Electric Fields CHAPTER KEY CONCEPTS. What Effect Does the Electric Force Have on the Motion of Charges?

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1 CHAPTER 7 Electric Fields KEY CONCEPTS After completing this chapter you will be able to describe properties of electric charges and the electric force, and describe how electric forces affect the motion of charges describe properties of electric fi elds solve problems related to electric force and electric fi elds analyze the operation of technol ogies that use electric fi elds and assess their social and environmental impact solve problems related to electric potential and electric potential energy conduct laboratory inquiries to examine the behaviour of particles in a fi eld What Effect Does the Electric Force Have on the Motion of Charges? Although physicists have answered many questions about our world, many puzzles still remain. For one thing, researchers still do not fully understand the phenomenon of lightning. We do know that lightning is an electrical effect in which an imbalance of electric charge forms in storm clouds and suddenly causes a discharge to try to cancel the imbalance. The violent energy of the storm somehow causes this charge imbalance, but researchers do not yet understand the exact cause. They also do not yet know how to predict when and where lightning will strike. Researchers do know that when the charge imbalance becomes great enough, a current runs from cloud to ground, cloud to cloud, or even ground to cloud. In fact, a rare form of lightning called ball lightning forms as a sphere of charge that can travel along the ground with a ghostly, dangerous glow. A bolt of lightning can provide up to a million times the voltage of a normal wall receptacle. The flow of charge can heat the air around a lightning bolt to seven times the temperature of the surface of the Sun. This sudden immense heating causes the air to expand rapidly, leading to the sound wave we know as thunder. Astrophysicists have recently discovered another surprising phenomenon. The imbalance of electric charge in storm clouds can cause a burst of antimatter! Antimatter is the opposite of regular matter in some specific properties such as charge. For example, the antimatter version of an electron is called the positron. A positron has all of the properties of an electron except that it is positively charged. The storm releases electrons upward from the top of the clouds, where the electrons bump into air molecules. The collision leads to bursts of particles high above the clouds, including positrons. This phenomenon is depicted in an artist s rendering on the facing page. When antimatter and regular matter collide, they annihilate each other to produce a significant amount of energy. You will learn more about antimatter in Unit 5. To understand this effect and other puzzles of lightning, we must understand the electric force and how it affects electrons and other charged particles. You will be learning about the properties of electric charges and electric fields in this chapter. STARTING POINTS Answer the following questions using your current knowledge. You will have a chance to revisit these questions later, applying concepts and skills from the chapter. 1. What might cause an imbalance of charges inside an object? 2. Can the path of an electron be directed? 3. How do electric fi elds compare to gravitational fi elds? 4. Is it possible to do work with electric charges? 5. What are some applications of electric fi elds? 318 Chapter 7 Electric Fields NEL

2 Mini Investigation The Van de Graaff Generator Skills: Performing, Observing, Analyzing, Communicating Your teacher will fi rst demonstrate how a Van de Graaff generator works. You will then have a chance to touch the generator and observe its effect on your hair. The Van de Graaff generator produces a very high voltage and can give you a dangerous shock, especially if you have a heart condition. Remove metal jewellery. Follow your teacher s instructions. Equipment and Materials: Van de Graaff generator; grounding electrode; tall wooden stool 1. Observe as your teacher turns on the Van de Graaff generator and slowly brings the grounding electrode close to the generator sphere. Record your observations. 2. Your teacher will discharge the generator before having you touch it and then turning it on. Sit on the stool, which is insulated from the ground, and touch the generator SKILLS HANDBOOK A2.1 with both hands. Do not let go when the generator is turned on. Have your classmates observe what happens to your hair during the demonstration. Note any sensations that you feel when you touch the generator. The generator must be discharged before you let go. Observe what happens when other students touch the generator. Record your observations. A. Describe what happened when your teacher brought the grounding electrode near the Van de Graaff generator. K/U B. Suppose the spark represented a lightning strike. Explain how to apply what you observed to protecting buildings from lightning damage. T/I A C. Describe what you observed in Step 2 when you were touching the generator. Describe what you observed in Step 2 while watching your classmates. Did you note any differences in the effect on students with long hair compared to the effect on students with short hair? K/U C A NEL Introduction 319

3 7.1 Properties of Electric Charge A visit to a science museum can be, literally, a hair-raising experience. In Figure 1, the device that the child is touching is a Van de Graaff generator, which produces a large amount of electric charge. When you touch the generator, the electric charge spreads along your skin and onto each individual hair. The charges repel each other, so the hairs spread out and stand on end. How do electric charges move from one place to another and charge an object? In this section, you will learn the answers to this and other questions related to the properties of electric charges. Electric Charge Figure 1 Electric charges on each strand of hair exert a repulsive force on the other strands, causing the hair to rise and spread out. About 2500 years ago, Greek scholars first reported that rubbing amber with a piece of animal fur caused the amber to attract dust particles. You can demonstrate this effect with modern materials, such as a plastic rod and small bits of paper (Figure 2). Note that neither the fur nor the rod attracts the paper pieces under normal conditions, but rubbing the two materials together seems to create an attractive force on each object. An even more remarkable feature is that the rod attracts the pieces of paper without making direct contact with them. (a) (b) Figure 2 When a plastic rod is rubbed with fur (a), the rod acquires an electric charge. (b) The charged rod attracts small bits of paper and other objects. By the early 1900s, physicists had identified the subatomic particles called the electron and the proton as the basic units of charge. All protons carry the same amount of positive charge, e, and all electrons carry an equal but opposite charge, 2e. Charges interact with each other in very specific ways governed by the law of electric charges (Figure 3). Law of Electric Charges Like charges repel each other; unlike charges attract. Like charges repel. Unlike charges attract. Figure 3 The electric force between two like charges (charges with the same sign) is repulsive, whereas the electric force between unlike charges (charges with opposite signs) is attractive. An atom or a molecule is considered to have a charge of zero because the number of protons is equal to the number of electrons. Atoms and molecules can become charged to form ions. Ions can be positively or negatively charged. In a positive ion, or a cation, the number of protons must be greater than the number of electrons. Electrons are removed from an atom or a molecule to become a cation. Conversely, in a negative ion, or an anion, the number of electrons must be greater than the number of protons. Electrons are added to an atom or a molecule to make an anion. 320 Chapter 7 Electric Fields NEL

4 When looking at objects larger than atoms and molecules, we consider the total charge. The total charge on an object is the sum of all the charges in that object and can be positive, negative, or zero, just as in the case for atoms and molecules. A deficit of electrons means that the object is positively charged, and an excess of electrons means that the object is negatively charged. An object is said to have a total charge of zero when the number of negative charges equals the number of positive charges. An object with a charge of zero is said to be neutral. In addition, we know that charge can move from place to place, and from one object to another, but the total charge of the universe does not change. This is stated clearly in the law of conservation of charge. Law of Conservation of Charge Charge can be transferred from one object to another, but the total charge of a closed system remains constant. What Is Electric Charge? In the SI system, the basic unit of charge is called the coulomb (C), in honour of the French physicist Charles de Coulomb ( ). The charge of a single electron, 2e, is C, and the charge of a single proton, 1e, is C. The symbol e often denotes the magnitude of the charge on an electron or a proton. In this text, e will have the positive quantity ( C). The symbol q denotes the amount of charge, such as the total charge on a small piece of paper. To say that a particle has a charge q or that it carries a charge q is simply a way of saying that the total charge of the particle is q. For example, an alpha particle is a helium nucleus (a helium atom with no electrons). The helium nucleus has two protons, so its charge is 2e, or C ( C). coulomb the SI unit of electric charge; symbol C Conductors and Insulators To understand how researchers observe electric forces, you need to understand the various ways that charge can be transferred from one material to another. Conductors In most metals, each individual atom is electrically neutral, with equal numbers of protons and electrons. When these neutral atoms come together to form a large piece of metal, one or more electrons from each atom are able to escape from the parent atom and move freely through the entire piece of metal. These electrons are called conduction electrons, and the metal is a conductor. A conductor is a substance in which electrons can move easily among atoms. Copper is a good example of a conductor. The conduction electrons leave behind positively charged ions, which are bound in place and do not move. By adding or removing electrons from a conductor, the conductor can acquire a net negative or positive charge. Insulators On the other hand, electrons in insulators cannot move freely from atom to atom or escape from the molecules, so no conduction electrons are available to carry charge through the solid. Extra electrons placed on the surface of an insulator do not move about freely. Instead, these added electrons stay where they were initially placed. Examples of insulators are amber, plastic, and quartz. conductor any substance in which electrons are able to move easily from one atom to another insulator any substance in which electrons are not free to move easily from one atom to another NEL 7.1 Properties of Electric Charge 321

5 Conductivity of Liquids and Gases In the case of liquids, the atoms and molecules are free to move throughout the substance. Certain substances, such as table salt (NaCl), dissolve in water to form separately charged atoms, or ions. The sodium ion (Na 1 ) and the chloride ion (Cl 2 ) provide an electric charge in the salt water solution. This property of salt water makes it a good conductor, which is especially important for marine animals that rely on bioelectricity for survival. The conductivity of gases is similar to that of liquids. The constituent atoms and molecules of a gas are mostly neutral, but a few are present as ions and are able to carry charge from place to place. Some free electrons are usually present in a gas as well. Placing Charge on an Insulator To understand the electrical behaviour of an object, you must understand what happens when you add or remove charge from the object. Take, for instance, the case of an insulator such as quartz, when it is dry and surrounded by dry air. If you place a few electrons at a particular location on this insulator s surface, these electrons will stay in that location (Figure 4(a)). This is because there are no free electrons in an insulator, and the insulator does not allow the extra electrons to move about easily. In reality, excess charge will not stay on an insulator indefinitely. If an insulator contains some excess electrons, they will attract the stray positive ions that are usually part of the surrounding air. These stray ions will combine with the electrons on the insulator and cause the net charge on the insulator to become zero. Placing Charge on a Conductor Metals are excellent conductors because electrons can move easily through a metal. We have to refer to the law of electric charges to understand what happens when an excess of electrons is present in a conductor. Electrons naturally repel each other. In a metal object, the electrons are free to move from atom to atom, but they will never end up moving toward other electrons. The electrons do not concentrate at the centre of the piece of metal; rather, they move as far away from one another as possible. Also, the electrons do not spontaneously leave the metal to get away from each other. There are attractive forces from the protons in the atoms preventing the electrons from leaving. Since the electrons must repel each other while staying within the metal, excess electrons on a piece of metal must all spread evenly on the metal s surface (Figure 4(b)). (a) insulator conductor (b) Figure 4 (a) When excess electrons are placed on an insulator, they generally stay where they are placed for a period of time. (b) Excess electrons placed on a conductor, however, spread out on the surface of the material immediately. This does not mean that there are no charges within the metal. The interior of a neutral, uncharged metal contains equal numbers of positive and negative charges. These charges are present inside the metal at all times. Only the excess charge resides at the metal s surface. In Figure 4(b), excess charge on the metal surface distributes as negative charge, by adding electrons to the conductor, or as positive charge, by removing electrons from the conductor. 322 Chapter 7 Electric Fields NEL

6 Charging an Object by Friction Electricity was discovered when a piece of amber was rubbed with fur. The act of rubbing caused electrons to move from the fur to the amber. The amber thus acquired a net negative charge (an excess of electrons), whereas the fur was left with a net positive charge (an excess of positive ions). This process is called charging by friction. You can also charge a glass rod by rubbing it with a silk cloth, in which case electrons leave the glass and move to the silk, so the glass acquires a net positive charge. Through investigation, scientists have found that some materials have a stronger hold on their electrons than others. For example, glass has a weaker hold on its electrons than does silk. If you were to let glass and silk touch each other, some electrons would leave the glass and travel to the silk. Rubbing allows more surface to come into contact, and the friction it produces rips more electrons off the glass to go to the silk. Figure 5 shows the electrostatic series, which indicates the relative hold on electrons that different materials have when being charged by friction. acetate glass wool cat fur, human hair calcium, lead silk aluminum cotton paraffin wax ebonite polyethylene (plastic) carbon, copper, nickel sulfur platinum, gold weak hold on electrons increasing tendency to gain electrons strong hold on electrons Figure 5 The electrostatic series Charging an Object by Induced Charge Separation If you look back at the example of using a charged plastic rod to attract pieces of paper, you may notice that something seems wrong. You now know that rubbing the rod with fur charges the rod. However, the pieces of paper start out neutral with a total charge of zero. How is the neutral paper attracted to the charged rod? Although the paper is electrically neutral, the presence of the rod nearby causes some slight movement of the charges in the paper. The negatively charged rod repels electrons in the paper, which are then redistributed throughout the material. The electrons thus move a short distance away from the rod, as shown in Figure 6. A net positive charge remains on the portion of the paper nearest the rod, and a net negative charge stays on the paper opposite the rod. This process is called charging by induced charge separation. Figure 6 The negative charges in the paper have been repelled by the negative charges in the rod, leaving the area of the paper nearest to the rod positively charged and causing attraction between the paper and the rod. NEL 7.1 Properties of Electric Charge 323

