Chapter 17 ELECTRIC POTENTIAL

Transcription

1 Chapter 17 ELECTRIC POTENTIAL Conceptual Questions 1. (a) The electric fiel oes positive work on q as it moves closer to +Q. (b) The potential increases as q moves closer to +Q. (c) The potential energy of q ecreases. () If the fixe charge instea has a value Q, the electric fiel oes negative work, the potential ecreases, an the potential energy increases.. Such a capacitor can be built by replacing the air between the capacitor plates with a ielectric material. This change not only increases the maximum possible voltage across the capacitor but also increases the amount of charge on the capacitor plates for a given potential ifference. 3. While staning on a high voltage wire, the magnitue of a bir s electric potential varies between 100 kv an +100 kv. Important for the bir is the fact that although its boy is at a non-zero potential, the potential ifference across its boy is small. If a large potential ifference existe across its boy, the bir woul be electrocute. 4. A positive charge in an electric fiel moves towar a position of lower potential. A negative charge in this situation moves towar a position of higher potential. 5. Zero work is require to move a charge between two points at the same potential. An external force may nee to be applie to move the charge but the work one to start the charge in motion will be negate by the work one to stop it. 6. If the charge of a point particle is negative, its electric potential energy ecreases as it is move towars a region of higher electric potential. 7. If all parts of a conuctor in electrostatic equilibrium were not at the same potential, electric fiels woul exist within the conuctor an charges woul not remain stationary. The assumption of electrostatic equilibrium woul therefore be invali. 8. There is no physical significance to zero potential only potential ifferences have physical consequences. The potential of the earth is often taken to be zero an therefore an object that is groune has zero potential. This is only a reference value however an the potential of the earth coul be taken to be any other quantity as long as other values were appropriately offset by the same potential. 9. If the electric fiel is zero throughout a region of space, the electric potential must be constant throughout that region. 10. The woman s hea has acquire a net charge. The microscopic charges (electrons or ions) istribute themselves so as to maximize their separation from one another, as a result of their mutual repulsion. This is why the charges move out onto the woman s hair, which then spreas out in response to the repulsive electrical forces. The charge strans of hair orient themselves parallel to the electric fiel lines, which emanate raially outwar from the woman s hea, as though it were a charge conucting sphere. 11. If the potential is constant throughout a region of space, the electric fiel must be zero throughout that region. 68

2 College Physics Chapter 17: Electric Potential 1. If a uniform electric fiel exists throughout a region of space, the potential must be linearly changing in the irection parallel to the fiel an unchanging in the irections perpenicular to the fiel. 13. It oesn t matter which points we choose because the potential on each plate is constant over the whole plate. 14. The factor of 1/ appears because the average height of the water in the pool is (1/)h. The work require to fill the pool is exactly equal to the potential energy of the water in the pool, (1/)Mgh. This is analogous to the charging of a capacitor. The charge Q on the capacitor is like the mass M of water in the pool, an the electrical potential energy QV of the charge is like the gravitational potential energy Mgh. Thus, by analogy, the total potential energy of a charge capacitor is (1/)QV, which is correct. 15. Nothing happens to the capacitance, which epens only on the geometry an electrical properties of the materials in the capacitor. Since C = Q/V, if the charge oubles then the voltage oubles as well. 16. Cow A is more likely to be kille because the potential ifference between its front an hin legs will be greater than that for cow B. 17. We can t say anything about the electric fiel if all we know is the potential at a single point. The electric fiel tells us how the potential changes if we move from one point to another. 18. As long as the person touching the ome is isolate from the groun, there is no complete circuit for current to flow through, so she is safe. 19. The electric fiel points from regions of higher potential to regions of lower potential. Therefore, the upper atmosphere is at a higher potential than the Earth. 0. Since the capacitor is connecte to a battery the whole time, its voltage V remains constant. The capacitance is proportional to, so when the ielectric is remove the capacitance ecreases by a factor of 3. Since Q = CV, an V remains constant, the charge on the capacitor Q ecreases by a factor of 3 as well. The electric fiel remains constant, since V is constant. The energy store ecreases by a factor of In this case the capacitor plates are isolate, so it is the charge Q that remains constant. Again the capacitance ecreases by a factor of 3, so V must increase by a factor of 3 to keep Q = CV the same. Thus, the electric fiel increases by a factor of 3 an the energy store in the capacitor increases by a factor of 3 as well.. The plates are isolate so the charge remains constant. The capacitance is given by C e0 A/. As the plates are move closer, all these quantities remain constant except, which ecreases. Therefore, the capacitance increases. Since Q = CV, an Q is constant, V must ecrease. The electric fiel remains constant. The energy store in the capacitor ecreases as well. 3. The potential close to the positive charge must be positive, while close to the negative charge it must be negative. Therefore, there is a point in region B where the potential is zero. If we move very far away from the two charges, they will look like a single point of charge 3 C, so the potential very far away must be negative. Thus, there must be a point in region A with a potential of zero as well. The electric fiel can only be zero in region A, an this oes not occur at the same point where the potential is zero. 683

