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OpenStax College Physics Instructor Solutions Manual Chapter 18 CHAPTER 18: ELECTRIC CHARGE AND ELECTRIC FIELD 18.1 STATIC ELECTRICITY AND CHARGE: CONSERVATION OF CHARGE 1. Common static electricity involves charges ranging from nanocoulombs to microcoulombs. (a) How many electrons are needed to form a charge of –2.00 nC? (b) How many electrons must be removed from a neutral object to leave a net charge of 0.500 C ? Solution Q 2.00 10 9 C 1.25 1010 electrons (a) N 19 qe 1.60 10 C (b) N 0.500 10 6 C 3.13 1012 electrons 19 1.60 10 C 2. If 1.80 10 20 electrons move through a pocket calculator during a full day’s operation, how many coulombs of charge moved through it? Solution Q Nqe 1.80 10 20 1.60 10 19 C 28.8 C 3. To start a car engine, the car battery moves 3.75 10 21 electrons through the starter motor. How many coulombs of charge were moved? Solution Q Nqe 3.75 10 21 1.60 10 19 C 600 C 4. A certain lightning bolt moves 40.0 C of charge. How many fundamental units of charge qe is this? Solution Q N qe N Q 40.0 C 2.50 10 20 19 qe 1.60 10 C 18.2 CONDUCTORS AND INSULATORS 5. Suppose a speck of dust in an electrostatic precipitator has 1.0000 1012 protons in it and has a net charge of –5.00 nC (a very large charge for a small speck). How many electrons does it have? OpenStax College Physics Instructor Solutions Manual Chapter 18 Q 5.00 10 9 C 1.0000 1012 1.03 1012 qe 1.60 10 19 C Solution Q N p N e q e N e N p 6. An amoeba has 1.00 1016 protons and a net charge of 0.300 pC. (a) How many fewer electrons are there than protons? (b) If you paired them up, what fraction of the protons would have no electrons? Solution (a) Q N p N e q e N p N e (b) N p Ne Np Q 0.300 10 12 C 1.875 10 6 1.88 10 6 19 qe 1.60 10 C 1.875 10 6 1.88 10 10 16 1.00 10 7. A 50.0 g ball of copper has a net charge of 2.00 C . What fraction of the copper’s electrons has been removed? (Each copper atom has 29 protons, and copper has an atomic mass of 63.5.) Solution The number of moles of copper is n m 50.0 g so the number of protons, N p , A 63.5 g/mol is N p nN A 29 protons/at om 50.0 gm 6.02 10 23 atoms 29 protons 1.375 10 25 protons mol 63.5 g/mol cu atom Since there is the same number of electrons as protons in a neutral atom, before we remove the electrons to give the copper a net charge, we have 1.375 10 25 electrons. Next we need to determine the number of electrons we removed to leave a net charge of 2.00 C . We need to remove 2.00 C of charge, so the number of electrons to be removed is given by Q 2.00 10 6 C N e,removed = 1.25 1013 electrons removed. 19 qe 1.60 10 C Lastly, the fraction of copper’s electrons removed is given by: N e, removed 1.25 1013 9.09 10 13 N e,initially 1.375 10 25 8. What net charge would you place on a 100 g piece of sulfur if you put an extra electron on 1 in 1012 of its atoms? (Sulfur has an atomic mass of 32.1.) OpenStax College Physics Solution Instructor Solutions Manual Chapter 18 The number of sulphur atoms, N s , is: 100 g m 6.02 10 23 1.875 10 24 N A M 32.1g Nq 1.875 10 24 1.60 10 19 C Q s 12e 3.00 10 7 C 12 10 10 N s nN A 9. How many coulombs of positive charge are there in 4.00 kg of plutonium, given its atomic mass is 244 and that each plutonium atom has 94 protons? Solution Given Z= atomic number of an element = number of protons in elements: m Q nN A q e Z N A q e Z M 4.00 10 3 gm 6.02 10 23 /mol 1.6 10 19 C 94 1.48 10 8 C 244 gm/mol 18.3 COULOMB’S LAW 10. Solution What is the repulsive force between two pith balls that are 8.00 cm apart and have equal charges of – 30.0 nC? F k q1 q 2 9.00 10 9 N m 2 /C 2 30.0 10 9 C r2 0.0800 m 2 2 1.266 10 3 N 1.27 10 3 N 11. Solution (a) How strong is the attractive force between a glass rod with a 0.700 C charge and a silk cloth with a 0.600 C charge, which are 12.0 cm apart, using the approximation that they act like point charges? (b) Discuss how the answer to this problem might be affected if the charges are distributed over some area and do not act like point charges. 6 6 (a) F k q1 q 2 9.00 10 9 N m 2 /C 2 0.700 10 C 0.600 10 C r2 0.120 m2 0.2625 N F 0.263 N (b) Assuming the center of mass is separated by 12.0 cm, the answer in part (a) is only true if the charges are concentrated at the center of mass. If, however, the charges are distributed over some area, there will be a concentration of charge along the side closest to the oppositely charged object. This effect will increase the net force. OpenStax College Physics 12. Solution Instructor Solutions Manual Chapter 18 Two point charges exert a 5.00 N force on each other. What will the force become if the distance between them is increased by a factor of three? 