7 Unlike charges attract, and the positive side of the paper is closer than the negative side of the paper to the negatively charged rod. This makes the paper attractive, overall, to the charged rod. In this way, an electric force pulls on an electrically neutral object once the object has undergone induced charge separation. Another example of induced charge separation is shown in Figure 7. Ordinarily, the stream of water flows straight downward due to the force of gravity. However, if you place a charged balloon near the stream, it exerts a deflecting force on the water. The charged balloon induces a charge separation in the water molecules, producing an attractive force similar to that produced by the charged rod on the pieces of paper. Figure 7 A stream of water is defl ected by a nearby charged balloon. Grounding Suppose you place a charged rubber rod on a table. If you watch carefully with sensitive electronic measuring equipment, you will find that the excess electrons originally on the rod move to the table and then to other adjacent objects. Grounding occurs whenever any charge imbalance is cancelled out by either sending excess electrons into the ground or moving excess electrons from the ground into the object. Grounding works because Earth is so large that any extra electrons going into or out of the ground have an insignificant effect on Earth. The concept of grounding is used by electrical engineers to ensure that buildings and appliances are safe. Ground wires direct excess charges away from users, protecting them from electric shock. CAREER LINK Charging by Contact and by Induction How would you transfer some of the electrons on a negatively charged rubber rod to a piece of metal? You could do this by simply touching the rubber rod to the metal. The rod contains an excess of electrons, and some of these electrons will move to the metal upon contact (Figure 8). This is an example of charging by contact, or charging by conduction. before contact contact after contact Figure 8 If one charged object touches a second object, the second object will usually acquire some of the excess charge. Hence, the second object is charged by contact. The stand in this illustration is an insulating stand. 324 Chapter 7 Electric Fields NEL

8 Now suppose you want to give the metal a net positive charge using the same negatively charged rubber rod. This might seem an impossible task, but an approach called charging by induction will accomplish it. This approach uses the properties of induced charge separation and an electrical ground. First, bring the negatively charged rod near the metal. This polarizes the conductor by moving conduction electrons to the side opposite the rod and leaving the side of the metal near the rod with a net positive charge (Figure 9(a)). Now, connect the negatively charged portion of the metal to an electrical ground using a metal wire. This allows electrons to move even farther from the charged rod by travelling into the electrical ground region (Figure 9(b)). Finally, after removing the grounding wire, the original piece of metal has a net positive charge (Figure 9(c)). Notice that no positive charges are transferred to the metal; instead, electrons are removed and the charge left on the metal is positive. charged rod ground (a) (b) (c) Figure 9 An object can be charged by the process of induction. (a) The object is fi rst brought near a charged rod, separating the charges on the object. (b) The object is then connected to a ground; some electrons fl ow between it and the ground. (c) The object is left with an opposite excess charge when it is disconnected from the ground. Mini Investigation Observing Electric Charge Skills: Controlling Variables, Performing, Observing, Analyzing, Communicating SKILLS HANDBOOK A2.1 In this investigation, you will explore how different objects become charged. You will draw conclusions based on how these charged objects affect small pieces of paper. Equipment and Materials: plastic pen; pencil; glass stirring rod; tissue paper; fabric 1. Tear a piece of tissue paper into several small pieces and gather them into a pile. 2. Charge the plastic pen by rubbing it on a piece of fabric, such as part of an old fl annel shirt. 3. Bring the charged object near the pieces of paper. 4. Observe how the pieces of paper behave relative to the charged object used, and record your observations. 5. Gather the pieces of paper into a pile again, and repeat the steps, fi rst using the pencil, and then using the glass rod. Record your observations. A. Why does each charged object attract the pieces of paper differently? K/U T/I B. Why do some pieces of paper fall off your charged objects after a short while? K/U T/I C. If you used a metal sphere with a large charge, the pieces of paper would jump off instead of falling. Explain why this occurs. K/U T/I NEL 7.1 Properties of Electric Charge 325

9 7.1 Review Summary Electric charges are either positive or negative. Objects that have more negative charge than positive charge are negatively charged. Objects that have more positive charge than negative charge are positively charged. Objects that have an equal number of positive and negative charges are neutral and have a total charge of zero. Electrons are the only subatomic particle capable of being transferred from one object to another. Protons are bound to the atomic nucleus. The law of electric charges states the following: Unlike, or opposite, electric charges attract each other. Like, or similar, electric charges repel each other. The law of conservation of charge states that charge can be created or destroyed, but the total charge of a closed system remains constant. Objects can be charged by friction, by induced charge separation, by contact, and by induction. Questions 1. When you rub a glass rod with silk, the rod becomes positively charged; when you rub a plastic rod with fur, the rod becomes negatively charged. Suppose you have a charged object but do not know whether it carries a positive or a negative charge. Explain how you could use a glass rod and a piece of silk to determine the sign of the charge on the unknown object. K/U T/I 2. Explain how two objects attract one another due to an electric force, when one object has zero net charge. K/U C 3. Suppose children at a party rub balloons on their hair and then place the balloons on the wall. If the rubbing process puts excess electrons on the balloons, how do the balloons stay attached to the wall? K/U A 4. The end of a charged rubber rod attracts small pellets of foam plastic that, having made contact with the rod, quickly move away from it. Explain why this happens. T/I C 5. When two objects, such as a glass rod and a silk cloth, are rubbed together, electrons move from one object to the other. Can protons also move from one object to the other? Explain why or why not. K/U T/I 6. If you walk across a thick carpet on a cold, dry day, electrons will move from the carpet to your body. K/U (a) How does the charge on your body compare to the charge on the rug? (b) Which of the three methods of charging is demonstrated in this example? 7. A student upgrades the memory in her computer by replacing the memory chip. Before handling the chip, she first touches the metal casing of the computer. Why is this a good precaution? K/U A 8. Provide an example of how to give a neutral object a positive charge using only a negatively charged object. K/U C 9. In the following examples, describe the change in charge on each rod and charging material or object in terms of the movement of electrons. K/U (a) A glass rod is rubbed with a wool rag. (b) A plastic rod is rubbed with a silk scarf. (c) A platinum rod with a negative charge is touched with a similar rod with a positive charge. (d) A small metal rod touches a large positively charged metal sphere. 10. A student shakes hands with his father, who is wearing a wool sweater. As soon as they shake hands, a spark jumps between their hands. Explain what caused the spark. K/U 11. Fabric softener sheets are supposed to reduce the static cling between the clothes in a dryer. Research fabric softener sheets, and answer the following questions. K/U T/I (a) Why do clothes cling to each other when removed from a dryer? (b) How does a fabric softener sheet reduce the static cling among clothes? WEB LINK 326 Chapter 7 Electric Fields NEL

10 Coulomb s Law Recall that charged objects attract some objects and repel others at a distance, without making any contact with those objects. Electric force, F E, or the force acting between two charged objects, is somewhat similar to gravity. Both are non-contact forces. Similar to the force of gravity, the electric force becomes weaker as the distance, r, between the charged objects increases (Figure 1). Electric force becomes stronger as the amount of charge on either object increases, in the same way that the force of gravity becomes stronger with an increase in mass of either object. In this section, you will learn more about how the electric force depends on charge and distance. You will also learn how to solve problems related to the electric force. 7.2 electric force (F E ) a force with magnitude and direction that acts between two charged particles (a) F E F E 1 r 4 F 1 E 2r 4 F E (b) Figure 1 The electric force of repulsion between two identical charges decreases as the separation increases. The Electric Force If you drop a tennis ball, the force of gravity is responsible for its fall. The tennis ball will take approximately 1 s to fall from a height of 5 m. How long do you think the tennis ball will take to stop as it hits the ground? It will take a lot less time to stop than it took to fall. What force is responsible for making the tennis ball stop? The electric force of repulsion between the protons in the tennis ball and the protons in the ground stop it. This electric force must be significantly stronger than gravity. Consider two charged objects so tiny that you can model them as point particles. These objects have charges q 1 and q 2 and are separated by a distance r (Figure 2). The magnitude of the electric force between q 1 and q 2 is expressed by the equation F E r q 1 q 2 Figure 2 The electric force between two point charges q 1 and q 2 is given by Coulomb s law. F E F E 5 k q 1q 2 r 2 This equation represents Coulomb s law. Coulomb s Law The force between two point charges is inversely proportional to the square of the distance between the charges and directly proportional to the product of the charges. The constant k, which is sometimes called Coulomb s constant, has the value k N# m 2 /C 2. Do not confuse this constant k with the spring constant in Hooke s law. Most physicists use the same symbol for both. The direction of the electric force on each of the charges is along the line that connects the two charges, as illustrated in Figure 2 for the case of two like charges. As mentioned earlier, this force is repulsive for two charges with the same type of charge. Strictly speaking, the value of F E in the equation for Coulomb s law applies only to two point charges. However, it is a good approximation whenever the sizes of the particles are much smaller than their distance of separation. Coulomb s constant (k) the proportionality constant in Coulomb s law; k N# m 2 /C 2 NEL 7.2 Coulomb s Law 327

11 Investigation Coulomb s Law (page 366) Now that you have learned how to calculate the electric force between two charges with a given separation, perform Investigation to determine how electric force varies with distance r and 1 r 2. You will use this information with two different charge values to determine Coulomb s law. Coulomb s law has several important properties: You have already seen that the electric force is repulsive for like charges and attractive for unlike charges. Mathematically, this property results from the product q 1 q 2 in the numerator of the Coulomb s law equation. The magnitude of the electric force, F E, is always positive. In physics, it is convenient to use a negative sign to show direction. Including the negative sign of a negatively charged object would imply a direction. You never include the sign of the charge when solving problems related to Coulomb s law. The electric force is inversely proportional to the square of the distance between the two particles. The magnitude of the electric force is the magnitude of the force exerted on each of the particles. That is, a force of magnitude F E is exerted on charge q 1 by charge q 2, and a force of equal magnitude and opposite direction is exerted on q 2 by q 1. This pairing of equal but opposite forces is described by Newton s third law, the action reaction principle. Using the equation for Coulomb s law, we can show that electric forces can be extremely large. Suppose you have two boxes of electrons, each with a total charge of q T C separated by a distance r of 1.0 m. For simplicity, assume that each box is so small that it can be modelled as a point particle. The magnitude of the total electric force is F E 5 k q 1q 2 r 2 5 a N # m 2 F E N C 2 b C C m2 2 This is an extremely large force, all from just two small containers of electrons. Note that the negative sign for charge was not included. Why, then, does the electric force not dominate everyday life? The fact is that it is essentially impossible to obtain a box containing only electrons. Recall from earlier science studies that a neutral atom contains equal numbers of electrons and protons. If our two point-like boxes had contained equal numbers of electrons and protons, their total charges, q T, would have been zero, and so would the force calculated using Coulomb s law. Ordinary matter consists of equal, or nearly equal, numbers of electrons and protons. The total charge is therefore either zero or very close to zero. At the atomic and molecular scales, however, it is common to have the positive and negative charges (nuclei and electrons) separated by a small distance. In this case, the electric force is not zero, and these electric forces hold matter together. Comparing Coulomb s Law and Universal Gravitation The equation for Coulomb s law may seem familiar to you, even though you have only just learned it. This is because it is similar to the universal law of gravitation, which you learned in Chapter 6. Both of these laws describe the force between two objects. In addition, both depend on certain properties of the objects involved. For gravitation, this property is mass. For the electric force, this property is electric charge. Both mass and charge can be considered to be concentrated at a central point. If you think of the mass or charge as a solid sphere, you can treat this same mass or charge as if it were a point at the centre of that sphere. Another similarity between the two laws is that both the gravitational force and the electric force decrease as the distance between the two interacting objects increases. If the distance between the objects is great (compared to the size of either object), the actual size and shape of the objects involved become mathematically irrelevant. The theoretical mass and charge at the centre point of each object are then used in calculations. 328 Chapter 7 Electric Fields NEL