3 Chapter 17: Electric Potential College Physics Problems 1. Strategy Use Eq. (17-1). Solution Compute the electric potential energy qq 1 ( N m C )(5.010 C)(.010 C) UE k 18 mj r 5.0 m. (a) Strategy Use Eq. (17-1). Solution Compute the electric potential energy qq 1 ( N m C )( C)( C) 18 UE k J r m (b) Strategy an Solution The negative sign signifies that the force between the two charges is attractive; the potential energy is lower than if the two were separate by a larger istance. 3. Strategy The work one by the applie force is positive, since the irection of the applie force was in the irection of motion. (The force between the two charges is repulsive.) The potential energy of the charges is positive, so the work one on the charges is equal to their potential energy. Use Eq. (17-1). Solution Compute the work one on the charges. qq ( N m C )(6.510 C) W UE k r m 8.4 J 4. Strategy The work one by the external agent is positive since the potential energy increases. Use Eq. (17-1). Solution Fin the work one by the external agent ke ( N m C )( C) 13 W U Uf Ui U J r 15 f m 5. Strategy The work one on the charges is equal to their potential energy. Let the upper charge by 1, the lower left-han charge be, an the right-han charge be 3. Also, let a 0.16 m an b 0.1 m. Use Eq. (17-). Solution Compute the work one on the charges. qq 1 qq 1 3 qq 3 W UE k b a a b (5.510 C)( C) (5.510 C)(.510 C) ( N m C ) 0.1 m (0.16 m) (0.1 m) 6 6 ( C)(.510 C) 3.0 J 0.16 m 6. Strategy Let q1 q q 10.0 nc an = 4.00 cm. Use Eq. (17-). Solution Fin the total electric potential energy for the two charges. U E 9 9 kq1q kq ( N m C )( C) 11. J r ( m) 684

4 College Physics Chapter 17: Electric Potential 7. Strategy Let q1 q q 10.0 nc, = 4.00 cm, an q3 4. nc. Use Eq. (17-). Solution Fin the total electric potential energy of the three charges at point a. qq 1 qq 1 3 qq3 q qq3 qq 3 kq q 1 UE k k q3 1 r1 r13 r ( N m C )( C) C ( 4.10 C) 17.5 J m 3 8. Strategy Let q1 q q 10.0 nc, = 4.00 cm, an q3 4. nc. Use Eq. (17-). Solution Fin the total electric potential energy of the three charges at point b. qq 1 qq 1 3 qq3 q qq3 qq 3 kq UE k k r1 r13 r ( N m C )( C) 11. J ( m) 9. Strategy Let q1 q q 10.0 nc, = 4.00 cm, an q3 4. nc. Use Eq. (17-). Solution Fin the total electric potential energy of the three charges at point c. qq 1 qq 1 3 qq3 q qq3 qq 3 kq UE k k r1 r13 r ( N m C )( C) 11. J ( m) 10. Strategy Use Eq. (17-). Solution Fin the electric potential energy. qq 1 qq 1 3 qq3 UE k r1 r13 r C C ( N m C ) (4.010 C) (4.0 m) (3.0 m) 3.0 m 6 6 (3.010 C)( C).8 mj 4.0 m 11. Strategy Use Eqs. (6-8) an (17-1). Solution Compute the work one by the electric fiel. y (m) x (m) kq1q 3 kqq W 3 fiel U Ui Uf U1 ( U1 U13 U3) U13 U3 r13 r C C ( N m C )(.0010 C).70 J m m m 685

5 Chapter 17: Electric Potential College Physics 1. Strategy Use Eqs. (6-8) an (17-1). Solution Compute the work one by the electric fiel. kq1q 3 kqq W 3 fiel U Ui Uf U1 ( U1 U13 U3) U13 U3 r13 r ( N m C )(.0010 C)( C) 1.80 J m m 13. Strategy Use Eqs. (6-8) an (17-1). Solution Compute the work one by the electric fiel. Wfiel U Ui Uf U1 U13i U3i ( U1 U13f U3f ) U13i U13f U3i U3f kq1q3 qq3 r13i r13f r3i r3f ( N m C ) ( C)(.0010 C) m m ( C)(.0010 C) 4.49 J m m m 14. Strategy Use Eqs. (6-8) an (17-1). Solution Compute the work one by the electric fiel. Wfiel U Ui Uf U1 U13i U3i ( U1 U13f U3f ) U13i U13f U3i U3f kq1q3 qq3 r13i r13f r3i r3f ( N m C )( C)(.0010 C) m 0.10 m m 0.10 m 1.80 J 15. Strategy The potential ifference is the change in electric potential energy per unit charge. Use UE q V. Solution Fin the change in the electric potential energy. UE qv (3.0 nc)(5 V) 75 nj 16. Strategy The potential ifference is the change in electric potential energy per unit charge. Use UE q V. Solution Fin the change in the electric potential energy. E f i U qv q( V V) ev ( V ) ( C)[ 360 V ( 40 V)] J B A 686

6 College Physics Chapter 17: Electric Potential 17. Strategy Use the principle of superposition an Eq. (17-9). Solution Sum the electric fiels at the center ue to each charge. EEa Eb Ec E Ea Eb Ea Eb 0 Do the same for the potential at the center. 9 6 kqi 4kQ 4( N m C )(9.010 C) 7 V.3 10 V r r i (0.00 m) (0.00 m) 18. Strategy Use the principle of superposition an Eq. (17-9). Solution Sum the electric fiels at the center ue to each charge. kqa kqc EEa Eb Ec E Ea Eb Ec Eb Ea Ec towar c r r N m C ( C C) towar c N C towar c (0.00 m) (0.00 m) Do the same for the potential at the center. kqi k V (3 q q q q) 0 r r i 19. Strategy Use Eqs. (17-7), (17-8), (17-9), an (6-8). Solution Fin the potentials, potential ifference, change in electric potential energy, an work one by the electric fiel. (a) (b) (c) 9 9 kq ( N m C )( C) V 1.5 kv r 0.30 m 9 9 ( N m C )( C) V 900 V 0.50 m V kq ( N m C )( C) 600 V r B r A 0.50 m 0.30 m V 0, so the potential increases. () 9 7 UE qv ( C)(6.010 V) J U E 0, so the potential energy ecreases. (e) 7 Wfiel UE J 687

