their sum is zero and the total force F on Q has only an x-component Fx = 0.23 N + 0.23 N = 0.46 N. The total force on Q is in the +x-direction, with magnitude 0.46 N. evaLuate: The total force on Q points neither directly away from q1 nor directly away from q2. Rather, this direction is a compro- mise that points away from the system of charges q1 and q2. Can you see that the total force would not be in the + x-direction if q1 and q2 were not equal or if the geometrical arrangement of the changes were not so symmetric?  21.4 electric Field and electric Forces 695 test Your understanding of section 21.3 Suppose that charge q2 in 21.15 A charged body creates an electric Example 21.4 were - 2.0 mC. In this case, the total electric force on Q would be field in the space around it. (i) in the positive x-direction; (ii) in the negative x-direction; (iii) in the positive (a) A and B exert electric forces on each other. y-direction; (iv) in the negative y-direction; (v) zero; (vi) none of these. ❙ S q0 S −F0 F0 B 21.4 eLectric fieLd and eLectric forces A When two electrically charged particles in empty space interact, how does each (b) Remove body B... one know the other is there? We can begin to answer this question, and at the... and label its former position as P. same time reformulate Coulomb’s law in a very useful way, by using the concept of electric field. P A electric field S (c) Body A sets up an electric field E at point P. To introduce this concept, let’s look at the mutual repulsion of two positively S charged bodies A and B (Fig. 21.15a). Suppose B has charge q0 , and let F0 be Test charge q0 the electric force of A on B. One way to think about this force is as an “action- at-a-distance” force—that is, as a force that acts across empty space without S F0 S needing physical contact between A and B. (Gravity can also be thought of as an E= q A 0 “action-at-a-distance” force.) But a more fruitful way to visualize the repulsion S between A and B is as a two-stage process. We first envision that body A, as a E is the force per unit charge exerted by A result of the charge that it carries, somehow modifies the properties of the space on a test charge at P. around it. Then body B, as a result of the charge that it carries, senses how space has Sbeen modified at its position. The response of body B is to experience the force F0. To clarify how this two-stage process occurs, we first consider body A by itself: We remove body B and label its former position as point P (Fig. 21.15b). BIO application sharks and the We say that the charged body A produces or causes an electric field at point P “sixth sense” Sharks have the ability to (and at all other points in the neighborhood). This electric field is present at P even locate prey (such as flounder and other if there is no charge at P; it is a consequence of the charge on body SA only. If a bottom-dwelling fish) that are completely point charge q0 is then placed at point P, it experiences the force F0. We take hidden beneath the sand at the bottom of the ocean. They do this by sensing the weak elec- the point of view that this force is exerted on q0 by the field at P (Fig. 21.15c). tric fields produced by muscle contractions in Thus the electric field is the intermediary through which A communicates its their prey. Sharks derive their sensitivity to presence to q0. Because the point charge q0 would experience a force at any point electric fields (a “sixth sense”) from jelly-filled in the neighborhood of A, the electric field that A produces exists at all points in canals in their bodies. These canals end in the region around A. pores on the shark’s skin (shown in this photograph). An electric field as weak as We can likewise say that the point charge q0 produces an electric S field in the 5 * 10-7 N>C causes charge flow within the space around it and that this electric field exerts the force −F0 on body A. For canals and triggers a signal in the shark’s each force (the force of A on q0 and the force of q0 on A), one charge sets up an nervous system. Because the shark has canals electric field that exerts a force on the second charge. We emphasize that this is with different orientations, it can measure an interaction between two charged bodies. A single charge produces an electric different components of the electric-field vector and hence determine the direction of field in the surrounding space, but this electric field cannot exert a net force on the field. the charge that created it; as we discussed in Section 4.3, a body cannot exert a net force on itself. (If this wasn’t true, you would be able to lift yourself to the ceiling by pulling up on your belt!) The electric force on a charged body is exerted by the electric field created by other charged bodies. To find out experimentally whether there is an electric field at a particular point, we place a small charged body, which we call a test charge, at the point (Fig. 21.15c). If the test charge experiences an electric force, then there is an electric field at that point. This field is produced by charges other than q0. Force is a vector quantity, so electric field is also a vector quantity. (Note the use of vector signs as well as boldface letters and plus, minus, S and equals signs in the following discussion.) We define the electric field E at a point as the electric 696 Chapter 21 electric Charge and electric Field S force F0 experienced by a test charge q0 at the point, divided by the charge q0. That is, the electric field at a certain point is equal to the electric force per unit charge experienced by a charge at that point: S Electric force Electric field = S F0 on a test charge q0 electric force E= due to other charges (21.3) per unit charge q0 Value of test charge In SI units, in which the unit of force is 1 N and the unit of charge is 1 C, the unit S S of electric-field Smagnitude is 1 newton per coulomb 11 N>C2. 