12 Although there are similarities between electric and gravitational forces, there are also important differences. Gravitational forces are always attractive. The direction of electric forces depends on the types of charge: unlike charges attract, and like charges repel. Also, the magnitude of the electric force is much greater than the magnitude of the gravitational force over the same distance. The Superposition Principle So far, you have learned about the electric force between two point particles. Now consider how to use Coulomb s law with more than two charges. Suppose you have three particles. One particle has a charge of q 1. The second particle has a charge of 2q 1 and is a distance 2r from the first particle, as shown in Figure 3. What is the electric force on a third charge, q 2, placed midway between these two charges? Solve this problem by first using Coulomb s law to calculate the force exerted by charge q 1 on q 2. Then use Coulomb s law a second time to calculate the force exerted by the charge 2q 1 on q 2. The total force on q 2 equals the vector sum of these two separate contributions. This combining of two forces is an example of the superposition principle. The superposition principle says that the total force acting on q 2 is the sum, or superposition, of the individual forces exerted on q 2 by all the other charges in the problem. Remember that force is a vector, so be careful to add these forces as vectors. This means that, for charges not on a straight line, the solution requires some trigonometry involving triangles or the use of vector components. When all charges lie in a straight line, the superposition principle simplifies so that you can add or subtract the various individual forces to or from one another to obtain the resultant, or net, force. As shown in Figure 3, the separation between q 2 and q 1 is r. Coulomb s law for this pair of charges is therefore F E2 q 1 2q 1 r q 2 FE1 Figure 3 The total force on q 2 equals the sum of the forces from charges q 1 and 2q 1. superposition principle the resultant, or net, vector acting at a given point equals the sum of the individual vectors from all sources, each calculated at the given point r x F E1 5 kq 1q 2 r 2 Using the coordinate system in Figure 3, this corresponds to a force on q 2 by q 1 in the positive x-direction because like charges repel. So, F E1 is the component of the force along the positive x-axis. The charges line up along the x-axis, so the component of the force along the y-axis is zero. Now consider the force on q 2 by the charge 2q 1 in a similar way. The separation of the charges is again r, so, using the equation for Coulomb s law, you get F E2 5 k 12q 1 2q 2 r 2 F E2 5 2kq 1q 2 r 2 From Figure 3 you can see that this corresponds to the force on q 2 by 2q 1 in the negative x-direction, again because like charges repel. This is because the electric force acts along the line connecting the two charges, which in this case is the x-axis. For this reason, the component of the force along the y-axis is again zero. So, the force F E2 exerted by 2q 1 on q 2 is in the negative x-direction, or to the left. The total force on q 2 is the sum of the two electric forces, F E1 and F E2. Be careful to take direction into account for the calculation. In this case, the sign of each force indicates its direction. F > E net 5 F > E 1 1 F > E kq 1q 2 2 2kq 1q 2 r 2 r 2 F > E net 5 2 kq 1q 2 r 2 The negative sign means that, for the positive product q 1 q 2, the force exerted on q 2 is along the negative x-direction. NEL 7.2 Coulomb s Law 329

13 Tutorial 1 Using Coulomb s Law This Tutorial shows how to use Coulomb s law to solve charge distribution problems in both one and two dimensions. Sample Problem 1: Applying Coulomb s Law in One Dimension An early model of the hydrogen atom depicted the electron revolving around the proton, much like Earth revolving around the Sun. The proton and electron both have mass, so they exert a gravitational force upon each other. They also have charge, so they exert an electric force on each other. (a) The distance between the electron and the proton in a hydrogen atom is m, the charge of each is C, the mass of the electron is kg, and the mass of the proton is kg. Calculate the ratio of the electric force F E to the gravitational force F g. (b) Determine the accelerations of the electron caused by both the electric force and the gravitational force. Solution (a) Given: r m; q C; m e kg; m p kg; k N.m 2 /C 2 ; G N.m 2 /kg 2 Required: F E F g Analysis: Use Coulomb s law to calculate F E, and use the equation for universal gravitation to calculate F g. F g 5 Gm em p and F r 2 E 5 kq eq p, where q r 2 e 5 q p. From these results, calculate F E. F g Solution: F E 5 kq eq p r 2 a N # m 2 C b C m2 2 F E N 1two extra digits carried2 The equation for Coulomb s law is more accurately represented by 0F > E 0 5 k0q 1 0 0q 2 0. r 2 Here, the vertical bars surrounding the electric force vector, 0F > E0, represent the magnitude of the electric force, and the vertical bars surrounding the charges q 1 and q 2, q 1 and q 2, represent the absolute values of the electric charges. This representation is more accurate because the magnitude of the electric force is a strictly positive quantity. If you just use the equation F E 5 kq 1q 2, you will get a negative value for r 2 the magnitude of the electric force when one of the charges is negative. The absolute value bars keep the values of the charges positive. However, many people find the more accurate version of the equation cumbersome and confusing and prefer to use F E 5 kq 1q 2. This is what we will do when solving problems. Always remember to r 2 drop the negative signs on charges when using this equation. The following Tutorial will give you some practice in solving problems using Coulomb s law. F g 5 Gm em p r 2 a N # m kg b kg kg m2 2 F g N 1two extra digits carried2 F E N F g N F E F g Statement: The electric force F E between the electron and the proton of a hydrogen atom is times the gravitational force F g between these same particles. (b) Given: m e kg; F g N; F E N Required: a E ; a g Analysis: Use the equation for electric force to calculate a E using F E and m e. Likewise, use the equation for gravitational force to calculate a g using the values of F g and m e. a E 5 F E m e and a g 5 F g m e. Use 1 N 5 1 kg# m/s 2. Solution: a E 5 F E m e kg # m s kg a E m/s Chapter 7 Electric Fields NEL

14 a g 5 F g m e kg # m s kg a g m/s 2 Statement: The acceleration of the electron caused by the electric force of the proton is m/s 2, and the acceleration of the electron caused by the gravitational force is m/s 2. Sample Problem 2: Determining Electrostatic Equilibrium Two charges, q C and q C, are separated by a distance, L, of 4.00 m. A third charge, q C, is placed somewhere between q 1 and q 2, as shown in Figure 4, where the net force exerted on q 3 by the other two charges is zero. Determine the location of q 3. q 1 q 3 q 2 x L Figure 4 Given: L m; q C; q C; q C; k N.m 2 /C 2 Required: x, the location of q 3 Analysis: Use the superposition principle, and use the equation for Coulomb s law to calculate the forces F E1 exerted on q 3 by q 1 and F E2 exerted on q 3 by q 2. The force equations are F E1 5 kq 1q 3 and F x 2 E2 5 kq 2q 3 1L 2 x2 2. Since the net force on q 3 is zero, F E1 2 F E2 5 0, or F E1 5 F E2. Solution: F E1 5 F E2 kq 1 q 3 5 kq 2q 3 x 2 1L 2 x2 2 Divide both sides of the equation by the common terms k and q 3, and then simplify. q 1 x 5 q 2 2 1L 2 x2 2 q 1 1L 2 x2 2 5 q 2 x 2 q 1 1L 2 2 2Lx 1 x q 2 x 2 q 1 L 2 2 2q 1 Lx 1 q 1 x 2 5 q 2 x 2 1q 2 2 q 1 2x 2 1 2q 1 Lx 2 q 1 L q 2 2 q 1 2 x 2 1 2Lx 2 L q 1 Substitute the values for q 1, q 2, and L into the equation, noting that x is in metres with three significant digits. Solve for x C C2 x C x a C C bx x x x x 2 1 x x x x 5 1 or x 5 22 Distance is always a positive quantity, so x 5 1. Statement: The location of q 3, such that the electric forces from q 1 and q 2 cancel, is x m, or 1.00 m to the right of q 1. Sample Problem 3: Applying Coulomb s Law in Two Dimensions Two point particles have equal but opposite charges of 1q 1 and 2q 1. The particles are arranged as shown in Figure 5. Suppose a charge q 2 is placed on the x-axis as shown. q C, q C, and the distance between 1q 1 and 2q 1 is 8.0 m measured vertically along the y-axis. Calculate the magnitude and the direction of the net electric force on q 2. Given: q C; 2q C; q C; k N.m 2 /C 2 Required: net electric force on q 2 Analysis: As shown in Figure 5, there are symmetrical right triangles above and below the x-axis, so we can use the Pythagorean theorem to calculate r (the distance separating 1q 1 and q 2 ). Then use r to calculate the electric force between 1q 1 and q 2 and that between 2q 1 and q 2. Use trigonometry to calculate the horizontal and vertical components of these two forces to determine the magnitude and direction of the net electric force. Use Coulomb s law to calculate the force between the charges. 4.0 m 4.0 m y q 1 r 3.0 m q 1 r q 2 x Figure 5 NEL 7.2 Coulomb s Law 331

15 Solution: First, calculate the magnitudes of the electric forces along the horizontal and vertical directions. Both of the forces along the x-axis act on the left side of the charge q 2, so the net horizontal force is the sum of the two horizontal force components. Use the Pythagorean theorem to calculate r, the distance between 1q 1 and q 2, and that between 2q 1 and q 2. r 5 "14.0 m m2 2 r m Calculate the force of 1q 1 on q 2, F E1. F E1 5 kq 1q 2 r 2 a N # m 2 9 b C C2 C m2 2 F E N 1two extra digits carried2 The force between a positive charge and another positive charge is repulsion, so this force is directed toward q 2 along the line connecting 1q 1 and q 2. Calculate the force of 2q 1 on q 2, F E2. Since the magnitude of 2q 1 is the same as that of 1q 1, the magnitude of F E2 is the same as that of F E1. F E2 5 F E1 F E N 1two extra digits carried2 The force between a negative charge and a positive charge is attraction, so this force is directed away from q 2 along the line connecting q 1 and q 2. Now calculate the x-component of F E1, F Ex1. Let u be the angle that F E1 makes with the x-axis. F Ex1 F E1 5 cos u F Ex1 5 F E1 cos u N2 a 3.0 m 5.0 m b F Ex N 1two extra digits carried2 F E1 is directed toward q 2, so F Ex1 is directed toward q 2 along the positive x-direction. By symmetry, the x-component of F E2, F Ex2, must have the same magnitude as the x-component of F E1 and is directed away from q 2 along the negative x-direction. F Ex2 5 F Ex1 F Ex N 1two extra digits carried2 Determine the vector sum of the two horizontal force components. F > Ex net 5 F > Ex 1 1 F > Ex N N F > Ex net 5 0 N There is no net force on q 2 along the x-direction. Now calculate the y-component of F E1, F Ey1. F Ey1 F E1 5 sin u F Ey1 5 F E1 sin u N2 a 4.0 m 5.0 m b F Ey N 1two extra digits carried2 F E1 is directed toward q 2 from 1q 1, so F Ey1 is directed toward q 2 downward. Since F E2 5 F E1, the y-component of F E2, F Ey2, has the same magnitude as the y-component of F E1, F Ey1. F Ey2 5 F Ey1 F Ey N 1two extra digits carried2 F E2 is directed away from q 2 from 2q 1, so F Ey2 is also directed downward. Determine the vector sum of the two vertical force components. Choose upward as positive, so downward is negative. F > Ey net 5 F > Ey 1 1 F > Ey N N2 F > Ey net N Statement: The net electric force acting on q 2 is N [down]. Practice 1. Determine the magnitude of the electric force between two charges of C and C that are separated by a distance of 2.00 m. T/I [ans: 2.25 N] 2. Two particles of charge q and 22q (with q positive) are located as shown in Figure 6. A third charge q 3 is placed on the x-axis to the left of q. The total electric force on q 3 is zero. The charges are separated by m. Determine where q 3 must be located. K/U T/I [ans: m [left of q]] q m Figure 6 2q x 3. Three point charges are placed at the following points on the x-axis: 12.0 mc at x 5 0, 23.0 mc at x cm, and 25.0 mc at x cm. Determine the force on the 23.0 mc charge. T/I [ans: 0.55 N toward the negative x-direction, or 0.55 N [left]] 332 Chapter 7 Electric Fields NEL

16 7.2 Review Summary According to Coulomb s law, the force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between the charges, given as F E 5 kq 1q 2. r 2 Coulomb s constant, k, is equal to N m 2 /C 2. Coulomb s law applies to point charges and to charges that can be concentrated equivalently in points located at the centre, when the sizes of the charges are much smaller than their distance of separation. There are similarities between the electric force and the gravitational force. The superposition principle states that the total force acting on a charge q is the vector sum, or superposition, of the individual forces exerted on q by all the other charges in the problem. Questions 1. Two charged objects have a repulsive force of N. The distance separating the two objects is tripled. Determine the new force. T/I 2. Two charged objects have an attractive force of N. Suppose that the charge of one of the objects is tripled and the distance separating the objects is tripled. Calculate the new force. T/I 3. Determine the magnitude of the electric force between two electrons separated by a distance of 0.10 nm (approximately the diameter of an atom). K/U T/I 4. Two point charges are separated by a distance r. Determine the factor by which the electric force between them changes when the separation is reduced by a factor of K/U T/I 5. Determine the distance of separation required for two 1.00 µc charges to be positioned so that the repulsive force between them is equivalent to the weight (on Earth) of a 1.00 kg mass. T/I 6. Consider an electron and a proton separated by a distance of 1.0 nm. K/U T/I (a) Calculate the magnitude of the gravitational force between them. (b) Calculate the magnitude of the electric force between them. (c) Explain how the ratio of these gravitational and electric forces would change if the distance were increased to 1.0 m. 7. Particles of charge q and 3q are placed on the x-axis at x and x 5 50, respectively. A third particle of charge q is placed on the x-axis, and the total electric force on this particle is zero. Determine the position of the particle. K/U T/I 8. Two charges of C and C are placed at a separation of 10 cm. Determine where a third charge should be placed on the line connecting the two charges so that it experiences no net force due to these two charges. K/U T/I A 9. Three charges with q C are located as shown in Figure 7, with L 5 25 cm. Determine the magnitude and direction of the total electric force on each particle listed below. T/I q Figure 7 L L y q L (a) the charge at the bottom (b) the charge on the right (c) an electron placed at the origin 10. Two pith balls, each with a mass of 5.00 g, are attached to non-conducting threads and suspended from the same point on the ceiling. Each thread has a length of 1.00 m. The balls are then given an identical charge, which causes them to separate. At the point that the electric and gravitational forces balance, the threads are separated by an angle of Calculate the charge on each pith ball. K/U A q x NEL 7.2 Coulomb s Law 333