2 2 r qq qq F r F1 k 1 2 2 , F2 k 1 22 , 2 1 2 1 or F1 r2 r1 r2 r2 2 r 1 F2 F1 1 5.00 N 0.556 N 3 r2 2 13. Two point charges are brought closer together, increasing the force between them by a factor of 25. By what factor was their separation decreased? Solution Using F1 k q1 q 2 r1 2 we see that force is inversely proportional to the separation distance squared, so that F1 k q1 q 2 2 , F2 k q1 q 2 2 r1 r2 Since we now the ratio of the forces, we can determine the ratio of the separation 2 F r r F1 1 1 . The separation decreased by a distances: 1 2 so that 2 F2 r1 r1 F2 25 5 factor of 5. 14. How far apart must two point charges of 75.0 nC (typical of static electricity) be to have a force of 1.00 N between them? Solution qq kq q F k 1 22 r 1 2 r F 1/ 2 9.00 10 9 N m 2 /C 2 75.0 10 9 C 2 1.00 N 15. Solution 1/ 2 7.12 10 3 m 7.12 mm If two equal charges each of 1 C each are separated in air by a distance of 1 km, what is the magnitude of the force acting between them? You will see that even at a distance as large as 1 km, the repulsive force is substantial because 1 C is a very significant amount of charge. q1q 2 9.0 10 9 N m 2 /C 2 1 C2 F k 2 9 10 3 N 2 3 r 110 m OpenStax College Physics 16. Instructor Solutions Manual Chapter 18 A test charge of 2 C is placed halfway between a charge of 6 C and another of 4 C separated by 10 cm. (a) What is the magnitude of the force on the test charge? (b) What is the direction of this force (away from or toward the 6 C charge)? Solution 9 2 2 6 6 (a) F2 4 9.00 10 N m /C 4 10 C 2 10 28.8 N 0.05 m 2 F2 6 9.00 10 9 N m 2 /C 2 6 10 6 C 2 10 6 0.05 m 2 43.2 N Ftot F2 6 F2 4 43.2 N 28.8 N 14.4 N 10 N (b) The direction of the force is away from the 6 μC charge. 17. Bare free charges do not remain stationary when close together. To illustrate this, calculate the acceleration of two isolated protons separated by 2.00 nm (a typical distance between gas atoms). Explicitly show how you follow the steps in the ProblemSolving Strategy for electrostatics. Solution F k q1 q 2 ma r2 kq 2 9.00 10 9 N m 2 /C 2 1.60 10 19 C a 2 mr 2 1.67 10 27 kg 2.00 10 9 m 2 3.45 1016 m/s 2 18. (a) By what factor must you change the distance between two point charges to change the force between them by a factor of 10? (b) Explain how the distance can either increase or decrease by this factor and still cause a factor of 10 change in the force. Solution r F1 r 1 F r 2 10 3.16 (a) F 2 1 2 , so that 2 r F2 r1 r1 F2 r1 (b) If the distance increases by 3.16, then the force will decrease by a factor of 10; if the distance decreases by 3.16, then the force will increase by a factor of 10. Either way, the force changes by a factor of 10. 19. Suppose you have a total charge qtot that you can split in any manner. Once split, the separation distance is fixed. How do you split the charge to achieve the greatest force? 2 OpenStax College Physics Instructor Solutions Manual Chapter 18 Solution q1q2 q1 qtot q1 q1qtot q1 . This represents a quadratic equation in q1 , which would graph as a parabola with its peak midway between the two values that cause q1 q tot q1 0 , namely q1 0 and 0 q tot q tot q1 q tot . Thus, the maximum value occurs at q1 . Therefore, splitting 2 2 the charge evenly produces the greatest force. 20. (a) Common transparent tape becomes charged when pulled from a dispenser. If one piece is placed above another, the repulsive force can be great enough to support the top piece’s weight. Assuming equal point charges (only an approximation), calculate the magnitude of the charge if electrostatic force is great enough to support the weight of a 10.0 mg piece of tape held 1.00 cm above another. (b) Discuss whether the magnitude of this charge is consistent with what is typical of static electricity. Solution 2 (a) F k q1 q 2 kq mg r2 r2 2 0.0100 m 2 10.0 10 6 kg 9.80 m/s 2 r 2 mg 9 q 1.04 10 C. 9 2 2 9.00 10 N m /C k (b) This charge is approximately 1nC, which is consistent with the magnitude of charge of typical static electricity. 1/ 2 21. Solution 1/ 2 (a) Find the ratio of the electrostatic to gravitational force between two electrons. (b) What is this ratio for two protons? (c) Why is the ratio different for electrons and protons? (a) Since FE kq1q2 kq2 GMm Gm2 FE kq2 and F 2 , . G r2 r2 r2 r FG Gm2 For electrons me 9.11 10 31 kg, so that 2 FE 9.00 10 9 N m 2 /C 2 1.60 10 19 C FG 6.67 10 11 N m 2 /kg 9.11 10 31 kg (b) For protons, m p 1.67 10 27 kg, so that 2 4.16 10 42 2 FE 9.00 10 9 N m 2 /C 2 1.60 10 19 C 1.24 10 36 2 11 2 27 FG 6.67 10 N m /kg 1.67 10 kg (c) The ratio is different for electrons and protons because the masses are different. 