Chapter 21 Solutions Potential Energy o a Battery Description: The electric potential and potential energy associated with a battery. Learning Goal: To understand electrical potential, electrical potential energy, and the relationship between them. Electric potential and electric potential energy are related but dierent concepts. Be careul not to conuse the terms. Electrical potential energy U is the potential energy that a charge q has due to its position relative to other charges. The electric potential V at a speciic position is a measure o the amount o potential energy per unit charge a particle o net charge q would have at that position. In other words, i a charge q has an electric potential energy U, the electric potential Vat the location o qis. Recall that the gravitational potential energy (Ug mgy) o an object o mass m depends on where you deine y0. The dierence delta U in gravitational potential energy between two points is the physically relevant quantity. Similarly, or electric potential energy, the important quantity is the change in electric potential energy: delta U q delta V. This is why we oten just measure the potential dierence delta V. When we say that the potential o a car battery is 12 V, we mean that the potential dierence between the positive and negative terminals o the battery is 12 V. Consider dropping a ball rom rest. This ball moves rom a state o high gravitational potential energy to one o low gravitational potential energy as it alls to the ground. Similarly, charges move rom a state o high electric potential energy to one o low electric potential energy. Part A Mustang Sally just inished restoring her 1965 Ford Mustang car. To save money, she did not get a new battery. When she tries to start the car, she discovers that the battery is dead and she needs a jump start. While unhooking the jumper cables, the positive and negative cables almost touch and a spark jumps between the ends o the cables. This spark is caused by the movement o electrons through the air between the battery terminals. In what direction are the electrons traveling? Hint A.1 Another way to think about the movement o charge You can think o the movement o charges in terms o Coulomb's orce. A positive (high) potential is created by positive charges and a low (negative) potential is created by negative charges. To understand which way electrons will low across a potential dierence, think about the orces on an electron. An electron will be repelled by a negative charge and attracted to a positive charge. The negative terminal o a battery can be viewed as having a negative charge. The electrons are traveling rom terminal. The positive terminal is at a higher potential than the negative terminal. Unless provided with energy, positive charges will low rom a high to a low potential, and negatively charged electrons will low rom a low to a high potential. The table below summarizes this movement. Direction o motion high to low potential low to high potential Since potential dierence is the energy per unit charge, it is measured in units o energy divided by charge. Speciically, potential dierence is generally measured in volts (whose symbol is V). One volt is
equal to one joule per coulomb: 1V 1J / 1C. Part B There is a 12 V potential dierence between the positive and negative ends o the jumper cables, which are a short distance apart. An electron at the negative end ready to jump to the positive end has a certain amount o potential energy. On what quantities does this electrical potential energy depend? Hint B.1 The expression or electric potential energy The electric potential energy dierence is given by the potential dierence times the charge:. the distance between the ends o the cables the potential dierence between the ends o the cables the charge on the electron the distance and the potential dierence the distance and the charge the potential dierence and the charge the potential dierence, charge, and distance Part C Assume that two o the electrons at the negative terminal have attached themselves to a nearby neutral atom. There is now a negative ion with a charge -2e at this terminal. What are the electric potential and electric potential energy o the negative ion relative to the electron? The electric potential and the electric potential energy are both twice as much. The electric potential is twice as much and the electric potential energy is the same. The electric potential is the same and the electric potential energy is twice as much. The electric potential and the electric potential energy are both the same. The electric potential is the same and the electric potential energy is increased by the mass ratio o the oxygen ion to the electron. The electric potential is twice as much and the electric potential energy is increased by the mass ratio o the oxygen ion to the electron. Part D What is the electric potential energy o an electron at the negative end o the cable, relative to the positive end o the cable? In other words, assume that the electric potential o the positive terminal is 0 V and that o the negative terminal is -12 V. Recall that e1.602 X10^-19 C. Enter your answer numerically in joules.