17 7.3 Electric Fields If you have typed a letter on a computer, heard musical tones from a cellphone, or even just pressed a floor button in an elevator, then you have applied an electric force. In fact, almost everywhere you go you will find a device that, in one way or another, uses electric fields. An electric field is what causes the electric force. At the dentist s office, the X-ray machine uses an electric field to accelerate electrons as part of the process for producing X-rays. At coal-burning power plants, electric fields in smokestack scrubbers remove soot and other pollutants before gases are released into the air. Even when you are speaking into a telephone or listening to the other person on the line, electric fields help convert sound to electricity and back to sound again. The liquid crystal display (LCD) is a device that uses electric fields. Nearly all computer monitors, digital cameras, and smart phones use LCDs for their visual components (Figure 1). LCDs consist of a liquid crystal between two transparent sheets of glass or plastic, with a thin conducting material on the outside of the sheets. An electric field across the crystal causes its molecular arrangement to change, so that light passing through it is made either lighter or darker, depending on the design of the device. In this way, small changes in the electric field can make nearly any pattern appear on the LCD. Figure 1 LCDs use changing electric fields to alter the crystal s optical properties, creating images. Properties of Electric Fields electric field (e$) the region in which a force is exerted on an electric charge; the electric force per unit positive charge; unit is N/C We can describe the electric force between a pair of charges using Coulomb s law, but there is another way to describe electric forces. Suppose there is a single isolated point charge that is far from any other charges. This charge produces an electric field, a region in which a force is exerted on an electric charge. The electric field is similar to the gravitational field near an isolated mass, as discussed in Chapter 6. Were another charge to enter this field, the electric field would exert a force on it, much as a gravitational field exerts a force on a mass. The electric field has both magnitude and direction, so electric field is a vector denoted by e >. Consider a particular point in space where there is a uniform electric field e > (Figure 2). A point charge or an arrangement of several charges may have produced this field. A charge q at this location in the field will be affected by the electric field and experience an electric force given by F > E 5 qe > The electric force, F > E, is thus parallel to e >, with direction depending on whether the charge is positive or negative. 334 Chapter 7 Electric Fields NEL

18 The charge q 1, or the charge affected by the field, in Figure 2 is called a test charge. As a convention, physicists use a positive test charge to determine direction. By measuring the force on a positive test charge, you can determine the magnitude and direction of the electric field at the location of the test charge. Since we are working with a positive test charge, q 1, the electric field points in the same direction as the force that the test charge experiences. If q 1 happens to be a negative charge, then the direction of the electric field is in the opposite direction of the force that the negative charge experiences. The units for the electric field can be determined from the equation relating electric force to electric field. Force is measured in newtons (N), and charge is measured in coulombs (C), so the electric field is expressed in newtons per coulomb (N/C). force on test charge F E q 1 e e q 1 test charge q 2 Figure 2 The electric field at a particular point in space is related to the electric force on a test charge q 1 at that location. The electric field e > also relates to Coulomb s law for electric force. Coulomb s law for electric force allows you to calculate the electric field using the amount of charge that produces the field, q 2, and the distance of the field from the charge. As an example, calculate the magnitude of the electric field at a distance r from a charge q 2. The test charge is, once again, q 1. According to Coulomb s law, the electric force exerted on q 1 has a magnitude of F E 5 kq 1q 2 r 2 kq 2 F E 5 q 1 r 2 Inserting this expression into the equation relating electric force to the electric field gives the result qe 5 F E qe 5 q 1 kq 2 r 2 In this equation q 5 q 1, so we get e 5 kq 2 r 2 This is the magnitude of the electric field at a distance r from a point charge q 2. The direction of e > lies along the line that connects the charge producing the field, q 2, to the point where the field is measured. The direction of the electric field is determined by a positive test charge. If q 2 is positive, then a positive test charge will be repelled away from it, and thus the electric field points in a direction away from a positive charge. If q 2 is negative, then a positive test charge will be attracted toward it and thus the electric field points in a direction toward a negative charge. As a result of the direction of the electric field depending on the type of charge and the location in relation to the charge, we do not include the signs of charges in the equations to avoid implying a direction. NEL 7.3 Electric Fields 335

19 Tutorial 1 Determining Electric Fields In Tutorial 1, you will solve problems related to the electric field in both one and two dimensions. When using the equation for electric field, the symbol q really means the absolute value of q, just like it does for Coulomb s law. This Tutorial shows how to determine the electric field due to charge distribution at a point some distance from the charge. Sample Problem 1: Electric Field Due to Two Point Charges in One Dimension Two point charges are 45 cm apart (Figure 3). The charge on q 1 is C, and the charge on q 2 is C. q 1 Figure 3 P r 1 r 2 (a) Calculate the net electric field at point P, 27 cm from the positive charge, on the line connecting the charges. (b) A new charge of C is placed at P. Determine the electric force on this new charge. Solution (a) Given: r cm; r cm; q C; q C; k N.m 2 /C 2 Required: net electric field e > net at point P Analysis: The net electric field at point P equals the vector sum of the electric fields from the charges producing the fields. Use the equation e 5 kq to calculate e for each 2 r of the charges q 1 and q 2 at point P. Then determine the direction of each field based on the signs of the charges. Combine the two vector quantities to calculate e > net. r 2 5 r 12 2 r cm 2 27 cm 5 18 cm m Solution: Calculate the magnitude of the electric field at a distance r 1 from charge q 1 and determine the field s direction. e 1 5 kq 1 r 1 2 a N # m 2 9 b C2 C m2 2 e N/C 1two extra digits carried2 Since q 1 is a positive charge, the electric field on a positive test charge at P will be directed away from q 1. e > N/C 3right4 q 2 Calculate the magnitude of the electric field at a distance r 2 from charge q 2 and determine the field s direction. e 2 5 kq 2 r a N # m 2 9 b C2 C m2 2 e N/C 1two extra digits carried2 Since q 2 is a negative charge, the electric field on a positive test charge at P will be directed toward q 2. e > N/C 3right4 Now determine the vector sum of the two electric fields. Choose right as positive, so left is negative. e > net 5 e > 1 1 e > N/C N/C N/C 1two extra digits carried2 e > net N/C 3right4 Statement: The net electric field is N/C to the right of point P. (b) Given: e > net N/C, directed to the right of point P; q C Required: net electric force F > E net Analysis: Use the equation F > E 5 qe > to calculate the net electric force for the test charge q at point P. Solution: Use the sign of the charge to determine the direction of the electric field. Since the charge is positive, the electric force is directed to the right. Choose right as positive, so left is negative. F > E net 5 qe > net C N/C N F > E net N 3right4 Statement: The electric force acting on the new charge at point P is N to the right of point P. 336 Chapter 7 Electric Fields NEL

20 Sample Problem 2: Electric Field Due to Two Point Charges in Two Dimensions Two point charges are arranged as shown in Figure 4. q C, q C, and r cm. Calculate the magnitude of the electric field at the origin. y r q 1 Figure 4 r q 2 x Given: r cm m; q C; q C; k N.m 2 /C 2 Required: magnitude of net electric field at the origin, e net Analysis: The electric field at the origin results from the electric fields of charges q 1 and q 2. Use the equation e 5 kq r 2 for a point charge q to calculate the magnitudes of the net electric fields at the origin along the horizontal and vertical directions due to the charges q 1 and q 2. Then use the equation e net 5 "e 2 x net 1 e 2 y net to calculate the magnitude of the net electric field at the origin. Solution: First, calculate the magnitudes of the electric fields along the horizontal and vertical directions. The electric field of q 2 has only an x-component. e x 5 e x e x 5 kq 2 r 2 a N # m 2 9 b C2 C m N/C 1two extra digits carried2 The electric field of q 1 has only a y-component. e ynet 5 e y 5 kq 1 r 2 a N # m 2 9 b C2 C m2 2 e ynet N/C 1two extra digits carried2 Then calculate the magnitude of the net electric field at the origin. e net 5 "e 2 x net 1 e 2 y net 5 " N/C N/C2 2 e net N/C Statement: The magnitude of the net electric field at the origin is N/C. Practice 1. An electric force with a magnitude of 2.5 N, directed to the left, acts on a negative charge of 25.0 C. T/I (a) Determine the electric field in which the charge is located. [ans: 0.50 N/C [toward the right]] (b) Calculate the electric field when the force is the same but the charge is C. [ans: 3.3 N/C [toward the right]] 2. Calculate the magnitude and direction of the electric field at a point 2.50 m to the right of a positive point charge q C. T/I [ans: N/C [toward the right]] 3. Calculate the electric field at point Z in Figure 5, due to the point charges q C at point X and q C at point Y. T/I [ans: N/C, or 50.3 N/C [toward the left]] q 1 q m m X Y Z Figure 5 NEL 7.3 Electric Fields 337

21 Electric Field Lines electric field lines the continuous lines of force around charges that show the direction of the electric force at all points in the electric field An electric field exists in a region around a charge. The field and its properties can be represented with electric field lines. These electric field lines can help us determine the direction of the force on a nearby test charge. As was stated earlier, electric fields point away from positive charges and toward negative charges. This convention is based on using a positive test charge to determine direction. To show what the electric field would look like around a positive point charge, we draw electric field lines that extend radially outward from the charge, as shown in Figure 6(a). For a negative charge, the field lines are directed inward, toward the charge (Figure 6(b)). As you may expect, the electric field lines are parallel to e >, and the density of the field lines is proportional to the magnitude of e >. In both parts of Figure 6, the field lines are densest near the charges. In both cases, the magnitude of e > increases as the distance to the charge decreases. e e q q (a) Figure 6 Electric field lines near a point charge placed at the origin. (a) If the charge is positive, the electric field lines are directed outward, away from the charge. (b) If the charge is negative, the electric field lines are directed inward, toward the charge. (b) According to Coulomb s law, the force between two point charges varies as 1 r 2, where r is the separation between the two charges. In a similar way, the electric field produced by a point charge also varies as 1 r 2. The electric field thus obeys an inversesquare law, just as the gravitational force does (Figure 7(a)). The magnitude of the electric force is proportional to the density of field lines that is, the number of field lines per unit area of space at some distance r from the charge or charges producing the field. For a point charge q (Figure 7(b)), the density of field lines decreases farther away from the charge. The field lines from q spread out through a surface area. Therefore, the number of field lines per unit area (A), and thus the respective strengths of the electric field and electric force, decreases as r 2 increases. r 1 r 1 A 1 r 1 2 A 2 r 2 2 M r 2 q r 2 portions of a spherical surface (a) (b) Figure 7 (a) The gravitational force lines from a mass M spread out as r increases in all directions. (b) The electric field lines from a charge q also spread out in all directions. Both types of field lines pass through larger spherical surface areas at greater distances. The surface areas (A) over which the fields act increase as r 2 increases, so the fields themselves exert inverse-square forces. 338 Chapter 7 Electric Fields NEL

22 Another interesting aspect of the electric force relates to the question of action at a distance. How do two point charges that interact through Coulomb s law know about each other? In other words, how does one charge transmit electric force to another charge? In terms of the electric field lines, every charge generates (or carries with it) an electric field, through which the electric force is transmitted. Electric Dipoles Consider the two point particles with equal but opposite charge in Figure 8. Charges 2q and 1q, where q is the positive magnitude of the charge, are separated by a small distance r. This charge configuration is called an electric dipole. electric dipole a pair of equal and opposite electric charges with centres separated by a small distance q r q Figure 8 Two opposite charges separated by a distance r form an electric dipole. The two charges in an electric dipole give rise to a more complicated electric field than the one associated with a single electric charge. This is because the electric fields around the individual charges interact most strongly with each other at close distances, such as those that are similar in size to the dipole separation. Initially, the fields at the negative charge radiate inward toward the charge, and the fields at the positive charge radiate outward from the charge (Figure 9(a)). As the fields extend into the space around the other charge, they interact with each other, producing field lines that bend toward the other charge (Figure 9(b)). field lines from q alone field lines from q alone q q q q (a) (b) Figure 9 The electric field lines around each individual charge of an electric dipole (a) are affected by the field lines from the other charge, causing them to bend (b). It is important to note that the field lines extend in three dimensions around the charges, and that the view depicted here is of the field lines in a plane perpendicular to the line of sight. Notice that, along the vertical axis midway between the two charges, the electric field is parallel to the line connecting the two charges. This remains true along the vertical axis at all distances from the dipole, although the magnitude of the electric field decreases at distances that are farther from the dipole. The direction of this electric field always points from the positive charge to the negative charge. Although the field lines of a dipole merge at the midpoint, field lines do not cross. Instead, the cumulative effect, or the vector sum, of the electric fields from both charges produces a net electric field. That net electric field is represented by the electric field line. NEL 7.3 Electric Fields 339