22. At what distance is the electrostatic force between two protons equal to the weight of one proton? OpenStax College Physics Solution Instructor Solutions Manual kq2 kq2 m g r p m g r2 p 1/ 2 Chapter 18 9.00 10 9 N m 2 /C 2 1.60 10 19 C2 1.67 10 27 kg 9.80 m/s 2 1/ 2 0.119 m 23. A certain five cent coin contains 5.00 g of nickel. What fraction of the nickel atoms’ electrons, removed and placed 1.00 m above it, would support the weight of this coin? The atomic mass of nickel is 58.7, and each nickel atom contains 28 electrons and 28 protons. Solution The number of electrons, N e , in the nickel is 5.00 g m 6.02 10 23 1.436 10 24 . N e 28nN A 28 N A 28 M 58.7 g Assume the number of electrons removed from the nickel is N. Then the charge of the removed electrons is Nqe , and the charge of the remaining positively charged nickel is kq1 q 2 k Nqe Nqe . Note that there is a minus r2 r2 sign in front of mg because the attractive force between the electrons and the positively charged nickel will be equal and opposite in direction to the gravitational mgr 2 mgr 2 2 force. Continuing N , so that kq2e kq2e Nq p Nqe . Thus, F mg 5.00 10 3 kg 9.80 m/s 2 1.00 m2 N 2 9.00 10 9 N m 2 /C 2 1.60 10 19 C N 1.458 1013 1.02 10 11 Therefore, 24 N E 1.436 10 24. Solution 1/ 2 1.458 1013 (a) Two point charges totaling 8.00 C exert a repulsive force of 0.150 N on one another when separated by 0.500 m. What is the charge on each? (b) What is the charge on each if the force is attractive? (a) F kqQ q Fr 2 Fr 2 2 2 so that qQ q , q Qq 0 k r2 k OpenStax College Physics Instructor Solutions Manual Fr 2 0.150 N 0.500 m 4.167 10 12 C 2 9 2 2 k 9.00 10 N m /C 2 a 1; b Q 8.00 10 6 C ; c b b 2 4ac q 2 8.00 10 6 c 4.00 10 6 8.00 10 C 3.44 10 6 6 Chapter 18 C 2 4 4.167 10 12 C 2 2 C C, and Q q 0.56 10 6 C or q 7.44 106 C and Q q 0.56 106 C 7.44 10 (b) F 6 kqQ q Fr 2 Fr 2 so that , 0, qQ q 2 and q 2 qQ 2 r k k a 1; b 8.00 10 6 C; c 4.167 10 12 C 2 q b b 2 4ac 2 8.00 10 6 C 6 6 2 C 4 4.167 10 12 C 2 2 C 12.98 10 6 C 6 C 8.98 10 6 C, Q q 4.98 10 6 C. 4.00 10 13.0 10 8.00 10 25. Point charges of 5.00 C and 3.00 C are placed 0.250 m apart. (a) Where can a third charge be placed so that the net force on it is zero? (b) What if both charges are positive? Solution (a) We know that since the negative charge is smaller, the third charge should be placed to the right of the negative charge if the net force on it is to be zero. So if qq we want Fnet F1 F2 0 , we can use F k 1 2 2 to write the forces in terms of r distances: q1 q2 Kq1q2 Kq2q 2 2 0 , or since r1 0.250 m d and r2 d , Kq r12 r22 r1 r2 5 106 C 3 106 , or d 3 0.250 m 3 d so that 2 2 d 0.250 m d 5 5 OpenStax College Physics Instructor Solutions Manual Chapter 18 3 0.250 m 3 3 5 d 1 0.859 m 5 0.250 m and finally, d 5 3 1 5 The charge must be placed at a distance of 0.859 m to the far side of the negative charge. (b) This time we know that the charge must be placed between the two positive charges and closer to the 3 C charge for the net force to be zero. So if we want qq Fnet F1 F2 0 , we can again use F k 1 2 2 to write the forces in terms of r Kq q Kq2 q q q Kq 12 22 0 distances: 12 2 2 r2 r1 r2 r1 Or since r1 = 0.250m r2 5 10 6 C 0.250 r2 2 3 10 6 r2 2 , or r2 2 3 0.250 m r2 2 , or 5 3 0.250 m 3 5 0.109 m 0.250 m r2 , and finally r2 r2 3 5 1 5 The charge must be placed at a distance of 0.109m from the 3 C charge. 26. Two point charges q! and q2 are 3.00 m apart, and their total charge is 20 C . (a) If the force of repulsion between them is 0.075 N, what are magnitudes of the two charges? (b) If one charge attracts the other with a force of 0.525 N, what are the magnitudes of the two charges? Note that you may need to solve a quadratic equation to reach your answer. Solution (a) Q1 Q2 20 C, r 3.0 m kQ Q F 12 2 0.075 N r 2 2 r F 3.0 m 0.075 N Q1Q2 7.5 10 11 C 2 9 2 2 k 9 10 N m / C 6 Q1 Q2 20 10 C Q1 20 10 6 C Q1 7.5 10 11 C 2 Q12 Q1 20 10 6 7.5 10 11 0 a 1, b 20 10 6 , c 7.5 10 11 Use the quadratic formula: Q1 15 C, 5 C and Q2 20 μC Q1 5 C or 15 C So the two possibilities are: Q1 15 C and Q2 5 C or Q1 5 C and Q2 15 C r 2 F 3.0 m 0.525 N 5.25 10 10 C 2 9 2 2 k 9 10 N m / C 2 (b) Q1Q2 OpenStax College Physics Instructor Solutions Manual Chapter 18 Same method as part (a): Q1 Q2 20 10 6 C Q1 20 10 6 C Q1 5.25 10 10 C 2 Q12 Q1 20 10 6 5.25 10 10 0 a 1, b 20 10 6 , c 5.25 10 10 Use the quadratic formula. The two possibilities are: Q1 15 C and Q2 35 C or Q1 35 C and Q2 - 15 C 18.4 ELECTRIC FIELD: CONCEPT OF A FIELD REVISITED 27. What is the magnitude and direction of an electric field that exerts a 2.00 105 N upward force on a 1.75 C charge? Solution F 2.00 10 5 N E 11.4 N/C downward q 1.75 10 6 C 28. What is the magnitude and direction of the force exerted on a 3.50 C charge by a 250 N/C electric field that points due east? Solution E 29. Calculate the magnitude of the electric field 2.00 m from a point charge of 5.00 mC (such as found on the terminal of a Van de Graaff). Solution 30. Solution 31. E F F qE 3.50 10 6 C 250 N/C 8.75 10 4 N q kQ 9.00 10 9 N m 2 /C 2 5.00 10 3 C 1.13 10 7 N/C 2 r2 2.00 m (a) What magnitude point charge creates a 10,000 N/C electric field