# Physics exercises_solution: Chapter 24

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## Physics exercises_solution: Chapter 24

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Bộ tài liệu bài tập tham khảo môn vật lý bậc đại học bằng tiếng anh Chapter 24

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## Nội dung Text: Physics exercises_solution: Chapter 24

1. 24.1: Q  CV  (25.0 V)(7.28 μF)  1.82  10 4 C. A 0.00122 m 2 24.2: a) C  ε0  ε0  3.29 pF. d 0.00328 m Q 4.35  10 8 C b) V    13.2 kV. C 3.29  10 12 F V 13.2  10 3 V c) E    4.02  10 6 V m . d 0.00328 m Q 0.148  10 6 C 24.3: a) V    604 V. C 2.45  10 10 F Cd (2.45  1010 F)(0.328  103 m) b) A    0.0091 m 2 . ε0 ε0 V 604 V c) E   3  1.84  10 6 V m . d 0.328  10 m σ d) E   σ  ε0 E  ε0 (1.84 106 V m)  1.63  105 C/m 2 . ε0 σ 24.4: V  Ed  d ε0 2 (5.60  10 12 C m )(0.00180 m)  8.85  10 12 C 2 Nm 2 =1.14 mV 24.5: a) Q  CV  120 μC b) C  ε0 A d d  d 2 means C  C 2 and Q  Q 2  60 μC c) r  2r means A  4 A, C  4C , and Q  4Q  480 μC
2. 24.6: (a) 12.0 V since the plates remain charged. (b) (i) V  Q C Q does not change since the plates are disconnected from the battery. εA C d If d is doubled, C  2 C , so V  2V  24.0 V 1 (ii) A  πr 2 , so if r  2r , then A  4 A, and C  4C which means that 1 V  V  3.00 V 4 24.7: Estimate r  1.0 cm ε0 A  πr 2 ε0 π (0.010 m) 2 C so d  0   2.8 mm d C 1.00  10 12 F The separation between the pennies is nearly a factor of 10 smaller than the diameter of a penny, so it is a reasonable approximation to treat them as infinite sheets. 24.8: (a) V  Ed 100 V  (10 4 N C)d d  10 2 m  1.00 cm ε A ε  R2 C 0  0 d d Cd 4Cd R  πε0 4πε0 12 2 Nm 2 R  4(5.00  10 F)(10 m)(9  10 ) 9 C2 R  4.24  10 2 m  4.24 cm (b) Q  CV  (5pF)(100 V)  500 pC C 2πε0 24.9: a)  L ln(rb ra ) (0.180 m)2πε0 C  4.35  1012 F ln (5.00 0.50) b) V  Q / C  (10.0  10 12 C) /(4.35  10 12 F)  2.30 V
3. C 2πε0 2πε0 2πε0 r 24.10: a)   ln(rb ra )   12  1.77  b  5.84. L ln(rb ra ) C L 31.5  10 F m ra Q C b)  V  (2.60 V)(31.5  10 12 F m)  8.19  10 11 C m . L L 2πε0 2πε0 24.11: a) C L    6.56  10 11 F/m. ln (rb ra ) ln(3.5 mm / 1.5 mm) b) The charge on each conductor is equal but opposite. Since the inner conductor is at a higher potential it is positively charged, and the magnitude is: 2ε0 LV 2ε0 (2.8 m)(0.35 V) Q  CV    6.43  10 11 C. ln(rb ra ) ln (3.5 mm 1.5 mm ) 24.12: a) For two concentric spherical shells, the capacitance is: 1 r r  kCra C   a b   kCrb  kCra  ra rb  rb  k  rb  ra    kC  ra k (116  10 12 F)(0.150 m)  rb   0.175 m. k (116  1012 F)  0.150 m b) V  220 V, and Q  CV  (116  1012 F)(220 V)  2.55  108 C. 1  rb ra  1  (0.148 m)(0.125 m)  24.13: a) C    r  r   k  0.148 m  0.125 m   8.94  10 F.    11 k b a   b) The electric field at a distance of 12.6 cm: kQ kCV k (8.94  1011 F)(120 V) E 2  2   6082 N/C. r r (0.126 m) 2 c) The electric field at a distance of 14.7 cm: kQ kCV k (8.94  1011 F)(120 V) E 2  2   4468 N/C. r r (0.147 m) 2 d) For a spherical capacitor, the electric field is not constant between the surfaces.