7 Chapter 17: Electric Potential College Physics 0. Strategy Use Eq. (17-9). Solution Fin the electric potential at point P ue to the charges. 3 3 kq i C C V ( N m C ) ri 4.0 m (4.0 m) (3.0 m).7 MV 1. Strategy an Solution (a) Since V is positive, q is positive. y (m) mc mc P x (m) (b) 1 V, so since the potential is ouble, the istance is halve or 10.0 cm. r. Strategy Just outsie the surface of the sphere, the electric potential is given by V Er, where r is the raius of the sphere. Solution Fin the electric potential. 5 5 V Er ( V m)(0.750 m) V kqi 3. (a) Strategy an Solution Since V, the minimum or most negative value of the potential is the case ri where the two negative charges are closer to x 0 than the two positive charges. y + + x (b) Strategy Let = 1.0 m an q = 1.0 C. Use Eq. (17-9). Solution Fin the potential at the origin. 9 6 kq i q q q q kq 4( N m C )(1.010 C) V k r 3 3 i m 36 kv 4. (a) Strategy Use Eq. (16-5). Solution Fin R E. kq k(3 Q ) E E, so R 3 R E 0 R0 RE (b) Strategy Use Eq. (17-8). Solution Fin R V. kq k(3 Q ) V V, so R 3 R V 0 R0 RV 688

8 College Physics Chapter 17: Electric Potential 5. Strategy Use Eq. (17-9). Solution Fin the electric potential at the thir corner, B. 9 kqi k N m C 9 9 V ( QAQB) (.010 C C) 9.0 V ri r 1.0 m 6. (a) Strategy Use Eq. (17-9). Solution Fin the electric potential at each point. 9 9 kq kq C C V a ( N m C ) 70 V r r m m C C V b ( N m C ) 160 V 0.10 m 0.10 m (b) Strategy Use Eq. (17-6). Solution Compute the potential ifference for the trip from a to b. V V V 160 V 70 V 430 V b a (c) Strategy The work one by an external agent is equal to the change in electric potential energy of the point charge when move from a to b. Use Eq. (17-7). Solution Compute the work one. 9 7 W qv ( C)( 430 V) J 7. (a) Strategy Use Eq. (17-9). Solution Fin the potential at the points. 9 9 kq1 kq C.5010 C Va ( N m C ) 300 V r1 r m m 9 kq If Q C, 1 kq kq kq Vb 0. r1 r r r (b) Strategy The work one by an external agent is equal to the change in electric potential energy of the point charge when move from infinity to b. Use Eq. (17-7). Solution Compute the work one. W qv q( V V ) q(00) 0 b 689

9 Chapter 17: Electric Potential College Physics 8. (a) Strategy Let 4.00 cm, r 1.0 cm, an q 8.00 nc. Use Eq. (17-9). Let V 0 at infinity. Solution Fin the potentials at points a an b. 9 9 kq1 kq kq k( q) 3kq 3( N m C )( C) Va r1 r 4 4 4( m) 9 9 kq k( q) kq ( N m C )( C) Vb 899 V ( m) 1350 V (b) Strategy The potential ifference is the change in electric potential energy per unit charge. Use UE q V. Solution Compute the change in electric potential energy. 9 UE qv (.0010 C)( 899 V 1348 V) 4.49 J 9. (a) Strategy Let 4.00 cm, r 1.0 cm, an q 8.00 nc. Use Eq. (17-9). Let V 0 at infinity. Solution Fin the potentials at points b an c. 9 9 kq1 kq kq k( q) kq ( N m C )( C) Vb 899 V r1 r ( m) kq k( q) Vc 0 r r (b) Strategy The potential ifference is the change in electric potential energy per unit charge. Use UE q V. Solution Compute the change in electric potential energy. 9 UE qv (.0010 C)(0 899 V) 1.80 J 30. Strategy Rewrite each unit in terms of kg, m, s, an C. Solution Show that 1 N C 1 V m. kg m s J kgm s kgm s 1 N C an 1 V m, C mc mc C therefore 1 N C 1 V m. 31. (a) Strategy Use Eq. (16-4b). Solution Fin the electric force that acts on the particle. 9 F qe(4.10 C)(40 N C to the right) 1.0 to the right (b) Strategy The work one on the particle is equal to the electric force times the isplacement of the particle. Solution Compute the work one on the particle. W F (1.0 N)(0.5 m) 0.5 J (c) Strategy The electric fiel points in the irection of ecreasing potential, so Va Vb an Va Vb 0. Use Eq. (17-10). Solution Compute the potential ifference. Va Vb E (40 N C)(0.5 m) 60 V 690

10 College Physics Chapter 17: Electric Potential 3. (a) Strategy an Solution Positive work is require to move an electron (negative charge) from low potential to high potential, so Y is at the higher potential. (b) Strategy Use Eqs. (6-8) an (17-5). Solution Fin the potential ifference. 19 Wfiel E fiel Y X J qv ev U W, so V V V 5.0 V. e J 33. Strategy V E for a uniform electric fiel. Solution Fin the istance between the equipotential surfaces. V 1.0 V 1.0 cm E N C 34. Strategy Equipotential surfaces are perpenicular to electric fiel lines at all points. For equipotential surfaces rawn such that the potential ifference between ajacent surfaces is constant, the surfaces are closer together where the fiel is stronger. The electric fiel always points in the irection of maximum potential ecrease. Solution Outsie the sphere, E is raially irecte (towar the sphere), an V r 1. The equipotential surfaces are perpenicular to E at any point, so they are spheres. Insie the sphere, E 0 an V is constant. 35. Strategy Equipotential surfaces are perpenicular to electric fiel lines at all points. For equipotential surfaces rawn such that the potential ifference between ajacent surfaces is constant, the surfaces are closer together where the fiel is stronger. The electric fiel always points in the irection of maximum potential ecrease. Solution Outsie the cyliner, E is raially irecte away from the axis of the cyliner. The equipotential surfaces are perpenicular to E at any point, so they are cyliners. 691