at a distance of 0.250 m? (b) How large is the field at 10.0 m? (a) E kQ r 2 E 0.250 10000 N/C Q 6.94 10 8 C 2 9 2 2 k r 9.00 10 N m /C (b) E kQ 9.00 10 9 N m 2 /C 2 6.94 10 8 C 6.25 N/C 2 r2 10.0 m 2 Calculate the initial (from rest) acceleration of a proton in a 5.00 106 N/C electric field (such as created by a research Van de Graaff). Explicitly show how you follow the steps in the Problem-Solving Strategy for electrostatics. OpenStax College Physics Instructor Solutions Manual Chapter 18 Solution qE 1.60 10 19 C 5.00 10 6 N/C F ma qE a 4.79 1014 m/sec 2 27 m 1.67 10 kg 32. (a) Find the direction and magnitude of an electric field that exerts a 4.80 10 17 N westward force on an electron. (b) What magnitude and direction force does this field exert on a proton? Solution F 4.80 10 17 C 300 N/C (eas t) q 1.60 10 19 C (b) F qE 1.60 10 19 C 300 N/C 4.80 10 17 N (east) (a) E 18.5 ELECTRIC FIELD LINES: MULTIPLE CHARGES 33. (a) Sketch the electric field lines near a point charge q . (b) Do the same for a point charge 3.00q . Solution (a) +q (b) – 3q 34. Sketch the electric field lines a long distance from the charge distributions shown in Figure 18.26(a) and (b). OpenStax College Physics Instructor Solutions Manual Chapter 18 Solution (a) – 2q (b) field of two opposite charges No net charge is seen from far away. 35. Solution 36. Figure 18.47 shows the electric field lines near two charges q1 and q2 . What is the ratio of their magnitudes? (b) Sketch the electric field lines a long distance from the charges shown in the figure. q1 23 1.9 : 1 q2 12 (b) A long distance away, the two charges appear as one charge, with total charge q q1 q2 (23 12)e 11e , where e is some basic unit of charge. The field will look like that of one of the point charges in Figure 18.23, but with 11 lines directed away. (a) Sketch the electric field lines in the vicinity of two opposite charges, where the negative charge is three times greater in magnitude than the positive. (See Figure 18.47 for a similar situation.) Solution 18.7 CONDUCTORS AND ELECTRIC FIELDS IN STATIC EQUILIBRIUM OpenStax College Physics 37. Instructor Solutions Manual Chapter 18 Sketch the electric field lines in the vicinity of the conductor in Figure 18.48, given the field was originally uniform and parallel to the object’s long axis. Is the resulting field small near the long side of the object? Solution – – – – – –– – + – + ++ + + + + + Yes, the field is smaller near the long side of the object. 38. Sketch the electric field lines in the vicinity of the conductor in Figure 18.49 given the field was originally uniform and parallel to the object’s long axis. Is the resulting field small near the long side of the object? Solution –– – –– – – – – + + ++ + + + + + Yes, the field is smaller near the long side of the object. 39. Solution 40. Sketch the electric field between the two conducting plates shown in Figure 18.50, given the top plate is positive and an equal amount of negative charge is on the bottom plate. Be certain to indicate the distribution of charge on the plates. + + + + + ++ – – – – – – – Sketch the electric field lines in the vicinity of the charged insulator in Figure 18.51 noting its nonuniform charge distribution. Solution + + + 41. + + + + + + ++ + + What is the force on the charge located at x 8.00 cm in Figure 18.52(a) given that q 1.00 μC ? OpenStax College Physics Solution FK Instructor Solutions Manual q3q8 qq K 82 11 2 r 3, 8 r 8,11 Chapter 18 2 1.00 10 6 C 2 2 1.00 10 6 C 2 9.00 10 N.m /C 2 0.0300 m 2 0.0500 m 12.8 N to the right. 9 2 2 42. Find the total electric field at x 1.00 cm in Figure 18.52(b) given that q 5.00 nC . (b) Find the total electric field at x 11.00 cm in Figure 18.52(b). (c) If the charges are allowed to move and eventually be brought to rest by friction, what will the final charge configuration be? (That is, will there be a single charge, double charge, etc., and what will its value(s) be?) Solution (a) Since E K q , and there is negative point charge at x 1.00 cm , the electric field r2 at x 1.00 cm will be negative infinity: E x 1.00cm q q q q (b) F K 2 1 K 2 5 K 2 8 K 2 14 and since q 5.00 109 C r 1,11 r 5,11 r 8,11 r 11,14 10.0 10 9 C 5.00 10 9 C 2 0.0600 m 2 9.00 10 9 N m 2 0.100 m E 2.12 10 5 N/C 2 9 9 C 15.0 10 C 5.00 10 C 2 0.0300 m 2 0.0300 m (c) The two charges will come together to be a charge of q and the two right charges will come together to form one charge of q , so eventually the charges will come together to form one charge of q . 