Part E At the negative terminal o the battery the electron has electric potential energy. What happens to this energy as the electron jumps rom the negative to the positive terminal? It disappears. It is converted to kinetic energy. It heats the battery. It increases the potential o the battery. Just as gravitational potential energy is converted to kinetic energy when something alls, electrical potential energy is converted to kinetic energy when a charge goes rom a high potential energy state to a low potential energy state. Part F I you wanted to move an electron rom the positive to the negative terminal o the battery, how much work would you need to do on the electron? Hint F.1 Formula or work The work done on a charge q is equal to the product q delta V. Enter your answer numerically in joules. Because moving a negative charge rom the positive to the negative terminal o the battery would increase its electric potential energy, it would take positive work to move the charge. This is simliar to liting a ball upward. You do positive work on the ball to increase its gravitational potential energy. PSS 21.1 Conservation o energy in charge interactions Description: Knight/Jones/Field Problem-Solving Strategy 21.1 Conservation o energy in charge interactions. Learning Goal: To practice Problem-Solving Strategy 21.1 Conservation o energy in charge interactions. An alpha particle, which is the same as a helium-4 nucleus, is momentarily at rest in a region o space occupied by an electric ield. The particle then begins to move. Find the speed o the alpha particle ater it has moved through a potential dierence o 3.45 10 3 V. The charge and the mass o an alpha particle are q 3.20 10 19 C and m 6.68 10 27 kg, respectively. PROBLEM-SOLVING STRATEGY Conservation o energy in charge interactions PREPARE Draw a beore-and-ater visual overview. Deine symbols that will be used in the problem, list known values, and identiy what you're trying to ind. SOLVE The mathematical representation is based on the law o conservation o mechanical energy: K + qv Ki + q Vi. Find the electric potential at both the initial and inal positions. You may need to calculate it rom a known expression or the potential, such as that o a point charge. Ki and K are the total kinetic energies o all moving particles. Some problems may need additional conservation laws, such as conservation o charge or conservation o momentum. ASSESS Check that your result has the correct units, is reasonable, and answers the question.
Prepare Identiy the system involving the alpha particle. Then examine the orces acting within this system to determine whether mechanical energy is conserved. Make certain to draw a beore-and-ater visual overview and to deine your symbols. Part A Mechanical energy is conserved in the presence o which o the ollowing types o orces? Select all that apply. electrostatic magnetic rictional gravitational Gravitational, electrostatic, and magnetic orces are the only orces acting on the system involving the alpha particle, so the mechanical energy o the system is conserved. Note that in this problem you can ignore gravitational and magnetic orces because the electric orce is much larger. Part B Which o the ollowing quantities are unknown? Select all that apply. the initial speed o the alpha particle the dierence in potential between the initial and inal positions o the alpha particle the value o the electric potential at the initial position o the alpha particle the charge o the alpha particle the mass o the alpha particle the value o the electric potential at the inal position o the alpha particle the inal speed o the alpha particle The unknown quantity that you have to ind is the inal speed o the alpha particle, v. Although you do not know the electric potential at the initial and inal positions, you do know the dierence between these values, delta V V Vi 3.45 10 3 V. Beore you start creating the equation you will use to solve or v, irst draw a beore-and-ater visual overview. Solve Use conservation o energy K + qv Ki + q Vi to solve or the inal kinetic energy. Then use this value to solve or the inal velocity o the alpha particle. Part C What is the value o the change in potential energy, delta U U - Ui, o the alpha particle? Hint C.1 Relationship between electric potential energy and potential At a point where the electric potential is V, the electric potential energy U o a charged particle q is U qv. Hence, the change in electric potential energy is related to the change in electric potential:
Express your answer in joules using three signiicant igures. Part D What is the inal velocity o the alpha particle, v? Hint D.1 Identiy the initial kinetic energy What is the initial kinetic energy Ki o the alpha particle? Express your answer in joules. Hint D.2 Find the inal kinetic energy Use the mathematical representation o the law o mechanical energy to ind the value o the inal kinetic energy K o the alpha particle. Express your answer in joules using three signiicant igures. Hint D.3 Kinetic energy relationship The kinetic energy o a particle with mass m and speed v is K1/2 mv^2. Express your answer in meters per second using three signiicant igures. Assess Check to see that your inal result makes sense. Part E I you had carried out the algebra using variables beore plugging numbers into your expressions, you would have ound that