4. 1 1 1 1 1 24.14: a)     Ceq C1  C2 C3 ((3.0  5.0)  10 F) (6.0  10 6 F) 6  Ceq  3.42  10 6 F. The magnitude of the charge for capacitors in series is equal, while the charge is distributed for capacitors in parallel. Therefore, Q3  Q1  Q2  VCeq  (24.0 V )(3.42  10 6 F)  8.21  10 5 C. Q1 Q2 C 5 Since C1 and C 2 are at the same potential,   Q2  2 Q1  Q, C1 C 2 C1 3 Q3  8 Q1  8.21  10 5 C  Q1  3.08  105 C, and Q2  5.13  10 5 C. 3 b) V2  V1  Q1 C1  (3.08  10 5 C) /(3.00  10 6 F)  10.3 V. And V3  24.0 V  10.3 V  13.7 V. c) The potential difference between a and d: Vad  V1  V2  10.3 V. 1 1 1 1 1 24.15: a)  1    Ceq ( C1  C2 )  C3 C4 (2.00 μF  4.0 μF) (4.0 μF) 1  Ceq  2.40 μF. Then, Q12  Q3  Q4  Qtotal  CeqV  (2.40  10 6 F)(28.0 V )  6.72  10 5 C and Qtotal 6.72  10 5 C 2Q12  Q3  Q12    2.24  10 5 C, and Q3  4.48  10 5 C. But 3 3 5 also, Q1  Q2  Q12  2.24  10 C. b) V1  Q1 C1  (2.24  10 5 C) (4.00  10 6 F)  5.60 V  V2 V3  Q3 C3  (4.48  10 5 C) (4.00  10 6 F)  11.2 V. V4  Q4 C4  (6.72  105 C) (4.00  106 F)  16.8 V. c) Vad  Vab  V4  28.0 V  16.8 V  11.2 V. 24.16: a) 1 1 1 1 1     Ceq C1 C 2 (3.0  10 F) (5.0  10 6 F) 6  5.33  105 F 1  Ceq  1.88  10 6 F  Q  VCeq  (52.0 V)(1.88  10 6 F)  9.75  10 5 C b) V1  Q / C1  9.75  105 C 3.0  106 F  32.5 V. V2  Q / C2  9.75  105 C 5.0  106 F  19.5 V.
5. 24.17: a) Q1  VC1  (52.0 V)(3.0  10 6 F)  1.56  10 4 C. Q2  VC2  (52.0 V)(5.0  106 F)  2.6  104 C. b) For parallel capacitors, the voltage over each is the same, and equals the voltage source: 52.0 V. 24.18: Ceq   1 C1  C12  1   d1 ε0 A  ε0 A d 2 1  ε0 A d1  d 2 . So the combined capacitance for two capacitors in series is the same as that for a capacitor of area A and separation (d1  d 2 ) . ε0 ( A1  A2 ) 24.19: Ceq  C1  C2    ε0 A1 ε0 A2 d d d . So the combined capacitance for two capacitors in parallel is that of a single capacitor of their combined area ( A1  A2 ) and common plate separation d. 24.20: a) and b) The equivalent resistance of the combination is 6.0 F, therefore the total charge on the network is: Q  CeqVeq (6.0 μF)(36 V)  2.16  10 4 C. This is also the charge on the 9.0 μF capacitor because it is connected in series with the point b. So: Q9 2.16  10 4 C V9    24 V. C9 9.0  10 6 F Then V3  V11  V12  V6  V  V9  36 V  24 V  12 V.  Q3  C 3V3  (3.0 F)(12 V)  3.6  10 5 C.  Q11  C11V11  (11 μF)(12 V)  1.32  104 C.  Q6  Q12  Q  Q3  Q11  2.16  10 4 C  3.6  10 5 C  1.32  10 4 C.  4.8  10 5 C. So now the final voltages can be calculated: Q6 4.8  105 C V6    8 V. C6 6.0  10 6 F Q12 4.8  10 5 C V12    4 V. C12 12  10 6 F c) Since the 3 μF, 11 μF and 6 μF capacitors are connected in parallel and are in series with the 9 μF capacitor, their charges must add up to that of the 9 μF capacitor. Similarly, the charge on the 3 μF, 11 μF and 12 μF capacitors must add up to the same as that of the 9 μF capacitor, which is the same as the whole network. In short, charge is conserved for the whole system. It gets redistributed for capacitors in parallel and it is equal for capacitors in series.