43. (a) Find the electric field at x 5.00 cm in Figure 18.52(a), given that q 1.00μC . (b) At what position between 3.00 and 8.00 cm is the total electric field the same as that for 2q alone? (c) Can the electric field be zero anywhere between 0.00 and 8.00 cm? (d) At very large positive or negative values of x, the electric field approaches zero in both (a) and (b). In which does it most rapidly approach zero and why? (e) At what position to the right of 11.0 cm is the total electric field zero, other than at infinity? (Hint: A graphing calculator can yield considerable insight in this problem.) Solution q3 q8 q11 (a) E K 2 2 2 r3,5 r5,8 r5,11 9.00 10 9 N m 2 C2 4.00 10 7 N/C 1.00 10 6 C 2.00 10 6 C 1.00 10 6 C 2 0.0300 m 2 0.0600 m 2 0.0200 m OpenStax College Physics Instructor Solutions Manual Chapter 18 (b) For the total electric field to have the same value as it would be for the 2q charge alone, we need to find the place where the two positive charges alone create a zero electric field. This point is halfway between the two points or at x 7.00 cm. (c) No, the electric field cannot be zero anywhere between 0.00cm and 8.00cm. (d) At very large positive or negative values of x, the electric field approaches zero because the distance from the charges approaches infinity. The electric field approaches zero more rapidly in Figure 18.52 because the net charge is zero. (e) For a position greater than 11.0 cm, we must find a value of x that satisfies: 1 2 1 where x has units of centimeters. We can then y 2 2 x 3 x 8 x 112 graph this function, using a graphing calculator or graphing program, to determine the values of x that yield y 0. From the graph we find that the electric field is zero at x 30.6 cm. 44. (a) Find the total Coulomb force on a charge of 2.00 nC located at x 4.00 cm in Figure 18.52(b), given that q 1.00μC . (b) Find the x-position at which the electric field is zero in Figure 18.52(b). Solution (a) The net force is given by q q q Kq q Kq q Kq q Kq q q F 21 25 28 14 Kq 12 25 82 142 , to the right. 2 r1 r5 r8 r14 r14 r1 r5 r8 Substituting in the values given: 9.00 10 9 N.m 2 2.00 10 9 C F 2 C 2.00 10 6 C 1.00 10 6 C 3.00 10 6 C 2 2 2 0.0500 m 0.0400 m 0.0800 m 0.0400 m 0.0400 m 0.0100 m 1.00 10 6 C 2 0.140 m 0.0400 m = - 0.252 N(right) or 0.252 N to the left (b) The only possible location where the total electric field could be zero is between 5.00 and 8.00 cm. For the total electric field to be zero between 5.00 and 8.00 cm, Kq Kq Kq Kq we know that E 2 1 2 5 2 8 214 0 . r1 r5 r8 r14 Dividing by common factors and ignoring units (but remembering x has a unit of cm), we can get a simplified expression: 2 1 3 1 y 2 2 2 x 1 x 5 x 8 x 142 OpenStax College Physics Instructor Solutions Manual Chapter 18 We can graph this function, using a graphing calculator or graphing program to determine the values of x that yield y 0 Therefore the total electric field is zero at 6.07 cm. 45. (a) Using the symmetry of the arrangement, determine the direction of the force on q in the figure below, given that q a qb 7.50 μC and qc qd 7.50 μC . (b) Calculate the magnitude of the force on the charge q , given that the square is 10.0 cm on a side and q 2.00 μC . Solution (a) Due to symmetry the net force on q will be straight down, since q a and q b are positive and q c and q d are negative and all have the same magnitude. q a and q b will force the charge downward and q c and q d will pull the charge downward also. (b) Since the square is 10.0 cm on a side, the distance to the charge q will be 10.0 cm 2 7.071 cm so 2 6 6 qqa 8.99 109 N m 2 2.00 10 C 7.50 10 C Fa k 2 26.97 N 2 r C2 0.07071 m 45 degrees below x-axis. The total force will be: Fy 4 Fa sin 45 4 26.97 N sin 45 76.3 N (downward) 46. (a) Using the symmetry of the arrangement, determine the direction of the electric field at the center of the square in Figure 18.53, given that q a qb 1.00 μC and qc qd 1.00 μC . (b) Calculate the magnitude of the electric field at the location of q , given that the square is 5.00 cm on a side. Solution (a) The electric field at the center of the square will be straight up, since q a and q b are positive and q c and q d are negative and all have the same magnitude. (b) Since the square is 5.00cm on a side, the distance to the center of the square is 5.00 cm Q 2 3.536 cm , so using E k 2 , the electric field due to upper left 2 r hand charge is: q 9.00 10 9 N m 2 /C 2 1.00 10 6 C E a K 2a 7.20 10 6 N/C, 2 r 0.03536 m 45 degrees below x-axis, or Ea 7.20 106 N/C, 135 above x-axis. The other three fields will have the same magnitude, only in different directions. From part (a) we know that the only component that we need to calculate is the y- OpenStax College Physics Instructor Solutions Manual Chapter 18 component, since the x-component will cancel. The total electric field at the center will then be: E y 4 Ea sin 45 4 7.20 106 N/C sin 135 2.04 107 N/C(upward ) 47. Find the electric field at the location of q a in Figure 18.53 given that qb qc qd 2.00 nC , q 1.00 nC , and the square is 20.0 cm on a side. Solution q 9.00 10 9 N m 2 /C 2 2.00 10 9 C E a K 2a 450 N/C in the x-dir; Ea = 450N/C 2 r 0.200 m in the +y direction. q 9.00 10 9 N m 2 /C 2 2.00 10 9 C Ed K d2 225 N/C, 135 above the x-axis. 2 r 2 0.200 m q 9.00 10 N m /C Eq K 2 r 9 2 2 1.00 10 9 C 2 0.100 m 2 450 N/C, 45 below the x-axis. Summing gives: E x E b cos 180 E c cos 90 E d cos 135 E q cos 45 450 N/C 0 159 N/C 318 N/C 291 N/C and E y E b sin 180 E c sin 90 E d sin 135 E q sin 45 0 450 N/C 159 N/C 318 N/C 291 N/C, So the electric field at the location of q is E 291 N/C2 291 N/C2 412 N/C, 45 degrees above the x-axis. 48. Find the total Coulomb force on the charge q in Figure 18.53, given that q 1.00 μC , qa 2.00 μC , qb 3.00 μC , qc 4.00 μC , and qd 1.00 μC . The square is 50.0 cm on a side. Solution 0.500 m r 2 0.3536 m. 2 6 qa q C 1.00 10 6 C 9 2 2 1.00 10 Fd k 2 9.00 10 N m /C 0.0720 N, 45 r 0.3536 m 2 above the x-axis. Fb 3 Fd 0.216 N, 45 above the +x-axis, Fc 4 Fd 0.288 N, 45 below the -x axis, and Fa 2 Fd 0.144 N, 45 below the +x-axis. Summing up the components gives: OpenStax College Physics Instructor Solutions Manual Chapter 18 Fx Fa cos 45 Fb cos 45 Fc cos 135 Fd cos135 0.102 N 0.153 N 0.204 N 0.0509 N 0.00 N. Fy Fa sin 45 Fb sin 45 Fc sin 135 Fd sin 135 0.102 N 0.153 N 0.204 N 0.0509 N 0.102 N So the total coulomb force on the charge q is 0.102 N, in the y direction . 49. (a) Find the electric field at the location of qa in Figure 18.54, given that qb 10.00 μC and qc 5.00 μC . (b) What is the force on qa , given that qa 1.50 nC ? Solution 6 qb C 9 2 2 10.0 10 1.44 10 6 N/C, 60 (a) E b k 2 9.00 10 N m /C 2 r 0.250 m above the -x-axis and 6 qc C 9 2 5.00 10 Ec k 2 9.00 10 N m 7.2 10 5 N/C, 60 below the –x-axis. 2 r 0.250 m Summing up the components gives: E x E b cos 120 E c cos 120 7.20 10 5 N/C 3.60 10 5 N/C 1.08 10 6 N/C, and E y E b sin 120 E c sin 120 1.25 10 5 N/C, So the total electric field at the location of qa is 1.25 10 6 N/C and 6.24 105 N/C 30.0 above the –x-axis, or E 1.25 10 6 N/C, 30.0 6 1.08 10 N/C above the –x-axis. tan 1 (b) Fa qa E 1.50 10 9 C 1.25 10 6 N/C 1.88 10 3 N, 30.0 above the –x-axis. 50. (a) Find the electric field at the center of the triangular configuration of charges in Figure 18.54, given that qa 2.50 nC , qb 8.00 nC , and qc 1.50 nC . (b) Is there any combination of charges, other than qa qb qc , that will produce a zero strength electric field at the center of the triangular configuration? Solution (a) To determine the electric field at the center, we first must determine the distance from each of the charges to the center of the triangle. Since the triangle is equilateral, the center of the triangle will be halfway across the base and 1/3 of OpenStax College Physics Instructor Solutions Manual Chapter 18 the way up the height. To determine the height use the Pythagorean theorem, or the height is given by h 25.0 cm 12.5 cm 21.7 cm . So the distance from each charge to the center of the triangle is 2/3 of 21.7 cm, or 2 Q r 21.7 cm 14.4 cm. Since E k 2 , 3 r 2.50 10 9 C q 1085 N/C at a 90 angle Ea k 2a 9.00 10 9 N m 2 /C 2 2 r 0 . 144 m below the horizontal, 8.00 10 9 C q 3472 N/C at a 30 angle E b k 2b 9.00 10 9 N m 2 /C 2 2 r 0.144 m below the horizontal, and 2 C 681.0 N/C at a 30 angle 0 . 144 m above the horizontal. Adding the vectors by components gives: q Ec k 2c 9.00 10 9 N m 2 /C 2 r 1.50 10 2 9 2 E x E a cos 90 E b cos 30 E c cos 30 E x 0 N/C 3472 N/C 0.860 681.0 N/C 0.8660 3597 N/C E y E a sin 90 E b sin 30 E c sin 30 E y 1085 N/C - 3472 N/C 0.5000 681 N/C 0.5000 2481 N/C So that the electric field is given by: E Ex E y 2 2 3597 N/C2 2481 N/C2 4370 N/C and 2481 N/C 34.6 , or E 4.37 10 3 N/C, 34.6 Ex 3597 N/C below the horizontal. tan 1 Ey tan 1 (b) No, there are no combinations other than qa q b qc that will produce a zero strength field at the center of the triangular configuration because of the vector nature of the electric field. OpenStax College Physics Instructor Solutions Manual Chapter 18 18.8 APPLICATIONS OF ELECTROSTATICS 51. Solution (a) What is the electric field 5.00 m from the center of the terminal of a Van de Graaff with a 3.00 mC charge, noting that the field is equivalent to that of a point charge at the center of the terminal? (b) At this distance, what force does the field exert on a 2.00 μC charge on the Van de Graaff’s belt? (a) E kQ 9.00 10 9 N m 2 /C 2 3.00 10 3 C 1.08 10 6 N/C 2 2 r 5.00 m (b) F qE 2.00 10 6 C1.08 10 6 N/C 2.16 N 52. Solution (a) What is the direction and magnitude of an electric field that supports the weight of a free electron near the surface of Earth? (b) Discuss what the small value for this field implies regarding the relative strength of the gravitational and electrostatic forces. (a) F mg qE E mg 9.11 10 31 kg 9.80 m/s 2 5.58 10 11 N/C 19 q 1.60 10 C Electric field is towards the surface of earth. (b) The coulomb force is extraordinarily stronger than gravity. 