, where delta V is measured in volts. To veriy that this expression or v has the correct units o velocity, you need to perorm some unit analysis. Begin by inding the equivalent o a volt in terms o basic SI units. What is a volt in terms o meters (m ), seconds (s), kilograms (kg), and coulombs (C)? Express your answer using the symbols or the units meters, seconds, kilograms, and coulombs. volt A volt is equivalent to a kg m^2/(c s^2). I you had known this beore using the expression given here or v, you could have carried out a unit analysis to determine that your expression would give you the desired units o velocity (m/s ). This process would look like Electric Potential and Potential Energy Description: Short quantitative problem on the motion o a charged particle along a line o varying potential and a line o constant potential. Based on Young/Geller Quantitative Analysis 18.2. A particle with charge 1.60 10 19 C is placed on the x axis in a region where the electric potential due to other charges increases in the +x direction but does not change in the y or z direction. Part A The particle, initially at rest, is acted upon only by the electric orce and moves rom point a to point b along the x axis, increasing its kinetic energy by 3.20 10 19 J. In what direction and through what potential dierence Vb Va does the particle move? Hint A.1 How to approach the problem Because no orces other than the electric orce act on the particle, the positively charged particle must move in the direction parallel to the electric ield, and the ield must do positive work on the particle. Recall that when the electric ield does positive work on a charged particle, the potential energy o the particle decreases. Thus, the particle must move in the direction in which its potential energy decreases (which is consistent with the act that the particle's kinetic energy increases as it moves rom a to b). Moreover, rom the deinition o potential and the energy conservation equation, you can directly calculate the potential dierence Va - Vb. Hint A.2 Electric potential The electric potential Vat any point in an electric ield is the electric potential energy U per unit charge associated with a test charge q at that point:. Hint A.3 Find the change in potential energy o the particle What is the change in potential energy o the particle, Ub Ua, as it moves rom a to b? Hint A.3.1 Energy conservation
Recall that the total mechanical energy (kinetic plus potential) is conserved. That is, Ka + Ua Kb + Ub, where the subscripts reer to points a and b, and K and U are the corresponding kinetic and potential energies. Hint A.3.2 Find the change in kinetic energy o the particle What is the change in kinetic energy o the particle, Kb - Ka, as it moves rom a to b? Recall that particle is initially at rest, and its kinetic energy at b is 3.20 10 19 J. Express your answer in joules. Express your answer in joules. The particle moves to the let through a potential dierence o Vb Va 2.00 V. The particle moves to the let through a potential dierence o Vb Va - 2.00 V The particle moves to the right through a potential dierence o Vb Va 2.00 V. The particle moves to the right through a potential dierence o Vb Va - 2.00 V. The particle moves to the let through a potential dierence o Vb Va 20.0 V. The particle moves to the right through a potential dierence o Vb Va - 20.0 V. In general, i no orces other than the electric orce act on a positively charged particle, the particle always moves toward a point at lower potential. Part B I the particle moves rom point b to point c in the y direction, what is the change in its potential energy, Uc - Ub? Hint B.1 How to approach the problem Recall that the electric potential increases in the +x direction but does not change in the y or z direction. 3.20 10 19 0 3.20 10 19 Every time a charged particle moves along a line o constant potential, its potential energy remains constant and the electric ield does no work on the particle.
Electric Potential Ranking Task Description: Short conceptual problem involving electrical potentials o point charges. (ranking task) In the igure there are two point charges, +q and -q. There are also six positions, labeled A through F, at various distances rom the two point charges. You will be asked about the electric potential at the dierent points (A through F). Part A Rank the locations A to F on the basis o the electric potential at each point. Rank positive electric potentials as higher than negative electric potentials. Hint A.1 Deinition o electric potential The electric potential surrounding a point charge is deined by, where q is the source charge creating the electric potential and r is the distance between the source charge and the point o interest. I more than one source is present, determine the electric potential rom each source and sum the results. Hint A.2 Conceptualizing electric potential Because positive charges create positive electric potentials in their vicinity and negative charges create negative potentials in their vicinity, electric potential is sometimes visualized as a sort o "elevation." Positive charges represent mountain peaks and negative charges deep valleys. In this picture, when you are close to a positive charge, you are "high up" and have a higher positive potential. Conversely, near a negative charge, you are deep in a "valley" and have a negative potential. The utility o this picture becomes clearer when we begin to think o charges moving through a region o space containing an electric potential. Just as particles naturally roll downhill, converting gravitational potential energy into kinetic energy, positively charged particles naturally "roll downhill" as well, toward regions o lower electric potential, converting electrical potential energy into kinetic energy.
Rank the locations rom highest to lowest potential. To rank items as equivalent, overlap them. View Change in Electric Potential Ranking Task Description: Short conceptual problem related to the electric potential dierence between pairs o points. (ranking task) In the diagram below, there are two charges o +q and q and six points (a through ) at various distances rom the two charges. rank changes in the electric potential along paths between pairs o points. Part A You will be asked to Using the diagram to the let, rank each o the given paths on the basis o the change in electric potential. Rank the largest-magnitude positive change (increase in electric potential) as largest and the largest-magnitude negative change (decrease in electric potential) as smallest. Hint A.1 Change in electric potential Determining the change in electric potential along some path involves determining the electric potential at the two end points o the path, and subtracting: Hint A.2 Determine the algebraic sign o the change in potential The path rom point d to point a results in a positive change in electric potential. Which o the other paths also involves a positive change in electric potential (i.e., electric potential that increases along the path)?