11 Chapter 17: Electric Potential College Physics 36. Strategy The rate at which work is one by the electric organs is equal to the rate of change of the electric potential energy. Use Eq. (17-7). The total amount of work one in one pulse is equal to the rate times the uration of the pulse. Solution (a) Compute the rate at which work is one. W q V q 3 V (0.010 V)(18 C s) 3.6 kw t t t (b) Compute the total amount of work one. W 3 W t (3.610 W)( s) 5.4 J t 37. (a) Strategy Equipotential surfaces are perpenicular to electric fiel lines at all points. For equipotential surfaces rawn such that the potential ifference between ajacent surfaces is constant, the surfaces are closer together where the fiel is stronger. The electric fiel always points in the irection of maximum potential ecrease. Solution The electric fiel lines are raial. They begin on the point charge an en on the inner surface of the shell. Then they begin again on the surface an exten to infinity. + (b) Strategy Use Eqs. (16-5) an (17-8), an the principle of superposition. Solution For r r1, E is that ue to the point charge, E kq r. For r 1 r r, E 0, since this is insie a conuctor. For r r, E once again is that ue to the point charge, kq r For r r 1, V kq r (point charge). For r 1 r r, V kq r 1, since V is continuous, an it is constant in a conuctor. For r r, kq kq kq V r1 r r E (to preserve continuity). The graphs of the electric fiel magnitue an potential: V. kq r 1 kq r kq r 1 0 r 1 r r 0 r 1 r r 69

12 College Physics Chapter 17: Electric Potential 38. Strategy Since the electric fiel is uniform, we can use Eq. (17-10). Solution Fin the magnitue of the charge on the rop in terms of the elementary charge e. 16 F Fe Fe Fe (0.16 m)(9.610 N) F qe an V E, so q e e. E Ee ( V ) e ev 19 ( C)(480 V) 39. Strategy Since the electric fiel is uniform, we can use Eq. (17-10). Use Newton s secon law. Solution Fin the magnitue of the charge on the rop. F qemg 0, so 15 mg mg mg (1.010 kg)(9.80 m s )(0.16 m) 19 q C e. E V V V qe mg 40. Strategy Use conservation of energy an Eq. (17-7). Solution Fin the change in kinetic energy K U qv ( C)( V V) J 41. Strategy Use conservation of energy an Eq. (17-7). Solution Fin the potential ifference mv ( kg)(7.610 m s) U ev K mv, so V 150 V. e 19 ( C) 4. (a) Strategy The electric fiel always points in the irection of maximum potential ecrease. Electrons, being negatively charge, move in the irection opposite the irection of the electric fiel; that is, in the irection of potential increase. Solution Since the spee of the electron ecrease, it must have travele in the irection of the electric fiel, so it move in the irection of potential ecrease; that is, to a lower potential. (b) Strategy The kinetic energy of the electron ecrease, so its potential energy increase. Use conservation of energy an Eq. (17-7). Solution Compute the potential ifference the electron move through. U K mv ( f vi ) ( kg)[( m s) (8.50 V 10 m s) ] 188 V q e e 19 ( C) 43. Strategy Accoring to Example 17.8, the spee of the electrons at the anoe is proportional to the square root of the potential ifference. Use a proportion. Solution Fin the spee of the electrons. v V V kv 7 v V, so an v v (6.510 m s) m s. v V V 1 kv

13 Chapter 17: Electric Potential College Physics 44. Strategy Accoring to Example 17.8, the spee of the electrons at the anoe is proportional to the square root of the potential ifference. Use a proportion. Solution Fin the potential ifference. 7 V v v m s v V, so an V V1 (1 kv).6 kv. V 7 1 v1 v m s 45. Strategy an Solution (a) Electrons travel opposite the irection of the electric fiel, so E is irecte upwar. ev ev vy vym (b) For a uniform electric fiel, Fy ee may, so ay. Thus, t. m a ev (c) Since the electron gains kinetic energy, its potential energy ecreases. 46. Strategy The fiel is uniform. Use conservation of energy an Eq. (17-10). Solution Fin the kinetic energy increase K U qv ee ( C)(500.0 N C)( m).4 10 J. 47. Strategy Use conservation of energy an Eq. (17-7). Solution Fin the final kinetic energy. K K K U qv ev, so f i Kf Ki ev J ( C)( V).810 J. 48. Strategy The force between the nuclei is repulsive, since they both have positive charge. Use conservation of energy an Eq. (17-1). Solution Fin the closest istance that a helium nucleus approaches the gol nucleus. kq1q 1 Uf Ki mhevi, so r kq 9 19 Auq He k(79 e)( e) 316( N m C )( C) 14 r m. 7 7 m v m v ( kg)( m s) He i He i 15 Note that the raius of the gol nucleus is about 7 10 m an the raius of the gol atom is about 1 10 m. 49. Strategy The electron must have enough kinetic energy at point A to overcome the potential ecrease between A an C. Use conservation of energy an Eq. (17-7). Solution Fin the require kinetic energy KA U ev ( C)( 60.0 V V) J 50. Strategy an Solution Since positive charges move through ecreases in potential, an since the potential an potential energy are greatest at A, the proton will spontaneously travel from point A to point E. So, K A 0. y

14 College Physics Chapter 17: Electric Potential 51. Strategy Use the efinition of capacitance, Eq. (17-14). Solution Fin the magnitue of the charge on each plate. Q CV (.0 μf)(9.0 V) 18 μc 5. (a) Strategy Use the efinition of capacitance, Eq. (17-14). Solution Fin the potential ifference between the plates. Q 0.75 μc Q CV, so V 50 mv. C 15.0 μf (b) Strategy an Solution The plate with the positive charge is at the higher potential, so the 0.75-C plate. 53. Strategy Use the efinition of capacitance, Eq. (17-14). Solution 6 4 Q CV (10.10 F)( 60.0 V) C 61 C of charge must be remove from each plate. 54. (a) Strategy Use Eq. (17-10). Solution Compute the maximum potential ifference across the capacitor. 6 Vmax Emax (310 V m)( m) 3 kv (b) Strategy Use the efinition of capacitance, Eq. (17-14). Solution Compute the magnitue of the greatest charge. 6 3 Q CV (.010 F)(310 V) 6 mc 55. Strategy an Solution (a) Since E oes not epen upon the separation of the plates ( E e 0), it stays the same. (b) Since V, V increases if increases. 56. Strategy Use the efinition of capacitance, Eq. (17-14), an the fact that the potential ifference epens upon the plate separation. Solution (a) Compute the potential ifference between the plates. 6 Q C V 667 V C F (b) Since V, V oubles if oubles. 695