6. 24.21: Capacitances in parallel simply add, so: 1 1  1 1    (11  4.0  x) μF  9.0 μF   (15  x) μF  72 μF  x  57 μF.  Ceq 8.0 μF   24.22: a) C1 and C 2 are in parallel and so have the same potential across them: Q2 40.0  106 C V   13.33 V C2 3.00  10 6 F Thus Q1  VC1  (13.33 V)(3.00  10 6 F)  80.0  10 6 C. Since Q3 is in series with the parallel combination of C1 and C2 , its charge must be equal to their combined charge: 40.0  106 C  80.0  106 C  120.0  106 C b) The total capacitance is found from: 1 1 1 1 1     Ctot C|| C3 9.00  10 F 5.00  10 6 F 6 Ctot  3.21 μF and Qtot 120.0  106 C Vab    37.4 V Ctot 3.21  10 6 F 24.23: V1  Q1 C1  (150 μC) (3.00 μF)  50 V C1 and C2 are in parallel, so V2  50 V V3  120 V  V1  70 V 24.24: a) V  Q / C  (2.55 μC) (920  1012 F)  2772 V. b) Since the charge is kept constant while the separation doubles, that means that the capacitance halves and the voltage doubles to 5544 V. c) U  1 CV 2  1 (920  10 12 F)(2772 V) 2  3.53  10 3 J. Now if the separation 2 2 is doubled, the capacitance halves, and the energy stored doubles. So the amount of work done to move the plates equals the difference in energy stored in the capacitor, which is 3.53  10 3 J. 24.25: E  V d  (400 V) (0.005 m)  8.00  104 V m. And u  1 ε0 E 2  1 ε0 (8.00  10 4 V m) 2  0.0283 J m 3 . 2 2
7. 24.26: a) C  Q V  (0.0180 μC) (200 V)  9.00  10 11 F. ε0 A Cd (9.00  1011 F)(0.0015 m) b) C   A   0.0152 m 2 . d ε0 ε0 c) E max  Vmax d  Vmax  E max d  (3.00  10 6 V m)(0.0015 m)  4500 V. Q 2 (1.80  10 8 C) 2 d) U   11  1.80  10 6 J. 2C 2(9.00  10 F) 24.27: U  1 CV 2  1 (4.50  10 4 F)(295 V) 2  19.6 J. 2 2 24.28: a) Q  CV0 . b) They must have equal potential difference, and their combined charge must add up to the original charge. Therefore: Q Q V  1  2 and also Q1  Q2  Q  CV0 C1 C 2 C Q Q2 Q C1  C and C2  so 1   Q2  1 2 C (C 2) 2 3 2 Q 2Q 2  Q  Q1  Q1  Q so V  1   V0 2 3 C 3C 3 1  Q1 Q2  1  ( 2 Q) 2 2( 1 Q) 2  1 Q 2 1 2 2 c) U    2  C  C   2  C  C   3 C  3 CV0 3 3 2 1 2    2 2 d) The original U was U  1 CV0  U  61 CV0 . 2 e) Thermal energy of capacitor, wires, etc., and electromagnetic radiation. Q 2 xQ 2 24.29: a) U 0   . 2C 2ε0 A ( x  dx ) Q 2 b) Increase the separation by dx  U  2ε0 A  U 0 (1  dx x). The change is Q2 then 2 ε0 A dx . c) The work done in increasing the separation is given by: dxQ 2 Q2 dW  U  U 0   Fdx  F  . 2 ε0 A 2 ε0 A d) The reason for the difference is that E is the field due to both plates. The force is QE if E is the field due to one plate is Q is the charge on the other plate.