23 Now consider an arrangement of charges slightly different from an electric dipole. In this case, a positive charge, 1q, replaces the negative charge, 2q, so that the two charges are equal and alike. Now the electric field lines extend outward from both charges. Instead of the field lines from a positive charge merging with the lines from a negative charge, the lines from similar charges do not connect at any point. This arrangement produces a disk-shaped region of zero electric field everywhere around the midpoint between the two charges (Figure 10). q q Figure 10 Two identical charges separated by a small distance produce this electric field pattern. Again, the field lines occupy all the space around the charges. This illustration shows how they appear in a plane. Note that the electric field is zero along the line that bisects the line connecting the charges. Notice that, farther away from the charges, the field behaviour starts to resemble that of a single charge. That is, the field lines appear to be radiating from a single point charge. This makes sense because, at a large distance from the two charges, the separation between them is not noticeable, and both charges have the same sign. Midway between the two charges, there is a gap where there are no field lines. This is expected because the vector sum of the two electric fields from both charges is zero at the midpoint. Note that this electric field pattern would be the same if the two charges were both negative. The difference would be in the direction in which the field lines were pointing but not the shape of the combined fields. Finally, consider a dipole-like arrangement of two charges that have different magnitudes and signs. If the positive charge 1q is replaced with a charge 12q, the symmetry of the dipole field is altered (Figure 11 on the next page). This is because the number of field lines for each charge is proportional to the magnitude of that charge. The number of field lines leaving the positive charge is therefore twice the number of field lines meeting at the negative charge. Half of these lines converge on the negative charge, while the other half emanate outward, as if there were only one charge (12q). At large distances, where r is much greater than the charge separation, the electric field radiates outward as it would for a charge of 12q. The field-line pattern for unequal and opposite charges includes regions near the charges where the density of field lines becomes very high. Note that this does not mean that the electric field is stronger in these regions. The electric field is still strongest along the line connecting the charges. Electric field patterns can become very complex, but many simulations exist to show what the electric field would look like in a multi-charge system. WEB LINK 340 Chapter 7 Electric Fields NEL

24 2q q Figure 11 The electric field lines around a dipole consisting of charges 2q and 12q. The number of lines from the larger charge is twice the number of lines from the smaller charge. Half of these lines extend from the positive charge to the negative charge. The other half extend outward, as if there were only one charge, 12q. Uniform Electric Fields So far, you have learned about electric fields that vary with distance from a charge. In the case of a dipole, for instance, the strength of the electric field varies with the number of charges, their placement, and the distance from the charges. A different electric field arises from a different type of dipole. Instead of point charges, suppose you have two parallel planes of charge. As with the dipole, one plane has a positive charge and the other plane has a negative charge. In both cases, the charge spreads uniformly along each plane. Just as the electric field along the line connecting two unlike charges extends straight from the positive to the negative charge, the electric field between the planes of charge extends from the positive plane of charge to the negative plane and is uniform. These field lines are straight, parallel to each other, and perpendicular to the planes of charge. At any location between the planes, the electric field has the same magnitude and direction. Outside the planes, the vector sum of the electric fields from all the individual charges in the two parallel planes yields a value of zero. This description of planes of charges involves infinite parallel planes carrying infinite amounts of charge and thus does not exist in real-world scenarios. However, you can create a close approximation by using two large conducting plates charged by dry cells. These plates are parallel and carry equal and opposite charges. As long as the separation between the plates is much smaller than their surface area, the electric field between the plates remains uniform (Figure 12). In fact, except near the edges of the plates, the magnitude of the electric field depends only on the amount of charge, the area of the plates, and the material between the plates. total charge q total charge q Figure 12 The electric field between two parallel conducting plates is uniform in direction and magnitude. NEL 7.3 Electric Fields 341

25 Earth s Electric Field UNIT TASK BOOKMARK You can apply what you have learned about electric fi elds to the Unit Task on page 422. Energy from the Sun bombards Earth s upper atmosphere. Some of this energy strips electrons from atoms, leaving a region of positively charged ions and free electrons. Some of the electrons recombine with the ions, but others travel into space and other regions of the atmosphere. This region, appropriately called the ionosphere, therefore has a positive charge and is able to conduct electricity. In contrast to the ionosphere, Earth s surface is more negatively charged. Both areas tend to stay charged, so that a permanent electric field exists throughout the atmosphere. Near Earth s surface, when the sky is clear of storms, the electric field has an average magnitude of about 120 N/C. This electric field varies seasonally, but the greatest and most sudden change occurs during thunderstorms, when the atmospheric electric field can reverse direction. Cloud-to-ground lightning, which often accompanies these storms, is both very common and potentially very destructive, so it is essential that scientists understand how Earth s electric field changes during storms, and what field conditions are likely to result in a lightning strike. One device, called an electric field mill, or just field mill, is widely used to measure Earth s electric field. A field mill makes use of the uniform electric field between two parallel conducting plates and detects changes in the field strength at a given location. The design of a field mill incorporates two circular conducting plates (Figure 13). The conducting plate at the front of the mill is the sensor plate. This part of the mill is exposed to the atmospheric electric field. A set of motor-driven, rotating shutter blades (the mill) exposes the sensor plate for a short time and then blocks the plate from the electric field for an identical length of time. During exposure, the plate becomes charged. During non-exposure, an analyzer circuit measures the electric field between the charged sensor plate and the uncharged detector plate. This process repeats continuously, providing information about variations in the field strength of Earth s electric field over a given time interval and in a particular region of the atmosphere. CAREER LINK Field mills have fairly simple designs and generally perform reliably. Although often set up in permanent positions on the ground, field mills located in balloons and aircraft measure field changes at different elevations. Field mills are used in cases where lightning could do extreme damage, such as to spacecraft before launch. shutter motor rotating shutter blade (the mill) conducting sensor plate detector plate electric field analyzer circuit Figure 13 A fi eld mill measures the electric fi eld between two parallel conducting plates to determine the changes in Earth s electric fi eld. 342 Chapter 7 Electric Fields NEL

26 Electrostatic Precipitators During the Industrial Age, heavily polluted air resulted from the smoke pouring out of chimneys and smokestacks. Most of these emissions, called flue gases because they passed through the flues of chimneys, consisted of gases such as nitrogen and carbon dioxide, both of which are clear substances. However, tiny particles of carbon, sulfur compounds, and dust produced by various chemical processes and combustion combined with the gases. These particles gave the air its smoky appearance. Today, industrial processes continue to pollute our air. Air containing polluting gases leads to many environmental concerns such as climate change and acid precipitation. Acid precipitation is known to harm both plant and aquatic life. The polluting gases also affect the respiratory health of not only the people who live nearby but also those who live where the prevailing winds tend to push the gases. In recent years, devices called electrostatic precipitators have reduced the numbers of these particles released into the atmosphere. Electrostatic precipitators use electric fields to remove extremely small particles of soot, dust, and ash from flue gases and other emissions produced by combustion, smelting, and refining. The exact arrangements of different electrostatic precipitators vary, but the basic principle is the same in all of them. In the design shown in Figure 14, the flue gas and particles it contains pass between a grid of negatively charged conducting wires and a conducting plate carrying a positive charge. The wires transfer electrons to the various particles that come into contact with the wires, making the particles negatively charged. The electric field between the wires and the plates is about N/C, so the force drawing these negatively charged particles toward the positively charged plates is very large. Shaking the plates from time to time loosens the particles that accumulate on them. A storage (collection) hopper below the precipitator collects this refuse. Repeating the procedure of passing the flue gas through several series of plates and wires removes about 99 % of the various particles from the gas. plate (+) wires ( ) flue gas collection hopper Figure 14 An electrostatic precipitator uses electric fi elds to remove particles from fl ue gases. Some of the devices used for home air purification use the basic principles of electrostatic precipitators. However, the results from these air cleaners have been mixed, partly because of the comparatively low levels of particles in household air, and partly because the electric fields are not as strong as in industrial precipitators. NEL 7.3 Electric Fields 343

27 Electric Fields in Nature Electric fields are also produced by animals. These fields are often weak and produced by ordinary actions, such as motion in the muscles. Some animals have organs that detect and respond to these weak electric fields. Hammerhead sharks, for instance, detect fields as low as 6 N/C in fish that hide beneath the sand or in tunnelled shelters along shallow ocean bottoms. The hammerhead shark swims close to the sandy ocean floor (Figure 15). It preys on goby, small fish that hide in sand-covered holes. A goby produces electric fields from muscular movements of its fins or gills. Although the shark cannot see the goby, it can detect these electric fields up to 25 cm above the sand. Having detected the goby, the hammerhead shark swims in a figure eight until it pinpoints the location of greatest electric field strength and then catches and consumes the goby. Figure 15 A hammerhead shark can detect the electric fields produced by the movements of its prey. Research This Fish and Electric Fields Skills: Researching, Communicating SKILLS HANDBOOK A4.1 Many fish use electric fields to detect or stun their prey, or to ward off predators. Some examples are electric eels, electric catfish, elephant fish, Nile knifefish, and torpedo fish. 1. Research one of these fish on the Internet, and determine how it detects or uses electric fields. 2. Compare this fish s abilities with those of the hammerhead shark. 3. Write a brief report of your findings that includes answers to the following questions. A. What types of behaviours that are related to electric fields are typical for the organism you chose? T/I B. Why are fish that stun prey with electric fields typically freshwater species? T/I C. Many fish are able to detect weak electric fields from prey that live in rivers with large amounts of silt and soil suspended in the water. Why would an adaptation such as electric-field detection be beneficial for these fish? T/I WEB LINK 344 Chapter 7 Electric Fields 8160_CH07_p indd 344 NEL 4/27/12 10:37 AM

28 7.3 Review Summary An electric field exists in a region of space when a test charge placed at any point in the region has a force exerted upon it. The electric field is a vector and is denoted by e >. A test charge q will experience an electric force given by F > E 5 qe >. The directions of the electric force and electric field are determined by a positive test charge. For a point charge q 2, the magnitude of the electric field at a distance r from the charge is e 5 kq 2 r. 2 Electric field lines are continuous lines of force that show the direction of electric force at all points in the electric field around a charge or charges. An electric dipole consists of two equal but opposite charges separated by some small distance. The electric field between two parallel plates of charge is uniform and perpendicular to the plates. The electric field outside the parallel plates is zero. One application of electric fields is in electrostatic precipitators, which use electric fields to remove extremely small particles of soot, dust, and ash from flue gases. Some organisms can detect the weak electric fields produced by the movement of other organisms. Questions 1. A proton and an electron are placed in a uniform electric field. Comment on the magnitudes of the forces experienced by both. K/U 2. Calculate the magnitude of the electric field at a distance of 1.5 m from a point charge with q C. K/U 3. A point particle of charge q C is placed on the x-axis at x 5210 cm. A second particle of charge q 2 is placed on the x-axis at x cm. The electric field at the origin is zero. Determine the charge q 2. K/U T/I 4. A ring with a radius of 25 cm and total charge C is centred at the origin as shown in Figure 16. The charge is distributed uniformly around the ring. Calculate the electric field at the origin. (Hint: Think of applying symmetry.) K/U T/I y Figure 16 r 5. When drawing electric field lines, what determines the number of lines originating from a charge? K/U x 6. Two point particles with charges q 1 and q 2 are separated by a distance L, as shown in Figure 17. The electric field is zero at point A, which is a distance L 4 from q 1. Determine the ratio q 1 :q 2. K/U T/I y L 4 A q 1 q 2 L Figure Five point charges, all with q C, are spaced equally along a semicircle with a radius of 2.3 m, as shown in Figure 18. Calculate the electric field at the origin. K/U T/I q q y q q q Figure Draw the electric field lines between the wires and plates of an electrostatic precipitator. T/I C x x NEL 7.3 Electric Fields 345