rom b to a rom to e rom c to d rom c to e rom c to b Hint A.3 Conceptualizing changes in electric potential Since positive charges create large positive electric potentials in their vicinity and negative charges create negative potentials in their vicinity, electric potential is sometimes visualized as a sort o elevation. Positive charges represent mountain peaks and negative charges deep valleys. In this picture, when you are close to a positive charge, you are high up and have a large positive potential. Conversely, near a negative charge you are deep in a valley and have a negative potential. Thus, changes in electric potential can be thought o as changes in elevation. The change is positive i you are moving uphill and the change is negative i you move downhill. The arther you travel either uphill or downhill, the larger the magnitude o the change in electric potential. Rank rom largest to smallest. To rank items as equivalent, overlap them. View Electric Fields and Equipotential Suraces Description: Find the work done to move a unit charge rom and to given points on a diagram showing equipotential suraces, and compare the magnitude o the electric ield at these points. The dashed lines in the diagram represent cross sections o equipotential suraces drawn in 1 V
increments. Part A What is the work Wab done by the electric orce to move a 1 C charge rom A to B? Hint A.1 Find the potential dierence between A and B What is the potential dierence Va Vb between point A and point B? Hint A.1.1 Equipotential suraces Recall that an equipotential surace is a surace on which the electric potential is the same at every point. Express your answer in volts. Hint A.2 Potential dierence and work Recall that the potential dierence (in volts) between a point a and a point b equals the work (in joules) done by the electric orce to move a 1 C charge rom a to b. Express your answer in joules. Part B What is the work Wad done by the electric orce to move a 1 C charge rom A to D? Hint B.1 Find the potential dierence between A and D What is the potential dierence Vd Va between point A and point D?
Express your answer in volts. Hint B.2 Potential dierence and work Recall that the electric potential energy dierence between any two points is equal to the negative o the work done by the electric orce as a charged object moves between those two points. I we combine this with the relationship between electric potential energy and electric potential we have: Express your answer in joules. Part C The magnitude o the electric ield at point C is Hint C.1 Electric ield and equipotential suraces Since the diagram shows equal potential dierences between adjacent suraces, equal amounts o work are done to move a particular charge rom one surace to the next adjacent one. It ollows then that i the equipotentials are closer together, the electric orce does the same amount o work in a smaller displacement than i the equipotentials were arther apart. Thereore, the electric orce, as well as the corresponding electric ield, has a larger magnitude. greater than the magnitude o the electric ield at point B. less than the magnitude o the electric ield at point B. equal to the magnitude o the electric ield at point B. unknown because the value o the electric potential at point C is unknown. Capacitance: A Review Description: Multiple-choice questions on concept o capacitance. Use ater Introduction to Capacitance. Some questions are repetitions o those in other problems. This act may get some students rustrated. Learning Goal: To review the meaning o capacitance and ways o changing the capacitance o a parallel-plate capacitor. Capacitance is one o the central concepts in electrostatics. Understanding its meaning and the dierence between its deinition and the ways o calculating capacitance can be challenging at irst. This tutorial is meant to help you become more comortable with capacitance. Recall the undamental ormula or capacitance:, where C is the capacitance in arads, Q is the charge stored on the plates in coulombs, and V is the potential dierence (or voltage) between the plates. In the ollowing problems it may help to keep in
mind that the voltage is related to the strength o the electric ield E and the distance between the plates, d, by V Ed. Part A What property o objects is best measured by their capacitance? the ability to conduct electric current the ability to distort an external electrostatic ield the ability to store charge Capacitance is a measure o the ability o a system o two conductors to store electric charge and energy. It is deined as C Q/V. This ratio remains constant as long as the system retains its geometry and the amount o dielectric does not change. Capacitors are special devices designed to combine a large capacitance with a small size. However, any pair o conductors separated by a dielectric (or vacuum) has some capacitance. Even an isolated electrode has a small capacitance. That is, i a charge Q is placed on it, its potential V with respect to ground would change, and the ratio Q/V is its capacitance C. Part B Consider an air-illed charged capacitor. How can its capacitance be increased? Hint B.1 What does capacitance depend on? Capacitance depends on the inherent properties o the system o conductors, such as its geometry and the presence o dielectric, not on the charge placed on the conductors. Speciically, capacitance depends on the area A o the conducting plates and the distance d between the plates and is given by where is a constant called the permittivity o ree space., Increase the charge on the capacitor. Decrease the charge on the capacitor. Increase the spacing between the plates o the capacitor. Decrease the spacing between the plates o the capacitor. Increase the length o the wires leading to the capacitor plates. Part C Consider a charged parallel-plate capacitor. How can its capacitance be halved? Check all that apply. Double the charge. Double the plate area. Double the plate separation. Halve the charge. Halve the plate area. Halve the plate separation.