53. A simple and common technique for accelerating electrons is shown in Figure 18.55, where there is a uniform electric field between two plates. Electrons are released, usually from a hot filament, near the negative plate, and there is a small hole in the positive plate that allows the electrons to continue moving. (a) Calculate the acceleration of the electron if the field strength is 2.50 104 N/C . (b) Explain why the electron will not be pulled back to the positive plate once it moves through the hole. Solution (a) F ma qE qE 1.60 10 19 C 2.50 10 4 N/C a m 9.11 10 31 kg 4.391 1015 m/sec 2 4.39 1015 m/sec 2 (b) There is no field outside the plates. 54. Earth has a net charge that produces an electric field of approximately 150 N/C downward at its surface. (a) What is the magnitude and sign of the excess charge, noting the electric field of a conducting sphere is equivalent to a point charge at its center? (b) What acceleration will the field produce on a free electron near Earth’s surface? (c) What mass object with a single extra electron will have its weight supported by this field? OpenStax College Physics Solution Instructor Solutions Manual Chapter 18 KQ r2E 6.37 10 6 150 N/C Q 6.76 10 5 C 2 9 2 2 K r 9.00 10 N m /C F ma qE (b) qE 1.60 10 19 C 150 N/C a 2.63 1013 m/s 2 (upwards) 31 m 9.11 10 kg (a) E 2 (c) F mg qE m qE 1.60 10 19 C 150 N/C 2.45 10 18 kg 2 g 9.80 m/s 55. Point charges of 25.0 C and 45.0 C are placed 0.500 m apart. (a) At what point along the line between them is the electric field zero? (b) What is the electric field halfway between them? Solution 9 9 K q1 Kq1 9 1 9 5 5 , so that 2 x d x, (a) q 2 q1 , E1 E 2 2 2 2 5 5 x x d x d x And finally solving for x gives: d 0.500 m x 0.214 m from 25.0 C charge. 9 9 1 1 5 5 (b) Kq1 Kq2 4 K q 2 q1 4 9.00 10 9 N m 2 /C 2 20.0 10 6 C E E1 E 2 2 2 d2 0.500 m 2 d d 2 2 E 2.88 106 N/C towards the larger charge. 56. What can you say about two charges q1 and q 2 , if the electric field one-fourth of the way from q1 to q2 is zero? Solution If the electric field is zero 1/4 from the way of q1 and q 2 , then we know from q 3x Q Kq Kq2 so that 2 2 9 E k 2 that E1 E2 21 2 q1 x x r 3x 2 The charge q2 is 9 times larger than q1 . 57. Integrated Concepts Calculate the angular velocity of an electron orbiting a proton in the hydrogen atom, given the radius of the orbit is 0.530 10 10 m . You may assume that the proton is stationary and the centripetal force is supplied by Coulomb attraction. OpenStax College Physics Solution Fc Instructor Solutions Manual kq 2 mv 2 m(r ) 2 mr 2 , so that 3 r r mr kq1 q 2 r2 Chapter 18 9.00 10 9 N m 2 / C 2 1.60 10 19 C 2 3 31 10 9.11 10 kg 0.530 10 m 12 (where q1 q 2 q ) 12 4.12 1016 rad s 58. Integrated Concepts An electron has an initial velocity of 5.00 10 6 m/s in a uniform 2.00 105 N/C strength electric field. The field accelerates the electron in the direction opposite to its initial velocity. (a) What is the direction of the electric field? (b) How far does the electron travel before coming to rest? (c) How long does it take the electron to come to rest? (d) What is the electron’s velocity when it returns to its starting point? Solution (a) The field is in the direction of the electron’s initial velocity. 2 (b) v 2 v02 2ax x v0 v 0. Also, F ma qE a qE 2a m 2 v2m 5.00 10 6 m s 9.11 10 31 kg x 0 3.56 10 4 m 2qE 2 1.60 10 19 C 2.00 10 5 N C v0 v0 m v 0 (c) v v0 at t a qE t 5.00 10 m s9.11 10 1.60 10 C2.00 10 6 19 (d) v 2ax 12 2qEx m 31 5 kg 1.42 10 10 s NC 12 2 1.60 10 19 C 2.00 10 5 N C 3.56 10 4 m 9.11 10 31 kg 6 5.00 10 m/s (opposite its initial velocity) 12 59. Integrated Concepts The practical limit to an electric field in air is about 3.00 10 6 N/C . Above this strength, sparking takes place because air begins to ionize and charges flow, reducing the field. (a) Calculate the distance a free proton must travel in this field to reach 3.00% of the speed of light, starting from rest. (b) Is this practical in air, or must it occur in a vacuum? Solution (a) F qE ma a qE v2 v 0 0 , and v 2 v02 2ax x m 2a 0.0300 3.00 108 m s 1.67 10 27 kg 0.141 m v2m x 2qE 2 1.60 10 19 C 3.00 10 6 N C (b) No, it is impractical in air. A proton would collide with an air molecule before it could travel this far. 2 OpenStax College Physics 60. Instructor Solutions Manual Chapter 18 Integrated Concepts A 5.00 g charged insulating ball hangs on a 30.0 cm long string in a uniform horizontal electric field as shown in Figure 18.56. Given the charge on the ball is 1.00 C , find the strength of the field. Solution Let T tension force in the string. Tsin FE qE Tcos mg tan E 61. Solution qE mg mgtan 5.00 10 3 kg 9.80 m s 2 tan 8.00 6.89 103 N C 6 q 1.00 10 C Integrated Concepts Figure 18.57 shows an electron passing between two charged metal plates that create an 100 N/C vertical electric field