29 7.4 Potential Difference and Electric Potential In the previous section, you learned how two parallel charged surfaces produce a uniform electric field. From the definition of an electric field as a force acting on a charge, it follows that, for a given uniform electric field, charge, and particle mass, the particle undergoes a uniform acceleration. Another way of thinking about the physics of this situation is that the electric field does work on the charged particle. This view works well if the charge stays constant, but in reality the work done by the field varies with charge. What if, instead of describing the electric field in terms of force per charge, you expressed it in terms of energy per charge? As it turns out, we can describe the field in this way, as you will learn in this section. No matter how you choose to describe the uniform electric field, its ability to accelerate charged particles with known conditions has proven useful to physicists and engineers. Devices such as particle accelerators, which can accelerate particles to speeds near the speed of light, can only work if the particles are moving in the first place. Electric fields cause the initial motion of these particles by accelerating them. Particle acceleration is important in some everyday devices as well. Inkjet printers accelerate charged ink particles toward specific parts of the paper (Figure 1). Old television sets and computer monitors have cathode-ray tubes. These tubes accelerate electrons toward a phosphor screen. Variations in the deflection of the accelerated electrons determine the brightness and colour of the screen. paper deflection plates electrode nozzle of print head gutter charging control pump instructions from computer ink supply Figure 1 Ink droplets from the print head are either charged or uncharged. The uncharged droplets move to the paper undefl ected, forming the letters. Charged droplets are defl ected into the gutter, leaving those parts of the paper blank. work and Electric Potential Difference A q d e F E qe Figure 2 The electric fi eld gives rise to an electric force, which moves in the same direction as the charge if the charge is positive. B Recall that in a region of space where the electric field e > is constant, e > has the same magnitude and direction at all points. A point charge q in this region experiences an electric force F > E 5 qe > The force is parallel to e >, as shown in Figure 2. Suppose this charge moves a certain distance Dd, starting at point A and ending at point B. For simplicity, assume this displacement is parallel to the electric force F > E. According to the definition of work (W), the work done by the electric force on the charge is W 5 F E Dd 346 Chapter 7 Electric Fields NEL

30 The electric force does work on the charge and is independent of the path it takes from A to B. We can now define the electric potential energy, E E, which is the energy stored in the system that can do work W on a positively charged particle. From your studies of work and energy you know that the change in the potential energy associated with this type of force is equal to 2W, where W is the work done by that force. A more detailed explanation of the meaning of the negative sign will follow. So, if the electric force does an amount of work W on a charged particle, the change in the electric potential energy is DE E 5 2W DE E 5 2F E Dd Combining this equation with the equation relating force to the electric field, the change in electric potential energy when the charged particle moves from A to B in Figure 2 is DE E 5 2W 5 2F E Dd electric potential energy (E E ) the energy stored in a system of two charges a distance Dd apart, or the energy stored in an electric fi eld that can do work on a positively charged particle DE E 5 2qeDd This equation gives the change in the potential energy as the charge moves through a displacement Dd, in a region where the electric field is parallel to the displacement. Note that the change in the potential energy depends on the starting and ending locations but not on the path taken. In Figure 2, the displacement is along a line, but the charge may move from A to B along many other paths without affecting DE E. Electric potential energy is stored through the potential effect of the electric field on an electric charge. This effect is illustrated in Figure 3(a), where the charge q is moved from point B to point A by an external force F > a directed to the left. Force F > a results from some external agent, which could be your hand. The electric force on q points to the right, assuming that q is positive, so the displacement is opposite to the direction of the electric force. The work done by the electric field on the particle is thus negative. Therefore, according to the electric potential energy equation, the change in the electric potential energy must be positive. In other words, a positive amount of energy has now been stored in the system composed of the charge q and the electric field. That energy came from the positive amount of work done by the external force F > a that moved the charge from B to A. As the charge moves from A back to B, the process is reversed (Figure 3(b)). Now the electric field does a positive amount of work on the particle because the electric force and the particle s displacement are parallel and the change in the electric potential energy is negative. Energy stored in the electric field and particle system is now taken out of the system. This energy can appear as an increase in the kinetic energy of the particle when it reaches B. Potential energy becomes kinetic energy. e e A d B A d B F a q F E q E E 0 E E 0 (a) (b) Figure 3 (a) To move q from B to A, the external force acting on a charged particle works against the electric fi eld and produces a positive change in DE E of the fi eld and particle system. (b) The charged particle takes energy stored in the electric fi eld and converts it to kinetic energy. This produces a negative change in DE E of the fi eld and particle system. NEL 7.4 Potential Difference and Electric Potential 347

31 In the following Tutorial, you will learn more about how to solve problems that involve electric potential energy. Tutorial 1 Solving Problems Involving Electric Potential Energy This Tutorial explains how to determine the change in electric potential energy for a charge in a uniform electric field, given the position and magnitude of that field. Sample Problem 1: Potential Energy Difference in an Electric Field A charged particle moves from rest in a uniform electric field. (a) For a proton, calculate the change in electric potential energy when the magnitude of the electric field is 250 N/C, the starting position is 2.4 m from the origin, and the final position is 3.9 m from the origin. (b) Calculate the change in electric potential energy for an electron in the same field and with the same displacement. (c) Calculate the change in electric potential energy for an electron accelerated in an electric field with the same magnitude but opposite direction as in (a) and (b), and with a starting position of 2.4 m from the origin and a final position of 5.0 m from the origin. Solution (a) Given: d i m; d f m; q C; e N/C Required: DE E Analysis: Use the equation for electric potential energy in terms of q, e, and Dd, where Dd 5 d f 2 d i : DE E 5 2qeDd 5 2qe1d f 2 d i 2. Note that DE E is negative for a positive charge travelling in the same direction as e. Thus, the sign of q takes care of whether there is a gain (for negative q) or loss (for positive q) of electric potential energy. Note that we include the sign of the charge because we are dealing with energy, which is not a vector quantity and does not have a direction. Solution: DE E 5 2qeDd 5 2qe1d f 2 d i C2 a250 N C b 13.9 m m N# m DE E J Statement: The change in electric potential energy due to the movement of the proton in the uniform electric field is J. The negative sign indicates that the electric field loses potential energy by doing work on the proton. (b) Given: d i m; d f m; q C; e N/C Required: DE E Analysis: Use the same equation for electric potential energy: DE E 5 2qeDd 5 2qe1d f 2 d i 2. Note that we include the sign of the charge to determine whether there is a gain or loss of electric potential energy. Solution: DE E 5 2qeDd 5 2qe1d f 2 d i C2 a250 N C b 13.9 m m N# m DE E J Statement: The change in electric potential energy due to the movement of the electron in the uniform electric field is J. The positive value indicates that the electric field gains potential energy as the electron is accelerated in the field. (c) Given: d i m; d f m; q C; e N/C Required: DE E Analysis: Note that the sign for the electric field is negative because the electron is accelerating in a direction opposite to e. Use the same equation for electric potential energy: DE E 5 2qeDd 5 2qe1d f 2 d i 2. Solution: DE E 5 2qeDd 5 2qe1d f 2 d i C2 a2250 N C b 15.0 m m N # m DE E J Statement: When the direction of the electric field is reversed, the change in the electric potential energy is J. The negative value indicates that the electric field loses potential energy by doing work on the electron. 348 Chapter 7 Electric Fields NEL

32 Sample Problem 2: Dynamics of Charged Particles (a) Using the law of conservation of energy, calculate the speed of the proton in part (a) of Sample Problem 1 for the given displacement. Assume that the proton starts from rest. (b) Determine the initial speed of the electron in part (b) of Sample Problem 1, assuming its speed has decreased to half of its initial speed after the same displacement, Dd. Solution (a) Given: d i m; d f m; q C; m kg; e N/C Required: v Analysis: The law of conservation of energy states that the total change in potential energy of the field and particle system and the change in kinetic energy of the particle equals zero: DE E 1 DE k 5 0. The kinetic energy of the particle is related to its speed v by the equation DE k mv 2. Use this equation and the equation DE E 5 2qeDd for the change in potential energy of the field and proton system. Note that Dd m m m and 1 N 5 1 kg# m/s 2. Solution: By the law of conservation of energy, DE E 1 DE k 5 0 2qeDd mv mv 2 5 qedd v 5 Å 2qeDd m kg # m s C m2 C v 5 ï kg v m/s Statement: The speed of the proton accelerated for a distance of 1.5 m by an electric field of 250 N/C is m/s. (b) Given: Dd m; q C; m kg; e N/C; v f 5 0.5v i Required: v i Analysis: For situations that do not involve an object at rest, the change in kinetic energy is given by the equation DE k mv f mv i 2. Use this equation and the equation DE E 5 2qeDd for the change in potential energy of the field and electron system. Solution: By the law of conservation of energy, DE E 1 DE k 5 0 2qeDd 1 a 1 2 mv f mv i 2 b m a1 2 v ib mv i 2 5 qedd mv i 2 5 qedd v i 5 Å 2 8qeDd 3m C2 a250 N C b 11.5 m2 v i 5 2 ã kg2 kg # m s C m2 C 5 ï v i m/s kg2 Statement: The initial speed of the electron before entering the electric field is m/s. Practice 1. An electron enters a uniform electric field of 145 N/C pointed toward the right. The point of entry is 1.5 m to the right of a given mark, and the point where the electron leaves the field is 4.6 m to the right of that mark. T/I A (a) Determine the change in the electric potential energy of the electron. [ans: J] (b) The initial speed of the electron was m/s when it entered the electric field. Determine its final speed. [ans: m/s] 2. Calculate the work done in moving a proton 0.75 m in the same direction as the electric field with a strength of 23 N/C. T/I A [ans: J] 3. An electron experiences a change in kinetic energy of J. Calculate the magnitude and direction of the electric field when the electron travels 0.18 m toward the right. T/I A [ans: N/C [toward the left]] NEL 7.4 Potential Difference and Electric Potential 349

33 Electric Potential electric potential (V ) the value, in volts, of potential energy per unit positive charge for a given point in an electric field; 1 V 5 1 J/C electric potential difference ( V ) the amount of work required per unit charge to move a positive charge from one point to another in the presence of an electric field Electric potential energy is a property of a system of charges or of a point charge in an electric field, where the field is created by other charges. In either case, this electric potential energy is not the property of a single charge alone. For this reason, the potential energy depends on the values of the charges and the electric field involved in the interaction. This leads to a new quantity called electric potential, V, which is a measure of how much electric potential energy is associated with a specific quantity of charge at a particular location in an electric field. Based on this definition, V 5 E E q Electric potential, or just potential, is a convenient measure because it is independent of the amount of charge at a particular location in the field. It depends only on the electric field strength at that location. For example, if you had 1 C of electrons at a particular location in a uniform electric field, you would possess a certain amount of electric potential energy. If you doubled the amount of electrons to 2 C at the same location in the electric field, you would have double the electric potential energy. In both these situations, you would have the same electric potential. The SI unit of electric potential is the volt (V), named in honour of physicist Alessandro Volta ( ). The volt relates to other SI units in the following equation: 1 V 5 1 J/C 5 1 N# m/c The volt is used in the measurement of many electrical quantities. Note in particular that the units of volts per metre are equivalent to units of newtons per coulomb, the units for electric field strength. These equivalent units give you several different ways to express the units of the electric field. The ones most commonly used are 1 V/m 5 1 N/C. Another convenient definition relating to electric potential energy is electric potential difference, DV. We return to the concept of the change in potential difference and the displacement of the particle: You can define the change in the potential, or potential difference, for a charge q that moves between two points: DV 5 DE E q For the case of a uniform electric field, the equation for electric potential difference becomes DV 5 DE E q 5 2W q 5 2qeDd q DV 5 2eDd This relationship shows how a non-uniform electric field varies with the change in electric potential (that is, electric potential difference) and the change in position in the field: e 5 2 DV Dd 350 Chapter 7 Electric Fields NEL

34 This equation states that the magnitude of the electric field is largest in regions where V is large and changes rapidly with small changes in displacement. Conversely, the electric field is zero in regions where V is constant. Notice that because of the negative sign in the equation, the electric field points from regions of high potential to regions of low potential (Figure 4). If we consider a circuit in which a battery is the source of electrical energy, a positive test charge will naturally move from the positive terminal where a high potential exists to the negative terminal where a low potential exists. Since the electric field points from positive to negative, the positive test charge will also move in the same direction as the field. Conversely, electrons will naturally travel from a region of low potential to a region of high potential, in a direction opposite to the direction of the electric field. UNIT TASK BOOKMARK You can apply what you have learned about electric potential to the Unit Task on page 422. e q d potential difference V V e d Figure 4 The magnitude and direction of the electric fi eld are related to how the electric potential V changes with position. The relation between the electric field and the changes in the potential and position involves the component of the electric field that is in the direction parallel to the displacement Dd. The electric field is a vector, so if you want to determine e > in a particular direction, you must consider how the potential V changes for a test charge moving along that direction. With this equation and knowledge of e >, you can, in principle at least, calculate how the electric potential changes as you move from place to place within the find. Strictly speaking, this relation only holds for small steps Dd in a constant field. You can, however, combine the potential changes DV from many such small steps to determine the change in the potential difference over a large distance. The following Tutorial will demonstrate how to solve problems involving electric potential. Tutorial 2 Solving Problems Related to Electric Potential This Tutorial shows how to use electric potential to solve problems related to a charge in a uniform electric fi eld. Sample Problem 1: TV Tubes and Particle Accelerators The cathode-ray tubes in old television sets and computer monitors work in a way that is similar to certain parts in particle accelerators. Both devices accelerate particles in a similar way, using the uniform electric fi eld between conducting plates. Looked at another way, the particles accelerate as they move through an electric potential difference. (a) An electron leaves the negative plate of a cathode-ray tube and travels toward the positive plate. The electric potential difference between the plates is V. Using the law of conservation of energy and the defi nition of electric potential difference, calculate the speed of an electron as it reaches the positive plate in a cathode-ray tube. Assume that the electron is initially at rest. The mass of an electron is kg. (b) Calculate the magnitude of the electric fi eld at a distance of 15 cm, which is at the end of the cathode-ray tube. NEL 7.4 Potential Difference and Electric Potential 351