8. 24.30: a) If the separation distance is halved while the charge is kept fixed, then the capacitance increases and the stored energy, which was 8.38 J, decreases since U  Q 2 2C. Therefore the new energy is 4.19 J. b) If the voltage is kept fixed while the separation is decreased by one half, then the doubling of the capacitance leads to a doubling of the stored energy to 16.76 J, using U  CV 2 2, when V is held constant throughout. 24.31: a) U  Q 2 2C Q  2UC  2(25.0 J)(5.00  109 F)  5.00  104 C The number of electrons N that must be removed from one plate and added to the other is N  Q e  (5.00  104 C) /(1.602  1019 C)  3.12  1015 electrons. b) To double U while keeping Q constant, decrease C by a factor of 2. C  ε0 A / d ; halve the plate area or double the plate separation. Q 8.20  10 12 C 24.32: C    3.417  10 12 farad V 2.40 V Since C  Kε0 A d for a parallel plate capacitor Kε0 A (1.00)(8.85  10 12 C 2 / N  m 2 )(2.60  10 3 m 2 ) d  C 3.417  10 12 farad  6.734  10 3 m The energy density is thus CV 2 1 1 (3.42  1012 farad)(2.40 V) 2 J u 2  2 3 3  5.63  10 7 3 Ad (2.60  10 m )(6.734  10 m) 2 m 1 2U 2(3.20  109 J ) 24.33: a) U  QV  Q    1.60  10 9 C. 2 V 4.00 V C 2πε0 r b)   a  exp(2πε0 L C )  exp(2πε0 LV Q) L ln(ra rb ) rb r  a  exp(2πε0 (15.0 m) (4.00 V) (1.60  10  9 C))  8.05. rb
9. 24.34: a) For a spherical capacitor: 1 ra rb 1 (0.100 m)(0.115 m) C   8.53  1011 F k rb  ra k (0.115 m  0.100 m)  V  Q C  (3.30  10 9 C) (8.53  1011 F)  38.7 V. 1 (8.53  10 11 F)(38.7 V) 2 b) U  CV  2  6.38  10 8 J. 2 2 2 2 1 ε  kq  ε  kVC  ε k 2 (120 V) 2 (8.94  10 11 F) 2 24.35: a) u  ε0 E 2  0  2   0  2   0 2 2r  2 r  2 (0.126 m) 4  u  1.64  104 J m3 . b) The same calculation for r  14.7 cm  u  8.83  10 5 J m 3 . c) No, the electric energy density is NOT constant within the spheres. 2 1 1  1 q 1 (8.00  10 9 C) 2 24.36: a) u  ε0 E 2  ε0     1.11  10 4 J m 3 . 2 2  4πε0 r 2    32π 2 ε0 (0.120 m) 4 b) If the charge was –8.00 nC, the electric field energy would remain the same since U only depends on the square of E. 24.37: Let the applied voltage be V. Let each capacitor have capacitance C. U  1 CV 2 2 for a single capacitor with voltage V. a) series Voltage across each capacitor is V 2. The total energy stored is 1  U s  2 C V 2 2   1 CV 2 4 2  parallel Voltage across each capacitor is V. The total energy stored is   U p  2 1 CV 2  CV 2 2 U p  4U s b) Q  CV for a single capacitor with voltage V . Qs  2C V 2  CV ; Qp  2(CV )  2CV ; Qp  2Qs c) E  V d for a capacitor with voltage V Es  V 2d ; Ep  V d ; Ep  2 Es
10. 24.38: a) C  Kε0 A d gives us the area of the plates: Cd (5.00  1012 farad)(1.50  103 m) A 12 2  8.475  10 4 m 2 Kε0 (1.00)(8.85  10 C / N  m ) 2 We also have C  Kε0 A d  Q V , so Q  K 0 A(V d ). V d is the electric field between the plates, which is not to exceed 3.00  10 4 N C. Thus Q  (1.00)(8.85  10 12 C 2 N  m 2 )(8.475  10 4 m 2 )(3.00  10 4 N C)  2.25  10 10 C b) Again, Q  Kε0 A(V d )  2.70ε0 A(V d ). If we continue to think of V d as the electric field, only K has changed from part (a); thus Q in this case is (2.70)(2.25  10 10 C)  6.08  10 10 C. 24.39: a) σ i  ε0 ((3.20  2.50)  105 V m)  6.20  10 7 C m 2 . The field induced in the dielectric creates the bound charges on its surface. E 3.20  105 V m b) K  0   1.28. E 2.50  105 V m 24.40: a) E0  KE  (3.60)(1.20  106 V m)  4.32  106 V m  σ  ε0 E0  3.82  105 C m 2 .  1 b) σ i  σ 1    (3.82  10 5 C m 2 )(1  1 3.60)  2.76  10 5 C m 2 .  K c) U  2 CV 2  uAd  1 Kε0 E 2 Ad 1 2  U  1 (3.60)ε0 (1.20  106 V m) 2 (0.0018 m)(2.5  10 4 m 2 )  1.03  10 5 J. 2 Kε0 A Kε0 AE CV (1.25  10 9 F)(5500 V) 24.41: C    A   0.0135 m 2 . d V Kε0 E (3.60)ε0 (1.60  10 7 V m) 24.42: Placing a dielectric between the plates just results in the replacement of  for  0 in the derivation of Equation (24.20). One can follow exactly the procedure as shown for Equation (24.11).