15 Chapter 17: Electric Potential College Physics 57. Strategy an Solution (a) The battery maintains a constant potential ifference between the plates: ΔV stays the same. (b) The electric fiel magnitue increases because the same potential ifference occurs over a shorter istance (E = ΔV/ for a uniform fiel). (c) To maintain a constant potential ifference while the plate spacing changes, the battery must change the charge on the plates. A larger electric fiel means that the charge increases. Check: Q = CΔV, C increases an ΔV oesn t change, so Q increases. 58. Strategy Use the efinition of capacitance, Eq. (17-14), an the fact that the potential ifference epens upon the plate separation. Solution (a) Fin the magnitue of the charge on each plate. Q CV (1.0 nf)(1 V) 14 nc (b) The capacitor remains connecte to the battery so the potential ifference stays the same. From E = ΔV/, increasing means that the electric fiel ecreases. The electric fiel is proportional to the charge per unit area on the plates, so the charge ecreases. 59. Strategy The capacitance of a parallel plate capacitor is irectly proportional to its area. Form a proportion. Solution Fin the capacitance for each situation. (a) (b) 1 A 1 C C1 C C A A 1, so (0.694 pf) pf. C A A A A 3 1 C C1 C C A A, so (0.694 pf) pf. C A A A Strategy Use Eq. (17-10). Solution Compute the plate separation. V 1.5 V 1500 km E V m 61. Strategy Use the efinition of capacitance, Eq. (17-14). Solution Fin the capacitance of the spheres. 14 Q 3.10 C Q CV, so C 8.0 pf. V V 6. Strategy Use the efinition of capacitance an the efinition of potential for a spherical conuctor. Solution Fin the capacitance of the Moon if wrappe in aluminum foil. Q Q r m Q CV, so C F. V kq k N m C r 6 696

16 College Physics Chapter 17: Electric Potential 63. Strategy The electrons are accelerate by the electric fiel between the plates of the capacitor. When they emerge from the positive plate, their spee will be greater than their spee when they entere the capacitor. Solution Fin the acceleration of the electrons while they are insie the capacitor. F ma ev V E, so a. e e m Fin the spee of the electrons as they emerge from the capacitor. ev ev vfx vix axx a, so m m 19 ev 6 ( C)(40.0 V) v 6 fx vix (.5010 m s) m s. m kg 64. Strategy Use the efinition of electric flux an Gauss s law. Solution (a) The Gaussian surface is a cyliner whose axis is parallel to a raius vector (of the sphere) through it. One en is just within the conuctor an the other is just outsie it. The ens have area A, with a raius much smaller than that of the sphere, so the electric fiel is approximately uniform. Fin E just outsie the conuctor. Q E EAcos e 0 Cylinrical surface: E EAcos90 EA(0) 0 En insie conuctor: E EAcos (0) Acos 0, since the electric fiel is zero insie a conuctor. Q Q En just outsie the conuctor: E EAcos0 EA, so E. e Ae e (b) Consier an area A of the surface of an arbitrary conuctor. If A is small enough such that its surface is approximately flat, then the electric fiel will be nearly uniform just outsie the surface. Comparing an area of the same size on a spherical conuctor with the same charge ensity to that of the arbitrary conuctor, we see that the electric fiel just outsie either conuctor shoul be e 0 ; as long as A is small enough that it is approximately flat, then this hols for any conuctor. 65. (a) Strategy Use Eq. (16-6). Solution The electric fiel between the plates is Q C 3 E V m. e A 1 [ C (N m )](0.06 m)(0.0 m) 0 (b) Strategy Use the efinition of the ielectric constant, Eq. (17-17). Solution Fin the electric fiel between the plates of the capacitor with the ielectric. E 3 0 E V m, so E V m. E Strategy Assume the fiel is uniform. Use Eq. (17-10). Solution Fin the maximum possible height for the bottom of the thunerclou. 8 V V 300 m. E V m 697

17 Chapter 17: Electric Potential College Physics 67. (a) Strategy Use Eq. (17-10). Solution Compute the magnitue of the average electric fiel between the cow s front an hin legs. 3 V (400 00) 10 V 5 E 10 V m 1.8 m Since the electric fiel always points in the irection of ecreasing potential, the average electric fiel is 5 10 V m towar the hin legs. (b) Strategy an Solution The front an hin legs of Cow B are nearly at the same potential, whereas those for Cow A span a potential ifference of approximately 00 kv, thus Cow A is more likely to be kille. 68. Strategy Use the efinition of capacitance, Eq. (17-14). Solution Compute the capacitance of the capacitor. 6 Q C C 83 pf. V 40 V 69. Strategy The spark flies between the spheres when the electric fiel between them excees the ielectric strength. The magnitue of the electric fiel is given by V, where is the istance between the spheres. Solution Fin. V V 900 V E, so 0.30 mm. E V m 70. Strategy Use Eq. (17-16). Solution (a) Fin the greatest, since A an e 0 are constant , 3.5, an Since 35 > 3.5 > 0., the paper is the best choice e0 A 3.5[ C (N m )](1010 m ) C 3.7 nf m (b) Compute the smallest capacitance [ C (N m )](1010 m ) C m 1 pf 71. Strategy Use Eq. (17-16). Solution Compute the capacitance of the capacitor. 1 e0 A.5[ C (N m )](0.30 m)(0.40 m) C m 89 nf 7. Strategy Use Eq. (17-16). Solution Fin the average ielectric constant of the tissue in the limb. 1 e0 A C (0.030 m)( F) C, so 5.0. e A 1 4 [ C (N m )](4.010 m ) 0 698