35 (a) Given: q C; m kg; DV V; v i 5 0 m/s Required: v f Analysis: Use the equation for the law of conservation of energy, DE E 1 DE k 5 0, and the equation for the electric potential difference, DE E 5 qdv. The kinetic energy of the particle initially at rest is related to its final speed v f by the equation DE k mv f mv i 2. Since v i = 0, DE k mv f 2. Note that 1 V 5 1 N m/c 5 1 kg m/s 2 m/c. Solution: By the law of conservation of energy, DE E 1 DE k 5 0 qdv mv f mv f 2 5 2qDV v f qDV m v f 5 Å 2 2qDV m Calculate the final speed v f of the electron as it reaches the positive plate. v f 5 Å C V kg 5 ã C2 a kg # m s 2 # m C b kg v f m/s Statement: The speed of the electron as it reaches the positive plate is m/s. (b) Given: Dd 5 15 cm m; DV V Required: e Analysis: Use the equation for the electric field in terms of potential difference and displacement: e 5 2 DV Dd Solution: e 5 2 DV Dd V 0.15 m e V/m Statement: The electric field has a magnitude of V/m. The negative sign indicates that the field extends from high to low potential. The electric field vector always points from regions of high V to regions of low V because e > is parallel to the direction that a positive test charge would move if it were placed at that location. Sample Problem 2: Calculating the Speed of a Deflected Charged Particle An electron moves horizontally with a speed of m/s between two horizontal parallel plates. The plates have a length of 12.5 cm, and a plate separation that allows a charged particle to escape even after being deflected (Figure 5). The magnitude of the electric field within the plates is 150 N/C. Calculate the final velocity of an electron as it leaves the plates. y x e Figure 5 v i Given: v i m/s; L cm m; q C; e N/C Required: v > f Analysis: The electric field is uniform and directed downward between the plates, so the electric force acting on a proton is also constant and directed downward. A constant downward force means a constant downward acceleration on a positive charge. However, because the charge is negative, the acceleration must be upward. To calculate the magnitude e v f of the final velocity, solve for each component of the velocity: the constant horizontal component and the accelerated upward component. Then use the equation v f 5 "v 2 x f 1 v 2 y f for the magnitude of the final velocity. The components of the velocity are given by v xf 5 v i 5 L Dt (Dt can be calculated from v i and L) and v yf 5 v yi 1 a y Dt 5 a y Dt 1since v yi By combining the equations for Newton s second law, F net 5 ma, and the electric force on a charge in an electric field e, F E 5 qe, we can determine the vertical acceleration a y 5 F net m 5 F E m 5 qe m. Solution: Calculate a y. a y 5 qe m kg # m s C C kg a y m/s 2 1two extra digits carried2 352 Chapter 7 Electric Fields NEL

36 Calculate Dt. v i 5 L Dt Dt 5 L v i m m s Dt s 1two extra digits carried2 Calculate v yf. v yf v yf 5 a y Dt 5 a m s 2b s m/s 1two extra digits carried2 Now calculate the magnitude of the net velocity. v f 5 "v 2 x f 1 v 2 y f 5 " m/s m/s2 2 v f m/s Calculate the angle u to determine the direction of the electron. u 5 tan 21 a v y f v xf b m 6 s 5 tan 21 ± m 6 s u Statement: The final velocity of the electron is m/s [E 528 N]. Practice 1. An old television cathode-ray tube creates a potential difference of V across the parallel accelerating plates. These plates accelerate a beam of electrons toward the target phosphor screen. The separation between the plates is 12 cm. K/U T/I (a) Using the principle of energy conservation and the definition of electric potential difference, calculate the speed at which the electrons strike the screen. [ans: m/s] (b) Calculate the magnitude of the electric field. [ans: N/C] 2. Four parallel plates are connected in a vacuum as shown in Figure 6. An electron at rest in the hole of plate X is accelerated to the right. The electron passes through holes at W and Y with no acceleration at all. It then passes through the hole at Y and slows down as it heads to plate Z. T/I 6.0 cm 6.0 cm 6.0 cm X W Y Z V V Figure 6 (a) Calculate the speed of the electron at hole W. [ans: m/s] (b) Calculate the distance, in centimetres, from plate Z to the point at which the electron changes direction. [ans: 5.7 cm [to the left of Z]] 3. An electron enters a parallel plate apparatus that is 8.0 cm long and 4.0 cm wide, as shown in Figure 7. The electron has a horizontal speed of m/s. The potential difference between the plates is V. Calculate the electron s velocity as it leaves the plates. K/U A [ans: m/s [E 3.38 N]] 8.0 cm m/s 4.0 cm Figure 7 NEL 7.4 Potential Difference and Electric Potential 353

37 7.4 Review Summary The change in electric potential energy depends on the electric field, the charge being moved, and the charge s displacement: DE E 5 2qeDd. For DE E 7 0, work is done against the electric field (2W), resulting in energy stored in the field. For DE E 6 0, work is done by the electric field (1W) on a particle moving in the field, which typically increases the kinetic energy of the particle. The electric potential is the electric potential energy per unit charge at a given point in an electric field: V 5 E E q. The magnitude of an electric field varies with the electric potential difference and the change in position in the field: e 5 2 DV Dd. Questions 1. An electron moves from an initial location between parallel plates where the electric potential is V i 5 30 V to a final location where V f V. K/U T/I (a) Determine the change in the electron s potential energy. (b) Determine the average electric field along a 10 cm long line segment that connects the initial and final locations of the electron. Be sure to give both the magnitude and the direction of e >. 2. The electric potential difference between two parallel metal plates is DV. The plates are separated by a distance of 3.0 mm and the electric field between the plates is e V/m. Calculate DV. K/U T/I 3. A proton of mass kg moves from a location where V i V to a spot where V f V. K/U T/I (a) Calculate the change in the proton s kinetic energy. (b) Replace the proton with an electron, and determine its change in kinetic energy. 4. An electron moves from a region of low potential to a region of higher potential where the potential change is 145 V. K/U T/I (a) Calculate the work, in joules, required to push the electron. (b) What is doing the work? 5. The electrons in an old TV picture tube are accelerated through a potential difference of V. K/U T/I A (a) Do the electrons move from a region of high potential to a region of low potential, or vice versa? (b) Calculate the change in the kinetic energy of one of the electrons. (c) Calculate the final speed of an electron when the initial speed is zero. 6. A pair of parallel plates has an electric field of N/C. Determine the change in the electric potential between points that are 2.55 m (initial) and 4.55 m (final) from the plates. K/U T/I A 7. An electron with a horizontal speed of m/s and no vertical component of velocity passes through two horizontal parallel plates, as shown in Figure 8. The magnitude of the electric field between the plates is 150 N/C. The plates are 6.0 cm long. K/U T/I e Figure 8 y (a) Calculate the vertical component of the electron s final velocity. (b) Calculate the final velocity of the electron. 8. An electric field of 20 N/C exists along the x-axis in space. Calculate the potential difference DV 5 V B 2 V A, where the points A and B are given by (a) A 5 0 m; B 5 4 m (b) A 5 4 m; B 5 6 m T/I A 9. Points A and B are at the same potential. Determine the net work done in moving a charge from point A to point B. K/U T/I 10. The potential at a point is 20 V. Calculate the work done in bringing a charge of 0.5 C to this point. K/U A x 354 Chapter 7 Electric Fields NEL

38 Electric Potential and Electric Potential Energy Due to Point Charges In the first section of this chapter, you saw how a Van de Graaff generator in a science museum causes the hair of anyone in contact with the device to stand on end. At that point, the discussion dealt simply with the properties of electric charge, and how the like charges (electrons) on individual hairs caused the hairs to repel each other and spread out in all directions. We can look at this phenomenon in another way: the charge on the Van de Graaff generator creates a high electric potential on the conducting sphere of the generator. This potential provides an electric field that exerts a force on charged particles near the generator. This force, in turn, accelerates these particles. In fact, when the Van de Graaff generator was invented, it was an early type of particle accelerator. As a particle accelerator, however, the Van de Graaff generator has limitations. For one thing, it can only accelerate charges like those on the sphere. Also, if the potential on the sphere becomes too high with respect to the generator s surroundings, the air around the generator will ionize. This causes large sparks to jump between the sphere and objects at lower electric potentials, resulting in artificial lightning (Figure 1). At a distance, the conducting sphere of a Van de Graaff generator resembles a point charge. In this section, you will learn about electric potentials, electric potential energies, and electric potential differences for point charges and their surroundings. 7.5 Figure 1 The charge on the Van de Graaff generator has a large electric potential difference with its surroundings. This can cause large sparks to appear between the generator and its surroundings. Electric Potential Due to a Point Charge Earlier you learned about the electric properties of a point charge. This simple example is useful in many situations. In Section 7.4, we defined electric potential V in terms of a test charge q and the electric potential energy E E. The electric potential energy, in turn, depends on the position d of the test charge in an electric field e >. If the charge producing the electric field has the same sign as the test charge q, the work done by the electric field on q is negative. This means that DE E. 0 for Dd. 0, and the sign of potential energy (and thus potential) is reversed from that of the uniform electric field. For a point charge q, the magnitude of the electric field is e 5 kq r 2 Setting the location in the field equal to the radius of the field ( d 5 r), the electric potential due to a point charge is inversely proportional to the distance from the charge and directly proportional to the amount of charge. So, for a point charge q producing an electric field, at a distance r from the charge, the electric potential is V 5 kqdd r 2 V 5 kq r electric potential due to a point charge the electric potential is inversely proportional to the distance from the charge and proportional to the amount of charge producing the field These results reinforce what you learned about potential energy and the behaviour of like and unlike charges. NEL 7.5 Electric Potential and Electric Potential Energy Due to Point Charges 355

39 Suppose you place a source charge (q. 0) at the origin of the coordinate system in Figure 2. The charge produces a potential given by kq. Then you place a positive r test charge nearby. Both the source and the test charge are positive, so they repel one another, and the test charge experiences a force that carries it downhill and away from the origin along the blue potential curve in Figure 2. When the source charge is negative, the potential is described by the lower, red curve in Figure 2. The source charge is now negative, so the attractive force carries a positive test charge along the red potential curve toward the origin. V V when q 0 r V when q 0 Figure 2 Electric potential near a point charge. The blue curve is for a positive point charge, and the red curve is for a negative point charge. Only changes in potential energy are important, so you must always be clear about choosing the reference point for potential energy. Electric potential is proportional to electric potential energy, so you must also pay attention to the reference point for electric potential. For example, when dealing with a point charge, follow the standard convention by choosing V 5 0 at an infinite distance from the source charge. Infinite distance represents a distance so far away that any potential produced by a charge is negligible and for all intents and purposes equal to zero. In many other problems, choose Earth as V 5 0 because Earth conducts charge well. Put another way, the electric ground is the point where V 5 0. For two or more charges, use the superposition principle. Remember that electric potentials of negative charges are negative, and electric potentials of positive charges are positive. However, because electric potential is not a vector, the total electric potential of several point charges equals the algebraic sum of the electric potentials resulting from each of the individual charges and their specific distances from the chosen location. Electric Potential Energy of Two Point Charges Now consider the electric potential energy of a pair of charges. The electric potential energy of a pair of charges is the potential energy possessed by each charge in the pair because of its position. For a pair of charges, this potential energy is inversely proportional to the distance between the charges and directly proportional to the product of the two charges. For a pair of charges q 1 and q 2 separated by a distance r, use the equation for the electric field for the point charge q 2 and the definition of electric potential energy for the point charge q 1, in terms of V, to determine the potential energy for the two charges: E E 5 q 1 V 5 q 1 kq 2 r E E 5 kq 1q 2 r The electric potential energy of a pair of charges is then given by E E 5 kq 1q 2 r 356 Chapter 7 Electric Fields NEL