11. 24.43: a) ε  Kε0  (2.6)ε0  2.3  10 11 C 2 Nm 2 . b) Vmax  Emax d  (2.0  107 V m)(2.0  10 3 m)  4.0  10 4 V. σ c) E   σ  εE  (2.3  10 11 C 2 Nm 2 )(2.0  10 7 V m)  0.46  10 3 C m 2 . Kε0  1 And σ i  σ 1    (0.46  10 3 C m 2 )(1  1 2.6)  2.8  10 4 C m 2 .  K 24.44: a) Q  Q  Q0  ( K  1)Q0  ( K  1)C0V0  (2.1)(2.5  10 7 F)(12 V)  6.3  106 C. b) Qi  Q1  K   (9.3  10 6 C)(1  1 3.1)  6.3  106 C. 1 c) The addition of the mylar doesn’t affect the electric field since the induced charge cancels the additional charge drawn to the plates. 1 2U 0 2(1.85  10 5 J) 24.45: a) U 0  C0V 2  V    10.1 V. 2 C0 (3.60  10 7 F) 1 U 2(2.32  10 5  1.85  10 5 J ) b) U  KC 0V 2  K    2.27. 2 C 0V 2 (3.60  10 7 F)(10.1 V) 2
12. 24.46: a) The capacitance changes by a factor of K when the dielectric is inserted. Since V is unchanged (The battery is still connected), Cafter Q 45.0 pC  after   K  1.80 C before Qbefore 25.0 pC b) The area of the plates is πr 2  π (0.0300 m) 2  2.827  10 3 m 2 , and the separation between them is thus Kε A (1.00)(8.85  1012 C 2 N  m 2 )(2.827  103 m 2 ) d 0  C 12.5  1012 farad  2.002  10 3 m Before the dielectric is inserted, Kε A Q C 0  d V Qd (25.0  1012 C)(2.00  10 3 m) V  Kε0 A (1.00)(8.85  1012 C 2 N  m 2 )(2.827  10 3 m 2 )  2.000 V The battery remains connected, so the potential difference is unchanged after the dielectric is inserted. c) Before the dielectric is inserted, Q 25.0  1012 C E  ε0 KA (8.85  1012 C2 N  m 2 )(1.00)(2.827  10 3 m 2 )  999 N C Again, since the voltage is unchanged after the dielectric is inserted, the electric field is also unchanged. 24.47: a) before: V0  Q0 C 0  (9.00  10 6 C) (3.00  10 6 F)  3.00 V after: C  KC 0  15.0 F; Q  Q0 V  Q C  0.600 V; V decreases by a factor of K b) E  V d , the same at all points between the plates (as long as far from the edges of the plates) before: E  (3.00 V) (2.00  10 3 m)  1500 V m after: E  (0.600 V) (2.00  10 3 m)  300 V m