18 College Physics Chapter 17: Electric Potential 73. (a) Strategy Use Eq. (17-18c). Solution Compute the capacitance. Q Q (8.010 C) U, so C 7.1 F. C U (450 J) (b) Strategy Use Eq. (17-18a). Solution Compute the potential ifference. 1 U (450 J) U QV, so V V. Q C 74. Strategy Use Eq. (17-19). The ielectric strength of air is 3 kv mm, which is equal to the maximum electric fiel. Solution Compute the maximum electric energy ensity in ry air. u 1 e 1 ( )[ C (N m )](3 10 V m) 40 J m 0E Strategy The capacitance of a capacitor is inversely proportional to the istance between the plates an U Q ( C). Solution Form a proportion to fin the ratio of the new capacitance to the ol. C C Form a proportion to fin the energy store in the capacitor in terms of the ol. U Q ( C) C , so U 1.50 U1. U1 Q ( C1 ) C Thus, the energy increases by 50%. 76. Strategy The energy store in the capacitor is given by U 1 C( V). When the plate separation is increase, the capacitance changes but the potential ifference stays the same, so the energy in the capacitor changes as well. The work one on the capacitor in separating the plates is negative the change in energy. Solution Form a proportion. The capacitance is inversely proportional to the plate separation. U f Cf( V) Cf i 1.00 cm U i Ci ( V) Ci f.00 cm Fin the work one on the capacitor. So, the energy is reuce by half A W U i Uf Ui 0.500Ui 0.500Ui Ci( V) ( V) i [ C (N m )](31410 m ) (0.0 V).78 nj ( m) e 699

19 Chapter 17: Electric Potential College Physics 77. (a) Strategy Use Eq. (17-15). Solution Fin the capacitance for the thunerclou. 1 e0 A [ C (N m )](4500 m)(500 m) C 0.18 F 550 m (b) Strategy Use Eq. (17-18c). Solution Fin the energy store in the capacitor. Q (18 C) 8 U 10 J C 6 ( F) 78. (a) Strategy The capacitance after the slab is remove is equal to the capacitance with the slab ivie by the ielectric constant. Solution Compute the capacitance. C 6.0 F C0.0 F 3.0 (b) Strategy Use Eqs. (17-10) an (17-17). Solution Fin the potential ifference across the capacitor. E0 E E 0 E V0 V 3.0(1.5 V) 4.5 V (c) Strategy Use the efinition of capacitance, Eq. (17-14). Solution Compute the charge on the plates. Q CV (.0 F)(4.5 V) 9.0 C () Strategy Use Eq. (17-18c). Solution Compute the energy store in the capacitor. 6 Q (9.010 C) U 0 J C 6 (.010 F) 79. (a) Strategy Use the efinition of capacitance, Eq. (17-14), an Eq. (17-15). Solution Fin the charge on the capacitor. 1 e0 A [ C (N m )](0.100 m) (150 V) Q CV V m 18 nc (b) Strategy Use Eq. (17-18a). Solution Compute the energy store in the capacitor. 1 1 ( U QV C)(150 V) 1.3 J 700

20 College Physics Chapter 17: Electric Potential 80. (a) Strategy Use Eq. (17-15). Solution Compute the capacitance. 1 e0 A [ C (N m )](0.100 m) C m m 59.0 pf (b) Strategy From Problem 63, Q 18 nc. Use Eq. (17-18c) an conservation of energy. Solution Compute the new energy store in the capacitor. 9 Q (18 10 C) U.7 J C 1 ( F) Work was one on the capacitor when the plates were separate; that work has been store in the capacitor as potential energy. 81. (a) Strategy U Pt where P 10.0 kw an t.0 ms. Use Eq. (17-18b). Solution Fin the initial potential ifference. 1 U (10.0 kw)(.0 ms) U C( V), so V 630 V. C F (b) Strategy Use Eq. (17-18c). Solution Fin the initial charge. Q 6 U, so Q CU ( F)(10.0 kw)(.0 ms) C. C 8. Strategy Use Eq. (17-18a). Solution Compute the energy store in the capacitor. 1 1 ( U QV C)(40 V).4 J 83. Strategy The work one by the external agent is equal to the change in potential energy of the capacitor. Use Eq. (17-18c) an the fact that the capacitance is inversely proportional to the plate separation. Solution Fin the work require to ouble the plate separation. 6 Q Q Q Ci Q f ( C) W U 1 1 ( 1) 0.7 mj C 9 f Ci Ci Cf Ci i (1.010 F) 84. Strategy Use Eq. (17-18b). Solution Fin the require potential ifference. 1 U (300 J) U C( V), so V 8 kv. C F 701

21 Chapter 17: Electric Potential College Physics 85. (a) Strategy Use the efinition of capacitance, Eq. (17-14). Solution Compute the charge that passes through the boy tissues. 6 3 Q CV (15 10 F)(9.010 V) 0.14 C (b) Strategy Use Eq. (17-18b) an the efinition of average power. Solution Fin the average power elivere to the tissues. P 6 3 E U C( V) (15 10 F)(9.010 V) t t t (.010 s) av MW 86. Strategy Assume the that thunerclou an Earth system behaves like a capacitor. Use Eq. (17-18a). Solution Fin the electric potential energy release by the lightning strike. 1 1 (5.0 C)( U QV V) 1.5 GJ. 87. (a) Strategy Assume that the thunerclou an Earth system acts like a capacitor. Use Eq. (17-18a). Solution Fin the electric potential energy release by the lightning strike. 1 1 (0.0 C)( U QV V) 10.0 GJ (b) Strategy Use the efinition of latent heat. Solution Fin the mass of sap that is vaporize J Q energy absorbe mlv 0.100( J), so m 443 kg.,56,000 J kg (c) Strategy Divie 10.0% of the total energy release from the lightning strike by the homeowner s monthly energy use. 9 Solution 0.100( J) t 3 ( W h month)(3600 s h) month 88. Strategy Use Eq. (17-9). Solution Fin the potential miway between the charges. 9 kq1 kq k N m C 9 9 V ( q q ) ( C.010 C) 873 V r r r m 70