Part D Consider a charged parallel-plate capacitor. Which combination o changes would quadruple its capacitance? Double the charge and double the plate area. Double the charge and double the plate separation. Halve the charge and double the plate separation. Halve the charge and double the plate area. Halve the plate separation and double the plate area. Double the plate separation and halve the plate area. Video Tutor: Charged Conductor with Teardrop Shape Description: A teardrop-shaped conductor is charged, and its blunt and pointed ends are touched by electrodes o equal surace area. Which electrode acquires a greater charge? First, launch the video below. You will be asked to use your knowledge o physics to predict the outcome o an experiment. Then, close the video window and answer the question at right. You can watch the video again at any point. Part A Two conducting spheres are each given a charge Q. The radius o the larger sphere is three times greater than that o the smaller sphere. I the electric ield just outside o the smaller sphere is E0, then the electric ield just outside o the larger sphere is Hint A.1 How to approach the problem. The electric ield just outside o a conductor is proportional to the conductor's surace charge density. The surace charge density o a sphere is calculated by dividing the total charge on the sphere by the sphere's surace area. Think about how changing the radius o a sphere changes the sphere's surace area. 9
3 1/9 1/3 The larger sphere has nine times the surace area o the smaller one, and this reduces the surace charge density by a actor o nine. Questions Q21.21. Reason: The entire top plate is at the same potential as the positive terminal o the battery and the bottom plate is at the same potential as the negative terminal o the battery (assuming ideal wires). Each capacitor plate is an equipotential surace, so the potential is the same anywhere on the top plate as it is at a. The potential dierence is the same between any point on the top plate and any point on the bottom plate: 6 V. The correct choice is A. Assess: The angle o the dotted line is completely irrelevant. The conclusion to this question would be valid even i the plates were not lat or parallel. Q21.22. Reason: Since the direction o the electric ield is in the direction o decreasing electric potential, the electric ield is directed rom the 300 V to the 100 V electric potential. Since positive charges move in the direction o the electric ield, the charge will move rom the site o the 300 V to the site o the 100 V electric potential. The correct choice is A. Assess. This question points out two important things: First, the relationship between electric potential dierence and the electric ield and second, the relationship between the charge on an object and the orce it experiences due to an electric ield. Q21.23. Reason: By examining the 0Vequipotential line and the100 V equipotential line we see that each line is 50 V dierent rom an adjacent equipotential line. To get rom the 0Vequipotential line to point C requires us to go in the opposite direction rom the positive equipotentials, so our answer must be negative. Point C is two lines over rom the 300 V equipotential line, so the potential at C must be 400 V. The correct choice is A. Assess: This can be thought o as a contour map showing elevation and the reasoning would lead to the same conclusion. Q21.24. Reason: Knowing the dierence in electric potential V and the change in position d or two equipotential lines, we can determine the electric ield near the midpoint between these two equipotential lines by E V/ d,(i the ield is nonuniorm, the smaller d, the more accurate the value o E). Using the equipotential lines on either side o the point o interest, we get or each case that Va Vb V c 100 V. We can do an eyeball comparison o each d (the separation between the two equipotential lines on either side o the point o interest) to get da db dc, which allows us to write Ec Eb Ea. The correct choice is C. Assess: This question emphasizes the relationship between electric potential dierence and electric ield. Q21.25. Reason: While the spacing o the equipotential lines varies over the entire igure, the spacing is reasonably constant near point C. The electric ield vector will point down the hill rom higher potential to lower potential, and it will be perpendicular to the equipotential line.