13.   Q free q q 24.48: a)  KE  A  ε0  KE 4πd 2   E  ε0 4πεd 2 .   q q  qb q  qb q  qb b)  E  dA  total  f  E 4πd 2  E ε0 ε0 ε0 4πε0 d 2  qtotal  q  qb  q / K . c) The total bound change is qb  q K  1. 1   24.49: a) Equation (25.22):  KE  dA  Q free Q Q Q ε0  KEA  ε0 E Kε 0 A  εA . Qd Qd b) V  Ed   . Kε0 A εA Q εA ε A c) C    K 0  KC0 . V d d ε0 A ε0 (0.16 m) 2 24.50: a) C    4.8  1011 F. d 4.7  10 3 m b) Q  CV  (4.8  10 11 F) (12 V)  0.58  10 9 C. c) E= V d =(12 V)/(4.7  10 3 m) =2553 V m . d) U  1 CV 2  1 (4.8  10 11 F)(12 V) 2  3.46  10 9 J. 2 2 e) If the battery is disconnected, so the charge remains constant, and the plates are pulled further apart to 0.0094 m, then the calculations above can be carried out just as before, and we find: a) C  2.41  10 11 F b) Q  0.58  10 9 C. Q 2 (0.58  109 C) 2 c) E  2553 V m d) U    6.91  10 9 J. 2C 2(2.41  1011 F) 24.51: If the plates are pulled out as in Problem 24.50 the battery is connected, ensuring that the voltage remains constant. This time we find: V 12 V V a) C  2.4  10 11 F b) Q  2.9  1010 C c) E    1.3  10 3 d 0.0094 m CV 2 (2.4  10 11 F) (12 V) 2 d) U    1.73  10 9 J. 2 2
14. 24.52: a) System acts like two capacitors in series so Ceq   1 C1  C2  1 1 ε0 L2 ε0 L2 1 Q2 1 Q2 Q 2d C1  C 2  d so Ceq  2d  U 2 C  ε L2  2 2d 0   ε0 L2 . b) After rearranging, the E fields should be calculated. Use superposition recalling E  2 εQ A for a single plate (not εQA since charge Q is only on one face). 0 0  Q   Q   Q   Q  Q between 1 and 3: E    2ε L   2     2     2     2  2  0 1  2ε0 L 3  2ε0 L  2  2ε0 L  4 ε0 L  Q   Q   Q   Q  2Q between 3 and 2: E    2ε L2    2ε L2    2ε L2    2ε L2   ε L2         0 1  0 3  0  2  0  4 0  Q   Q   Q   Q  Q between 2 and 4: E    2ε L2    2ε L2    2ε L2    2ε L2   ε L2         0 1  0 3  0 2  0  4 0 1  1  Q2 4Q 2 Q2  3Q 2 d U new   ε0 E 2  L2 d  ε0  2 4  2 4  2 4  L2 d  2  2  ε0 L  ε0 L ε0 L   ε0 L2 3Q 2 d Q 2 d 2Q 2 d U  U new  U    ε0 L2 ε0 L2 ε0 L2 This is the work required to rearrange the plates. 24.53: a) The power output is 600 W, and 95% of the original energy is converted.  E  Pt  ( 2.70  105 W ) (1.48  10 3 s)  400 J  E0  400 J  421 J. 0.95 1 2U 2(421 J) b) U  CV 2  C  2   0.054 F 2 V (125 V) 2 Aε0 (4.20  10 5 m 2 )ε0 24.54: C0   4  5.31  10 13 F d 7.00  10 m  C  C0  0.25 pF  7.81  10 13 F. Aε0 Aε (4.20  105 m 2 )ε0 But C   d  0   4.76  10 4 m. d C 13 7.81  10 F Therefore the key must be depressed by a distance of: 7.00  10 4 m  4.76  10 4 m  0.224 mm.