21.16 The force F0 = q0 E exerted on a S If the field E at a certain point is known, rearranging Eq. (21.3) gives the S point charge q0 placed in an electric field E. force F experienced by a point charge q placed at that point. This force is just 0 S 0 Q S equal to the electric field E produced at that point by charges other than q0 , E (due to charge Q) multiplied by the charge q0 : q0 S F0 S S (force exerted on a point charge q0 F0 = q0 E S (21.4) The force on a positive test charge q0 points by an electric field E) in the direction of the electric field. S The charge q0 can be either positive or negative. If qS0 is positive, the force S F0 ex- S Q perienced by the charge is in the same direction as E ; if q0 is negative, F0 and E S S F0 E (due to charge Q) are in opposite directions (Fig. 21.16). q0 While the electric field concept may be new to you, the basic idea—that one body sets up a field in the space around it and a second body responds to that The force on a negative test charge q0 points opposite to the electric field. field—is one that you’ve actually usedS before. Compare Eq. (21.4) to the familiar expression for the gravitational force Fg that the earth exerts on a mass m 0 : S S Fg = m 0 g (21.5) S In this expression, g is the acceleration due to gravity. If we divide both sides of Eq. (21.5) by the mass m 0 , we obtain S S Fg g= m0 S Thus g can be regarded as the gravitational force per unit mass. By analogy to S Eq. (21.3), we can interpret g as the gravitational field. Thus we treat the gravita- tional interaction between the earth and the mass m 0 as a two-stage process: The S earth sets up a gravitational field g in the space around it, and this gravitational field exerts a force given by Eq. (21.5) on the mass m 0 (which we can regard as S a test mass). The gravitational field g, or gravitational force per unit mass, is a useful concept because it does not depend on the mass of Sthe body on which the gravitational force is exerted; likewise, the electric field E, or electric force per unit charge, is useful because it does not depend on the charge of the body on which the electric force is exerted. S S Caution F0 = q0 E is for point test charges only The electric force experienced by a test charge q0 can vary from point to point, so the electric field can also be different at different points. For this reason, use Eq. (21.4) to find the electricS force on a point charge only. If a charged body is large enough in size, the electric field E may be noticeably dif- ferent in magnitude and direction at different points on the body, and calculating the net electric force on it can be complicated. ❙ Electric Field of a Point Charge If the source distribution is a point charge q, it is easy to find the electric field that it produces. We call the location of the charge the source point, and we call the point P where we are determining the field the field point. It is also useful to introduce a unit vector nr that points along the line from source point  21.4 electric Field and electric Forces 697 S 21.17 S The electric field E produced at point P by an isolated point charge q at S. Note that in both (b) and (c), E is produced by q [see Eq. (21.7)] but acts on the charge q0 at point P [see Eq. (21.4)]. S (a) (b) E (c) q0 q0 q0 S E P P P q nr q nr q nr r At each point P, the electric field set up by an isolated positive point At each point P, the electric S Unit vector nr points from S S field set up by an isolated negative point source point S to field point P. charge q points directly away from the charge in the same direction as nr. charge q points directly toward the charge in the opposite direction from nr. S to field point (Fig. 21.17a). This unit vector is equal to the displacement vector r from the source point to the field point, divided by the distance r = 0 r 0 between S S these two points; that is, nr = r >r. If we place a small test charge q0 at the field point P, at a distance r from the source point, the magnitude F0 of the force is given by Coulomb’s law, Eq. (21.2): 1 0 qq0 0 F0 = 4pP0 r 2 From Eq. (21.3) the magnitude E of the electric field at P is 1 0q0 E = (magnitude of electric field of a point charge) (21.6) 4pP0 r 2 21.18 A point S charge q produces an Using the unit vector nr, we can write a Svector equation that gives both the magni- electric field E at all points in space. The field strength decreases with increasing tude and direction of the electric field E: distance. Value of point charge (a) The field produced by a positive point Unit vector from point charge Electric field charge points away from the charge. S 1 q toward where field is measured due to a E= nr (21.7) point charge 4pP0 r 2 Distance from point charge Electric constant to where field is measured S E By definition, the electric field of a point charge always points away from a posi- tive charge (that is, in the same direction as nr ; see Fig. 21.17b) but toward a negative charge (that is, in the direction opposite nr ; see SFig. 21.17c).