perpendicular to the electron’s original horizontal velocity. (These can be used to change the electron’s direction, such as in an oscilloscope.) The initial speed of the electron is 3.00 106 m/s , and the horizontal distance it travels in the uniform field is 4.00 cm. (a) What is its vertical deflection? (b) What is the vertical component of its final velocity? (c) At what angle does it exit? Neglect any edge effects. y x + + + + vf d v0 – – – L – OpenStax College Physics Instructor Solutions Manual qE m (a) F ma y qE a y L v0 x t t L 1 1 , and since d v0 y t a y t 2 d a y t 2 v0 y 0 v0 x 2 2 1 1 qE L d ayt 2 2 2 m v0 x 2 1.60 10 C 100 N C0.0400m 29.11 10 kg 3.00 10 m s v a t v a t v 0 19 31 (b) v y 0y y v y at Chapter 18 y 2 6 y 0y 2 1.56 10 3 m aL qEL 1.60 10 19 C 100 N C0.0400m 3.00 106 m s9.11 1031 kg v0 x v0 x m 2.342 105 m s 2.34 105 m s (c) tan vy 2.342 105 m s 4.46 tan 1 tan 1 6 v0 x v 3 . 00 10 m s 0 x vy 62. Integrated Concepts The classic Millikan oil drop experiment was the first to obtain an accurate measurement of the charge on an electron. In it, oil drops were suspended against the gravitational force by a vertical electric field. (See Figure 18.58.) Given the oil drop to be 1.00 m in radius and have a density of 920 kg/m 3 : (a) Find the weight of the drop. (b) If the drop has a single excess electron, find the electric field strength needed to balance its weight. Solution Let = density of oil drop, and V = volume of oil drop. 3 4 4 (a) W mg Vg r 3 g 1.00 10 6 m 920 kg m 3 9.80 m s 2 3 3 14 3.777 10 N 3.78 10 14 N (b) W FE qE E W 3.777 10 14 N 2.36 10 5 N C q 1.60 10 19 C 63. Integrated Concepts (a) In Figure 18.59, four equal charges q lie on the corners of a square. A fifth charge Q is on a mass m directly above the center of the square, at a height equal to the length d of one side of the square. Determine the magnitude of q in terms of Q , m , and d , if the Coulomb force is to equal the weight of m . (b) Is this equilibrium stable or unstable? Discuss. Solution (a) Because of symmetry, the only force that does not cancel is the vertical OpenStax College Physics Instructor Solutions Manual Chapter 18 component of the force, so we need only calculate the vertical component effect of one of the corner charges and multiply by four to get the total force. The distance from any corner charge q to the charge Q is 2 2 3 d d r d2 d 2 2 2 qQ 2 qQ , k r2 3 d 2 acting along the line from q to Q. So, the component of the force acting in the d 2 vertical direction is: Fv , q F sin , where sin , so that the total force 3 3 d 2 2 2 qQ qQ in the vertical direction is given by Fv 4 F sin 4 k 2 2.177k 2 . 33 d d Now, if this force is balanced by the force of gravity, then Fv Fg and Therefore, the magnitude of the force due to one charge is: F k 2 2.177k 64. Solution mgd qQ . mg so that q 0.459 2 kQ d Unreasonable Results (a) Calculate the electric field strength near a 10.0 cm diameter conducting sphere that has 1.00 C of excess charge on it. (b) What is unreasonable about this result? (c) Which assumptions are responsible? kQ 9.00 10 9 N m 2 C 2 1.00 C 9.00 1011 N C 2 2 r 0.100 m (b) The field is too large to be in air. (c) 1.00 C is too great for an excess charge on the sphere's surface. (a) E 65. Unreasonable Results (a) Two 0.500 g raindrops in a thunderhead are 1.00 cm apart when they each acquire 1.00 mC charges. Find their acceleration. (b) What is unreasonable about this result? (c) Which premise or assumption is responsible? Solution (a) F ma kq1 q 2 r2 2 9.00 10 9 N m 2 C 2 1.00 10 3 C kq 2 2 1.80 1011 m s 2 2 3 mr 0.500 10 kg 0.0100 m (b) The resulting acceleration is unreasonably large; the raindrops would not stay together. (c) The assumed charge of 1.00 mC is much too great; typical static electricity is on the order of 1 C or less. a OpenStax College Physics 66. Solution Instructor Solutions Manual Chapter 18 Unreasonable Results A wrecking yard inventor wants to pick up cars by charging a 0.400 m diameter ball and inducing an equal and opposite charge on the car. If a car has a 1000 kg mass and the ball is to be able to lift it from a distance of 1.00 m: (a) What minimum charge must be used? (b) What is the electric field near the surface of the ball? (c) Why are these results unreasonable? (d) Which premise or assumption is responsible? (a) F mg kq 2 mg , so that q r 2 r k 12 1.04310 3 C 1.10 10 6 C 1000 kg 9.80 m s 2 1.00 m 9 2 2 9.00 10 N m C kQ 9.00 10 9 N m 2 C 2 1.10 10 6 C 2.45 10 5 N/C r2 (0.200 m) 2 (c) It would be impractical to place that much excess charge on a wrecking ball. Even so, the necessary E-field strength is 245 MV/m , much too great to be created in air. (d) It is unreasonable that one can lift heavy objects with static electricity. (b) E This file is copyright 2016, Rice University. All Rights Reserved.