The strength o the electric ield is indicated by how close the lines are together. We must make a scale measurement on the igure. Do this with a ruler, or simply take a piece o paper, put it next to either the x- or y- axis, and make a couple o marks on it showing a scaled distance o, say, 0.5 m. Then put the paper over point C perpendicular to the equipotential line with the two marks straddling it and count the number o equipotential lines. There are about eight lines (corresponding to 400 V) in the 0.5 m distance. The ield strength is E ( V)/ d 400 V/(0 5 m) 800 V/m. The correct choice is C. Assess: The equipotentials above point C are spaced arther apart than those below point C, but the average around C is easily obtained this way, especially since we aren t looking or a lot o signiicant igures. Q21.27. Reason: Looking at the equipotential lines on either side o point B tells us that each equipotential line is 50 V dierent rom an adjacent equipotential line. Point A is one equipotential line on the negative side o the 0Vequipotential, so the potential at A is 50 V. From Question 23 we know the potential at C is 400 V, so VCA VA V C 50 V ( 400 V) 350 V. The work done is equal to the change in electric potential energy. 6 W Uelec q V ( 10 nc)(350 V) 3 5 10 J The correct choice is A. 6 Assess: 3 5 10 J is not very much but the charge is small as well, so this is a reasonable answer. Notice the actual distance between points A and C is not important as long as we know the potential dierence between them. Also note the units: C V J. PROBLEMS P21.1. Prepare: The work done is equal to the change in electric potential energy: W Uelec q V. We are given that V 150 V 300 V 150 V. Solve: Solve the irst equation or q: q Ur kq 6 2 9 2 2 10 8 2 / 1 ( 4.0 10 J)(10 m)/(9.0 10 N m / C )(4.4 10 C) 1.0 10 C Assess: The answer is negative because it requires positive work to move a negative charge to a lower potential. As or units, J/V C. P21.3. Prepare: The work done is equal to the change in electric potential energy: W Uelec q V. Solve the equation or V: V U /. q elec Solve: As the charge is moved rom A to B, its electric potential changes by V A B ( Uelec) A B 3 0 J q 15 nc Thus B is at a higher potential than A. When the charges is moved rom C to B, its electric potential changes by V C B ( Uelec) C B 5 0 J q 15 nc 200 V 330 V This means B is at a lower potential than C. Thus i the charge were moved in the opposite direction, rom B to C, its electric potential would increase, so that VB C V C B 330 V. The total change in electric potential in moving rom A to C is then C A AC AB BC 200 V ( 330 V) 530 V. V V V V V Assess: The potential at B is between the potential at A and the potential at C. So i we move the charge rom A to B and then to C, it will take a total work o 3.0 J ( 5.0 J) 8.0 J.
V AC ( Uelec) AC 8.0 J q 15 nc 530 V P21.8. Prepare: Energy is conserved. The potential energy is determined by the electric potential. The igure shows a beore-and-ater pictorial representation o a He ion moving through a potential dierence. The ion s initial speed is zero and its inal speed is 1.0 10 6 m/s. A positive charge speeds up as it moves into a region o lower potential (U K). Solve: The potential energy o charge q is U qv. Conservation o energy, expressed in terms o the electric potential V, is K qv K qv q( V V ) K K V i i i i 4(1.67 10 kg)(1.0 10 m/s) 27 6 2 19 2(1.60 10 C) K K q 1 i 2 4 2.1 10 V 0J mv q Assess: This result implies that the helium ion moves rom a higher potential toward a lower potential. P21.9. Prepare: Energy is conserved. The potential energy is determined by the electric potential. The igure shows a beore-and-ater pictorial representation o an electron moving through a potential dierence. 2 Solve: (a) Because the electron is a negative charge and it slows down as it travels, it must be moving rom a region o higher potential to a region o lower potential. (b) Using the conservation o energy equation, (c) 1 1 1 2 K U Ki Ui K qv Ki qvi V Vi ( Ki K ) mv i 0 J q ( e) 2 V mv (9.11 10 kg)(5.0 10 m/s) 2e 2(1.60 10 C) 2 31 5 2 i 19 1 2 Ki mv i q V ( e )( 0.712 V) 0.7 ev 2 0.712 V 0.7 V Assess: The negative sign with V veriies that the electron moves rom a higher potential region to a lower potential region. P21.18. Prepare: Please reer to Figure P21.18. The net potential is the sum o the potentials due to each charge given by Equation 21.10. Solve: The potential at the dot is V 9 9 9 1 q1 1 q2 1 q3 9 2 2 2.0 10 C 2.0 10 C 2.0 10 C (9.0 10 N m /C ) 1400 V 4 r 4 r 4 r 0.040 m 0.050 m 0.030 m 0 1 0 2 0 3