22 College Physics Chapter 17: Electric Potential 89. (a) Strategy Let ql qr 10.0 nc. Use Eqs. (17-5) an (17-9). Solution Fin the potential energy of the point charge at each location. kql kqr 1 1 Ua qva q kqql rl rr rl rr ( N m C )( 4.10 C)( C) 6.3 J m m Ub qvb ( N m C )( 4.10 C)( C) m m Uc qvc ( N m C )( 4.10 C)( C) m m (b) Strategy The work one by the external force is negative the work one by the fiel. Use Eq. (6-8). Solution Fin the work require to move the point charge. W Wfiel U Ua Ub 6.3 J J 90. Strategy Use Eqs. (6-8) an (17-7). Solution Compute the work one by the electric fiel Wfiel U ( e) V ( C)[100.0 V ( V)] J 91. Strategy The potential at the surface of a conucting sphere is equal to the magnitue of the electric fiel times the raius of the sphere. Solution Compute the potential. 6 V Er (3.010 N C)(0.15 m) 450 kv 9. Strategy Let 1 q.0 10 C an r = 1.0 nm. Use Eq. (17-9). Solution Fin the potential at the soium ions ue to the other three ions. 9 1 kq1 kq kq3 q q q kq ( N m C )(.010 C) V k r 9 1 r r 3 r r r r (1.010 m) 9.0 mv 93. (a) Strategy Electric fiel lines begin on positive charges an en on negative charges. The same number of fiel lines begins on the plate an ens on the negative point charge. Fiel lines never cross. Use the principles of superposition an symmetry. Solution The electric fiel lines for the cyliner an sheet: 703

23 Chapter 17: Electric Potential College Physics (b) Strategy Equipotential surfaces are perpenicular to electric fiel lines at all points. For equipotential surfaces rawn such that the potential ifference between ajacent surfaces is constant, the surfaces are closer together where the fiel is stronger. The electric fiel always points in the irection of maximum potential ecrease. Solution The equipotential surfaces for the cyliner an sheet: 94. Strategy Use Newton s secon law an Eq. (4-9). Solution Fin the acceleration. ee Fy ee may, so ay. m Fin the time to reach the lower plate my ( kg)(0.040 m) y ay( t), so t 3.0 ns. ee 19 4 ( C)(5.010 N C) 95. Strategy Assume the fiel is uniform. Use Eq. (17-10). Solution Compute the magnitue of the electric fiel in the membrane. 3 V 9010 V 6 E 910 V m m 96. Strategy Use Newton s secon law, x vx t, an Eqs. (4-7) an (4-9). Solution (a) Since E points ownwar, the negatively charge electron s change in velocity is irecte upwar. Fin the acceleration. ee Fy ee may, so ay. m Fin the time interval. x t vx Fin the change in velocity ee x ( C)(.010 N C)(0.060 m) 6 vy ayt m s m v 31 7 x ( kg)(3.010 m s) 6 So, v m s upwar. (b) Fin the eflection of the electrons ee x ( C)(.010 N C)(0.060 m) y ay( t) 7.0 mm m v 31 7 x ( kg)( m s) 704

24 College Physics Chapter 17: Electric Potential 97. Strategy The negatively charge particle will accelerate towar the positively charge plate while it is between the plates of the capacitor. Solution The particle is between the plates for a time given by t x v x. During this time, the particle travels a vertical istance y m. Fin the acceleration of the particle. 1 1 x a( x) vx y y a( t) a, so a. v ( ) x vx x V The magnitue of the electrical force on the particle is NeE Ne, where is the plate separation an N is the number of excess electrons on the particle. Accoring to Newton s secon law, the acceleration of the particle is Ne V NeV a. We set the two expressions for the acceleration of the particle equal an solve for N. m m NeV v ( kg)( m)(35.0 m s) x y, so mv x y N ( m) 51. m 19 ( x) ev( x) ( C)(3.00 V)( m) 98. (a) Strategy Compute the electrical an gravitational forces on the particle an compare. Refer to Problem Solution The gravitational force on the particle is mg ( kg)(9.80 m s ) N. The electrical force on the particle is Compare the forces N N 3 NeV 51( C)(3.00 V) N m The electrical force is times larger than the gravitational force. (b) Strategy Use the results from Problem 83. The horizontal component of the velocity oesn t change. Solution The horizontal component of the velocity is vx 35.0 m s. Compute the y-component of the particle s velocity. vx yx vxy (35.0 m s)( m) vy at 7.00 m s ( x) v x x m 99. Strategy Fin the charge on the capacitor. Use Eqs. (17-14) an (17-15). Solution The charge on the capacitor is Q Ne, where N is the number of excess electrons. e 1 0A e0a V [ C (N m )]( m) (3.00 V) Q NeCV V, so N e 19 ( m)( C)