40 This is qualitatively similar to the equation for Coulomb s law, except that E E varies as 1 r, not 1 r 2. If the charges are initially separated by a distance r i and then brought together to a final separation r f, the change in the potential energy is DE E 5 E Ef 2 E Ei DE E 5 kq 1q 2 r f 2 kq 1q 2 r i Note that only changes in potential energy matter, and that E E approaches zero when the two charges are very far apart. The blue curve in Figure 3 shows the general behaviour of the electric potential energy for two like charges. In this case, E E is positive and increases as the charges are brought together. If the charges have opposite signs, the electric force is attractive (negative), and the potential energy is then also negative, as shown by the red curve in Figure 3. The following Tutorial will demonstrate how to solve problems involving electric potential and electric potential energy. E E like charges r f r i r unlike charges Figure 3 Electric potential energy as a function of the separation r between two charges q 1 and q 2. The blue curve is for a repulsive force, and the red curve is for an attractive force. Tutorial 1 Electric Potential and Electric Potential Energy Sample Problem 1 models how to calculate the total electric potential of two point charges and the work required to move a third charge into this potential. In Sample Problem 2, we will use the electric potential energy and conservation of energy to calculate the speed of a charge at a given location. Finally, in Sample Problem 3, we show how to calculate the minimum particle separation by using electric potential energy, conservation of energy, and conservation of momentum. Sample Problem 1: Calculating the Electric Potential A point charge with a charge of C is 4.00 m due west from a second point charge with a charge of C. (a) Calculate the total electric potential due to these charges at a point P, 4.00 m due north of the first charge. (b) Calculate the work required to bring a third point charge with a charge of C from infinity to point P. Solution (a) Given: q C; q C; r m; r 1P m; k N.m 2 /C 2 Required: V T NEL 7.5 Electric Potential and Electric Potential Energy Due to Point Charges 357

41 Analysis: The total potential at P is the sum of the potentials due to the two individual charges q 1 and q 2. We can use the equation V 5 kq to calculate the potential due to r each charge and the Pythagorean theorem to calculate the distance of point P from q 2 : r 2P 5 "r r 2 1P. Solution: V T is the sum of the potentials due to the individual charges q 1 and q 2. Calculate the potential V 1 at P due to q 1. V 1 5 kq 1 r 1P a N # m 2 9 b C2 C m N# m/c V J/C To calculate the potential V 2 at P due to q 2, first calculate the distance r 2P between P and q 2. r 2P 5 "r r 1P 5 "14.00 m m2 2 r 2P m 1two extra digits carried2 Then calculate V 2. V 2 5 kq 2 r 2P a N # m 2 9 b C2 C m N# m/c V J/C 1two extra digits carried2 Now calculate the total potential V T. V T 5 V 1 1 V J/C J/C2 V T J/C 1one extra digit carried2 Statement: The total electric potential at point P is J/C. (b) Given: V T J/C; q C Required: work required to move q 3 from r 5 ` to P Analysis: Use the definition of electric potential energy. The work done on q 3 is W 5 q 3 1V` 2 V T 2. At infinity, the electric potential is zero, so W 5 2q 3 V T. Solution: W 5 2q 3 V T J C2 a C b W J Statement: The work done on the third point charge as it is brought from infinity to point P is J. The positive value of W indicates that the electric field does work on the charge to increase the charge s kinetic energy as it approaches P. 358 Chapter 7 Electric Fields NEL

42 Sample Problem 2: Electric Potential Energy and Dynamics A point charge q 1 with charge C is initially at rest at a distance of 0.25 m from a second charge q 2 with charge C and mass kg (Figure 4). Both charges are positive. Charge q 1 remains fixed at the origin, whereas q 2 travels to the right upon release. Determine the speed of charge q 2 when it reaches a distance of 0.50 m from q 1. q m q 2 q m q 2 v i 0 v f? 0 0 Figure 4 initial state final state Given: q C; q C; m kg; r m; r m Required: v f Analysis: No external forces are applied to the system, so energy must be conserved. We can apply the law of conservation of energy, using electric potential energy: E Ei 1 E ki 5 E Ef 1 E kf. Charge q 1 remains fixed at all times, so E k equals the kinetic energy of charge q 2, and E E equals the electric potential energy of the system of two charges. Charge q 2 is initially at rest, so the initial kinetic energy of q 2 is zero. Note that r 1 and r 2 are related by r 2 5 2r 1. E Ei 5 kq 1q 2 ; E r Ef 5 kq 1q 2 ; E 1 2r ki mv 2 i 5 0; E kf mv 2 f Solution: E Ei 1 E ki 5 E Ef 1 E kf kq 1 q kq 1q r 1 2r 1 2 mv 2 f 1 2 mv 2 f 5 kq 1q 2 2r 1 v f 5 Å kq 1 q 2 mr 1 5 ï v f m/s kg # m s 2 # m 2 C C C kg m2 Statement: The speed of charge q 2 when it reaches a distance of 0.50 m from charge q 1 is m/s. Sample Problem 3: A Head-on Collision Two particles, a proton with charge C and mass kg and an alpha particle (helium-4 nucleus) with charge C and mass kg, are separated by an extremely large distance. They approach each other along a straight line with initial speeds of m/s each. Calculate the separation between the particles when they are closest to each other. Given: q C; q C; m kg; m kg; v i1 5 v i m/s; r i S ` m; k N.m 2 /C 2 Required: minimum separation of charges, r f NEL 7.5 Electric Potential and Electric Potential Energy Due to Point Charges 359

43 Analysis: We can use the conservation of energy to determine the separation of the charges at the moment when both particles have converted their kinetic energy into electric potential energy: E Ei 1 E ki 5 E Ef 1 E kf. At an extremely large distance, r i S `, the electric potential energy of the pair of charges will be zero. At minimum separation, the kinetic energy of the charges will be zero, so E Ei 5 kq 1q 2 5 0; E r Ef 5 kq 1q 2 ; E i r ki 5 1 f 2 m 1v 2 i m 2v 2 i 2 ; E kf m 1v 2 f m 2v 2 f Solution: E Ei 1 E ki 5 E Ef 1 E kf m 1v 2 i m 2v 2 i 2 5 kq 1q 2 r f 1 0 2kq 1 q 2 r f 5 m 1 v 2 i 1 1 m 2 v 2 i 2 kg # m # m s C C2 5 C kg2 a m s b kg2 a m s b 2 Figure 5 comb comb q 1 q 2 q 3 B A 3.0 m conducting sphere belt motor and pulley ground Figure 6 In a Van de Graaff generator, charge is carried to the dome by a moving belt. The charge is deposited on the belt at point A and conveyed to the dome at point B. r f m Statement: The minimum separation of the proton and the alpha particle is m. Practice 1. Three charges, q C, q C, q C, are located at the vertices of an equilateral triangle (Figure 5). K/U T/I (a) Calculate the electric potential at the midpoint of each side of the triangle. [ans: between q 1 and q 2 is V J/C; between q 1 and q 3 is V J/C; between q 2 and q 3 is V J/C] (b) Calculate the total electric potential energy of the group of charges. [ans: J] 2. Four point charges, each with q C, are arranged at the corners of a square of side length 1.5 m. Determine the electric potential at the centre of the square. K/U T/I [ans: J/C] 3. Two electrons start at rest with a separation of m. Once released, the electrons accelerate away from each other. Calculate the speed of each electron when they are a very large distance apart. K/U T/I [ans: m/s] 4. Two protons move toward each other. They start at infi nite separation. One has an initial speed of m/s, and the other has an initial speed of m/s. Calcu late the separation when the protons are closest to each other. K/U T/I [ans: m] The van de Graaff Generator The Van de Graaff generator, discussed earlier, produces an electric potential on the conducting sphere, or dome, by separating charge at one place, and moving the charge to the conductor. As you can see in Figure 6, a moving cloth or rubber conveyorstyle belt receives electrons from a lower conducting comb at point A. The negatively charged part of the belt moves upward toward the conducting sphere and gives up the negative charge to the upper comb at point B. The charge travels to the conducting sphere and gives the dome a net negative charge. As the generator continues to run, the amount of charge on the sphere builds. When the charge reaches high enough levels, the electrons jump from the sphere through the air to another conductor brought close to the sphere, producing the artificial lightning seen in Figure Chapter 7 Electric Fields NEL

44 7.5 Review Summary The electric potential at a distance r from a point charge q is V 5 kq r. The total electric potential at a point P for a system of charges equals the algebraic sum of the potentials at P due to each individual charge at a distance that separates the charge and P. For an electric potential from a point source charge q 1, the work done moving a test charge q 2 from an initial separation r i to a final separation r f equals the change in the electric potential energy: W 5 E Ef 2 E Ei W 5 kq 1q 2 2 kq 1q 2 r f r i For a system of two charges, q 1 and q 2, separated by a distance r, the electric potential energy is E E 5 kq 1q 2. r The potential energy difference of two point charges, q 1 and q 2, as the separation between them changes from r i to r f is DE E 5 E Ef 2 E Ei DE E 5 kq 1q 2 2 kq 1q 2 r f r i When the amount of charge on the dome of a Van de Graaff generator is high enough, the electrons may jump from the sphere, producing artificial lightning. Questions 1. A charged particle is released from rest and begins to move as a result of the electric force from a nearby proton. For the following particles, answer these questions. Does the particle move to a region of higher or lower potential energy? Does the particle move to a region of higher or lower electric potential? K/U T/I (a) The particle is a proton. (b) The particle is an electron. 2. Two particles are at locations where the electric potential is the same. Do these particles have the same electric potential energy? Explain. K/U T/I 3. How much work is required to move a charge from one spot to another with the same electric potential? Explain your answer. K/U T/I 4. Two point particles with charges q C and q C have a potential energy of 40.0 J. Calculate the distance between the charges, in centimetres. K/U T/I 5. Two point particles with charges q C and q C are initially separated by a distance of 2.5 m. They are then brought closer together so that the final separation is 1.5 m. Determine the change in the electric potential energy. K/U T/I 6. Two point charges with charges q C and q C are initially very far apart. They are then brought together, with a final separation of 2.5 m. Calculate how much work it takes to bring them together. K/U T/I 7. A simple model of a hydrogen atom shows the electron and proton as point charges separated by a distance of m. Calculate how much work is required to break apart these two charges and to separate them by a very large distance when the electron is initially at rest. T/I A 8. A Van de Graaff generator used for classroom demonstrations has a conducting spherical dome with a radius of 15 cm. The electric potential produced by the charged generator is V near its surface. The sphere has a uniform distribution of charge, so you may assume that all charge on the sphere is concentrated at the sphere s centre. Assume that the electric potential at a large distance and at Earth s surface is zero. K/U T/I A (a) Calculate the charge on the sphere. (b) Calculate the magnitude of the electric field near the surface of the sphere. (c) In which direction does the electric field point? NEL 7.5 Electric Potential and Electric Potential Energy Due to Point Charges 361

45 7.6 The Millikan Oil Drop Experiment Physicists now know that each fundamental particle has a characteristic electric charge that does not change. In fact, the amount of charge is part of the definition of the kind of particle. At the start of the twentieth century, though, physicists still had many questions about the charge of fundamental particles. A big question was, is there a smallest unit of charge that nature will allow, and if so, what is the value of this charge? In this section, you will read about a brilliant experiment aimed at answering this question that resulted in the first measurement of the charge of the electron. Millikan s Experiment Figure 1 Robert Millikan fundamental physical constant a measurable natural value that never varies and can be determined by experimentation elementary charge (e) the magnitude of the electric charge carried by a proton, equal to the absolute value of the electric charge of an electron F E m q v b F g e Figure 2 A charged oil drop will be suspended between two charged plates when the electric force on the drop balances the gravitational force. r Physicist Robert Millikan (Figure 1) set out in 1909 to examine the existence of fundamental charge using a series of experiments. Millikan s work demonstrated that the electron is a fundamental particle with a unique charge. This electric charge is considered one of a few fundamental physical constants measurable values that can be determined by demonstration and do not vary that define natural laws. Millikan hypothesized that an elementary charge, e, the smallest unit of charge in nature, did exist, and that the charge of the electron equalled this elementary charge. To measure the charge, Millikan used a fine mist of oil droplets sprayed from an atomizer similar to what you may find on perfume bottles. The droplets picked up electric charges due to friction when sprayed from the atomizer. Millikan further hypothesized that the amount of charge any one drop picked up would be a wholenumber multiple of the fundamental charge. To measure the charge on a drop, Millikan used a device called an electrical microbalance. He allowed the oil drops to fall into a region between two oppositely charged parallel plates. The charges on the plates mean that there is an electric field in the space between the plates, which creates a potential difference between the top plate and the bottom plate. Millikan connected the plates to a series of adjustable batteries so that he could adjust the magnitude of the electric field and, therefore, the electric force on the droplets. By adjusting the electric force to balance the downward gravitational force, Millikan could bring a charged drop of oil to rest in the region between the plates (Figure 2). We can understand how Millikan used the electrical microbalance to determine the charge on an oil drop by comparing the electric and gravitational forces. For a drop of charge q, the electric force from the field e > is F > E 5 qe > If we assume that the charge is positive, then we can charge the plates so that the electric field points upward and gives an upward force to the drop. We can then carefully adjust the potential difference so that the falling drop comes to rest between the plates. When this happens, the electric force balances the gravitational force, and the magnitudes will be equal: F E 5 F g qe 5 mg In Section 7.4, you learned that the electric field between two charged plates depends on the potential difference V as e 5 DV Dd where Dd is the plate separation distance. Therefore, we can solve for the electric charge of the drop as q 5 mg e q 5 mgdd DV b where V b is the special value of potential difference that balances the drop. 362 Chapter 7 Electric Fields NEL