Assess: Potential is a scalar quantity, so we ound the net potential by adding three scalar quantities. P21.19. Prepare: The electric potential dierence between the plates is determined by the uniorm electric ield in the parallel-plate capacitor as given by Equation 21.6. Solve: (a) The potential o an ordinary AA or AAA battery is 1.5 V. Actually, this is the potential dierence between the two terminals o the battery. I the electric potential o the negative terminal is taken to be zero, then the positive terminal is at a potential o 1.5 V. (b) I a battery with a potential dierence o 1.5 V is connected to a parallel-plate capacitor, the potential dierence between the two capacitor plates is also 1.5 V. Thus, V C 1.5 V V V Ed where d is the separation between the two plates. The electric ield inside a parallel-plate capacitor is Q Q A E d Q A A d 2 2 12 2 2 (1.5 V)( 0) (1.5 V) (2.0 10 m) (8.85 10 C /N m ) 12 1.5 V 8.3 10 C 3 0 0 2.0 10 m Thus, the battery moves 8.3 10 12 C o electron charge rom the positive to the negative plate o the capacitor. Assess: This is the charge on the positive plate. The other plate has a charge o 8.3 10 12 C. P21.20. Prepare: Please reer to Figure P21.20. In a region that has a uniorm electric ield, Equation 21.17 gives the magnitude o the potential dierence between two points. Solve: (a) The electric ield points downhill. So, point A is at a higher potential than point B. (b) The magnitude o the potential dierence between points A and B is V Exd (1000 V/m)(0.07 m) 70 V That is, the potential at point A is 70 V higher than the potential at point B. Assess: Electric ield points rom higher potential to lower potential. P21.28. Prepare: Knowing that the relationship between the capacitance o the capacitor, the charge on each plate o the capacitor, and the electric potential dierence across the plates o the capacitor is C Q/ V, we can determine the capacitance o the capacitor (which will not change), the charge on the plates or the second case, and then the dierence in charge. Solve: 6 6 C Q / V 6.0 10 C/3.0 V 2.0 10 F Q 1 1 CV 2 2 Q Q Q 2 1 6 6 (2.0 10 F)(5.0 V) 10 10 C 6 6 6 10 10 C 6.0 10 C 4.0 10 C 4.0 C Assess: Considering the other charges, this is a reasonable value. P21.35. Prepare: Equation 21.22 gives the capacitance o a parallel-plate capacitor with a dielectric (the paper). We are given A 2 2 (0 35 m) 0 1225 m and d C d 0 A 3 0 25 10 m. look up the dielectric constant o paper in Table 21.2: paper 3 0. Solve: C 0A d 12 2 (3 0)(8 85 10 F/m)(0 1225 m ) 3 0 25 10 m Assuming the science air is at 20 C, we also 13 nf 8 Assess: The answer could be expressed in scientiic notation as1 3 10 F but capacitances are usually given in pf (sometimes pronounced pu ), nf, or F. In our calculation the m 2 cancels, leaving F.
P21.39. Solve: From Equation 21.23, the energy stored in a capacitor is 1 2 2U UC C ( VC ) C 2(1.0 J) VC 6 2 C 1.0 10 F Assess: This potential dierence is not unusual or capacitors. 1400 V (P21.57. Prepare: Outside a charged sphere the electric potential is identical to that o a point charge at the center and is given by Equation 21.11. Solve: (a) For a proton, assumed to be a point charge, the electric potential is V 19 1 ( e) 9 2 2 1.60 10 C 9 0 r (9.0 10 N m /C ) 27.2 V 27 V 4 0.053 10 m (b) The potential energy o a charge q at a point where the potential is V is U electron in the proton s potential is U ( 1.60 10 19 C) 27.2 V) 4.4 10 18 J qv. The potential energy o the P21.62. Prepare: Equation 21.17 gives the relation between the electric ield and the potential dierence: E V/ d. We are given in the text that V 0 070V and in the problem that Solve: E V d d 9 8 10 m. 0 070 V 8 75 10 6 V/m 8 8 10 6 V/m 9 8 10 m Assess: This rounds to 10 7 V/m to one signiicant igure. This is very large considering that a spark in air requires a ield o only about 3 10 6 V/m P21.63. Prepare: Equation 21.1 gives the relation between the electric potential energy and the potential dierence: Uelec q V. For a Na ion q 1, e V (0 V) ( 70 mv) 70 mv. Solve: Uelec Uelec q V ( 1e)(70 mv) 0 070 ev Because is positive the energy increases. Assess: The ion is pushed uphill in the ion pump by energy derived rom ATP. P21.64. Prepare: I there is an electric potential dierence V across a cell membrane o thickness d, the electric ield in the membrane is E V/ d. This electric ield exerts a orce on a charged particle o magnitude F qe q V/ d. Solve: The orce on the molecular ion is 19 2 9 11 F q V/ d (10 e) V/ d (10)(1.6 10 C)(7 10 V)/5.0 10 m 2. 2 10 N Assess: We expect the orce to be small, because the electric ield and charge are both small. P21.66. Prepare: Since the only orces in this problem are the orces between the alpha particle and the antiproton, the total momentum o these two is conserved as they accelerate toward one another. In addition, the sum o their kinetic energies and the potential energy must be constant.