25 Chapter 17: Electric Potential College Physics 100. (a) Strategy Use the efinition of capacitance, Eq. (17-14). Solution Compute the capacitance. 6 Q C C 83 pf V 40 V (b) Strategy Use Eq. (17-15). Solution Fin the area of a single plate e0 A C ( m)( F) C, so A e C (N m ) m. (c) Strategy Use Eq. (17-10). 0 Solution Compute the voltage require to ionize the air between the plates. 3 V E (3.010 V mm)(0.40 mm) 1. kv 101. Strategy The energy in the capacitor is converte into heat in the water. Use Eqs. (14-4) an (17-18b). Solution Fin the temperature change of the water. 1 6 C( V) ( F)(1.0 V) Q mct U C( V), so T 3.44 mk. mc 3 3 (1.00 cm )(1.00 g cm )[4.186 J (g K)] 10. (a) Strategy Use Eqs. (17-16) an (17-18b). Solution Fin the energy store in the capacitor. 1 1 e0 A 5.[ C (N m )](1.010 m )( V) U C( V) ( V) 9 (7.510 m) J (b) Strategy Divie the total charge by the charge of one ion. Use the efinition of capacitance, Eq. (17-14), an Eq. (17-16). Solution Fin the number of ions outsie of the membrane Q CV e0 AV 5.[ C (N m )](1.010 m )( V) ions e e e 19 9 ( C ion)(7.510 m) 103. Strategy Use Eq. (17-16). Solution Fin the capacitance of the axon. 1 1 e0 A 5[ C (N m )](510 m ) 14 C 510 F m 706

26 College Physics Chapter 17: Electric Potential 104. Strategy Use conservation of energy, Wfiel U, an the fact that the fiel is uniform. Solution Fin the kinetic energy of each electron when it leaves the space between the plates. V K Kf Ki U Wfiel eey e y, so 19 3 V 15 ( C)( V)( m) 15 Kf Ki e y.0 10 J J m 105. (a) Strategy For a parallel plate capacitor, E e 0 an V E. Solution Fin the potential ifference between the plates. 6 (4.010 C m )( m) V E.7 kv e C (N m ) 0 (b) Strategy Use conservation of energy an the fact that U q V. Solution Fin the kinetic energy of each point charge just before it hits the positive plate. 9 K Kf Ki U qv, so Kf Ki qv 0 (.510 C)(711 V) 6.8 J Strategy Use conservation of energy, Wfiel U, an the fact that the fiel is uniform. Solution Fin the final kinetic energy of the alpha particle. K K K U W ee, so f i fiel Kf Ki ee 0 ( C)( V m)(0.010 m) 3.10 J Strategy Use Eqs. (6-8) an (17-7). Solution Fin the work one by the electric fiel Wfiel U qv ev ( C)( V) J 108. Strategy Electric fiel lines begin on positive charges an en on negative charges. The same number of fiel lines begins on the plate an ens on the negative point charge. Fiel lines never cross. Use the principles of superposition an symmetry. The electric fiel always points in the irection of maximum potential ecrease. Equipotential surfaces are perpenicular to electric fiel lines at all points. For equipotential surfaces rawn such that the potential ifference between ajacent surfaces is constant, the surfaces are closer together where the fiel is stronger. Solution The electric fiel lines an equipotential surfaces for the cube: 707

27 Chapter 17: Electric Potential College Physics 109. Strategy The energy store in the capacitor is irectly proportional to the capacitance. Use Eq. (17-16) an the fact that V is constant. Form a proportion. Solution Determine what happens to the energy store in the capacitor. U C e U C 1 1 0A f i i 1 e0 A 0 0 f 1.5 i , so the energy is reuce by 0.0% Strategy Use Eq. (17-18b) an form a proportion. V is constant. Solution Fin the energy store in the capacitor after the ielectric is inserte. 1 C( V) C C0 1 0 C 0( V) 0 0 U, so U U0 3.0 U0. U C C 111. (a) Strategy Treat the axon as a parallel plate capacitor. Use Eq. (17-16) an the fact that the area of the curve surface of a cyliner is ( rl ), where r an L are the raius an length of a cyliner, respectively. Solution Calculate the capacitance per unit length of the axon. e0a e0[( r) L] C, so 1 6 C e0 r (7.0)[ C (N m )](5.010 m) F m. L m (b) Strategy Use Eq. (17-10) an the fact that the magnitue of the electric fiel insie a parallel plate capacitor is given by 0. e Solution The outsie of the membrane has the positive charge, since the potential is higher outsie than insie. 1 E0 e0v 7.0[ C (N m )](0.085 V) 4 V E, so C m. e m (a) Strategy Use Eq. (17-18b), the efinition of capacitance, Eq. (17-14), an the relationships between the quantities (energy, potential ifference, capacitance) before an after the ielectric is inserte. Solution Calculate the initial energy store in the capacitor (without the ielectric) Ui Ci( Vi) ( F)(100.0 V) 0.0 mj C Cf Ci an Q CiVi Cf Vf. So, i V V i f Vi. Cf Calculate the final energy. Vi f f f i i i i mj U C ( V ) C C ( V) U 3.3 mj 6.0 (b) Strategy an Solution Since the energy of the capacitor increases when the ielectric is remove, an external agent has to o positive work to remove the ielectric. 708

28 College Physics Chapter 17: Electric Potential 113. (a) Strategy Use Eq. (17-10). Solution Compute the minimum thickness of the titanium ioxie. V 5.00 V 1.5 m E 6 10 V m (b) Strategy Use Eq. (17-16). Solution Fin the area of the plates. 6 e0 A C ( m)(1.0 F) C, so A e [ C (N m )] 1600 m Strategy Use the efinition of capacitance, Eq. (17-14), to fin the charge that moves through the membrane. Then ivie the charge by e. Solution (a) (b) C Q CV AV (110 F cm )(0.05 cm )[010 V ( 9010 V)] 6 nc A Q e C ions C ion 115. Strategy Let E0 0.0 V m, E1 be the fiel outsie of the ielectric after it is inserte, an E be the fiel insie the ielectric. Use the principle of superposition for the potential after the ielectric is inserte an the fact that E E1. Solution Fin the electric fiel insie the ielectric. Initially: V E0 E Finally: 1 E1 1 V E1 E E1 1 Solve for E 1 in terms of E 0. E 1 1 E 1 E 0 0, so E Calculate E. E1 E0 (0.0 V m) E 8.0 V m