Solve: The conservation o momentum may be written as: and the conservation o energy may be written as: m ( v ) i mp ( vp) i m ( v ) mp ( v p) 1 1 K( e)(2 e) 1 1 K( e)(2 e) m ( v ) m ( v ) m ( v ) m ( v ) 2 2 2 2 r 2 2 2 2 i p p i p p ri The let hand sides o both equations are zero because the initial velocities are zero and the initial separation is great which means that the potential energy term is practically zero. I we solve the momentum equation, mp ( vp) 0 m ( v ) mp ( v p), or ( v ), we get ( v ). Now i this expression is plugged into the energy m equation, with the let hand side set to zero, we get: 1 mp ( vp) 1 2 2Ke 0 m mp ( vp) 2 m 2 r 2 2 9 2 2 19 2 4 Ke ( m ) 4(8.99 10 N m / C )(1.60 10 C) (4) p 27 9 mp ( mp m ) r 5(1.67 10 kg)(2.5 10 m) 2 ( v ) 13.3 km/s In the last equation, we used the act that ( v ) m ( v ) p p m : m 4m. Now we can ind the velocity o the alpha particle using p 1(13.3 km/s) ( v ) 3.3 km/s 4 The minus sign tells us that the alpha particle and antiproton are traveling in opposite directions. Assess: It makes sense that the alpha particle is traveling slower than the antiproton, considering that it is more massive. P21.71. Prepare: From conservation o energy, the increase in kinetic energy o the protons equals their decrease in potential energy. In other words, K U q V. In the ollowing igure, the source o the potential dierence is a capacitor, with the negatively charged plate having a hole cut in the center to allow protons through. Notice that while the protons are between the plates, they are accelerating, but once they pass through the plate on the right, they maintain a constant speed. This is because there is no electric ield outside an ideal capacitor. Solve: I we solve the equation or q, we get:
q K V K V 0.10 J 6 10 10 V 8 1.0 10 C 10 nc Here we have used the act that since the protons are accelerating, they must be traveling through a decreasing potential. Consequently, V is negative and V V. Assess: Since the protons have a charge o e, their kinetic energy ater passing the potential dierence is 19 1.60 10 J 6 12 10 MeV (10 10 ev) 1.60 10 J. 1 ev the number o protons, N : protons is Since the total energy was 0.10 J, we can determine 12 10 N (0.10 J)/(1.60 10 J) 6.3 10. As a check, the charge o this quantity o 10 19 (6.3 10 )(1.60 10 C) 10 nc, as we ind in the statement o the problem. P21.73. Prepare: The proton is ired rom a distance much greater than the nuclear diameter, so r i and U i 0 J. Because the nucleus is so small, a proton that is even a ew atoms away is, or all practical purposes, at ininity. As the proton approaches the nucleus, it is slowed by the repulsive electric orce. At the end point, the proton has just reached the surace o the nucleus (r nuclear diameter) with v 0 m/s. (The proton won t remain at this point but will be pushed back out again, but the subsequent motion is not part o this problem.) Initially, the proton has kinetic energy but no potential energy. At the point o closest approach, where v 0 m/s, the proton has potential energy but no kinetic energy. Energy is conserved. Because the iron nucleus is very large compared to the proton, we will assume that the nucleus does not move (no recoil) and that the proton is essentially a point particle with no diameter. Solve: Because energy is conserved, K U K i U i. This equation is ( e)(26 e) 1 2 0 J mprotonvi 0 J 4 r 2 where r is hal the nuclear diameter. The initial speed o the proton is 19 19 9 2 2 2( e)(26 e) 2(1.6 10 C)(26 1.6 10 C)(9.0 10 N m /C ) i 27 15 4 0rm proton (1.67 10 kg)(4.5 10 m) v 0 7 4.0 10 m/s Assess: Extremely large deceleration o the proton occurs as the proton is brought to rest momentarily. P21.80. Prepare: The energy stored in a capacitor is related to the physical size o the capacitor and the electric ield in the capacitor by U AdE 2 /2. Solve: The energy stored in the air-illed capacitor may be determined by 2 12 2 2 2 4 6 2 4 Uair 0 AdE / 2 (8.85 10 C / (N m ))(0.15 m) (5.0 10 m)(3.0 10 V/m) / 2 4.5 10 J The energy stored in the Telon-illed capacitor may be determined by 2 12 2 2 2 4 6 2 UTelon AdE / 2 (2 8.85 10 C / (N m ))(0.15 m) (5.0 10 m)(60 10 V/m) / 2 0.36 J Assess: The Telon-illed capacitor can store 800 times the amount o energy as the air-illed capacitor.