15. 2πε0 L 2πε0 L 2πε0 L 2πra Lε0 ε0 A 24.55: a) d  ra : C      . ln(rb ra ) ln((d  ra ) ra ) ln(1  d ra ) d d b) At the scale of part (a) the cylinders appear to be flat, and so the capacitance should appear like that of flat plates. 24.56: Originally: Q1  C1V1  (9.0 μF) (28 V)  2.52  104 C; Q2  C2V2  (4.0 μF)  (28 V)  1.12  10 4 C, and Ceq  C1  C2  13.0 μF. So the original energy stored is U  1 CeqV 2  1 (13.0  106 F) (28 V) 2  5.10 103 J. Disconnect and flip the capacitors, 2 2 so now the total charge is Q  Q2  Q1  1.4  10 4 C, and the equivalent capacitance is still the same, Ceq  13.0 μF. So the new energy stored is : Q2 (1.4  10 4 C) 2 U   7.54  10 4 J 2Ceq 2(13.0  10 6 F)  U  7.45  10 4 J  5.10  10 3 J   4.35  10 3 J. 24.57: a) Ceq  4.00 μF  6.00 μF  10.00 μF, and Qtotal  Ceq V  (10.00 μF) (660 V)  6.6  10 3 C. The voltage over each is 660 V since they are in parallel. So: Q1  C1V1  (4.00 μF) (660 V)  2.64  10 3 C. Q2  C2V2  (6.00 μF) (660 V)  3.96  10 3 C. b) Qtotal  3.96  10 3 C  2.64  10 3 C  1.32  10 3 C, and still Ceq  10.00  F, so the voltage is V = Q/C = (1.32  10 3 C) (10.00 μF)  132 V, and the new charges: Q1  C1V1  (4.00 μF)(132 V)  5.28  104 C. Q2  C2V2  (6.00 μF)(132 V)  7.92  10 4 C.
16. 24.58: a) C eq  C 2   C. So the total capacitance is the same as each individual capacitor, and C 2 the voltage is spilt over each so that V  480 V. Another solution is two capacitors in parallel that are in series with two others in parallel. b) If one capacitor is a moderately good conductor, then it can be treated as a “short” and thus removed from the circuit, and one capacitor will have greater than 600 V over it. 1 1 1 1 24.59: a)     C1  C5  2C 2 and  Ceq C1 C 2  C3  C14 1 1 C5 1 2 2 5 3 C 2  C3  C 4 so    C 2  Ceq  C 2  2.52 μF. Ceq C1 3C 2 3 5 b) Q  CV  (2.52 μF)(220 V)  5.54  104 C  Q1  Q5  V1  V5  (5.54  10 4 C) / (8.4  10 6 F)  66 V. So V2  220  2(66)  88 V  Q2  (88 V)(4.2 μF)  3.70  10 4 C. Also V3  V4  1 2 (88 V)  44 V  Q3  Q4  (44 V)(4.2 μF)  1.85  10 4 C.
17. 24.60: a) With the switch open: Ceq   1 3 μF   1 1 6 μF   1 3 μF     4.00 μF 1 1 6 μF  Qtotal  CeqV  (4.00 μF) (210 V)  8.4  104 C . By symmetry, each capacitor carries 4.20  10 4 C. The voltages are then just calculated via V=Q/C. So: Vad  Q / C3  140 V, and Vac  Q / C6  70 V  Vcd  Vad  Vac  70 V. b) When the switch is closed, the points c and d must be at the same potential, so the equivalent capacitance is: 1  1 1  Ceq    (3  6) μF  (3  6) μF   4.5 μF.     Qtotal  CeqV  (4.50 μF) (210 V)  9.5  10 4 C, and each capacitor has the same potential difference of 105 V (again, by symmetry) c) The only way for the sum of the positive charge on one plate of C 2 and the negative charge on one plate of C1 to change is for charge to flow through the switch. That is, the quantity of charge that flows through the switch is equal to the charge in Q2  Q1  0. With the switch open, Q1  Q2 and Q2  Q1  0. After the switch is closed, Q2  Q1  315 μC; 315 μC of charge flowed through the switch. 1  1 1 1  24.61: a) Ceq    8.4 μF  8.4 μF  4.2 μF   2.1 μF     Q  CeqV  (2.1 μ F) (36 V)  7.50  10 5 C. b) U  1 CV 2  1 (2.1 μF) (36 V) 2  1.36  10 3 J. 2 2 c) If the capacitors are all in parallel, then: Ceq  (8.4 μF  8.4 μF  4.2 μF)  21 μF and Q  3(7.56  10 5 C)  2.27  10 4 C, and V  Q C  (2.27  10 4 C) / (21 μF)  10.8 V. d) U  1 CV 2  1 (21 F) (10.8 V) 2  1.22  10 3 J. 2 2
18. 1  1 1  24.62: a) Ceq    4.0 μF 6.0 μF   2.4  10 F   6    Q  CeqV  (2.4  10 F) (600 V)  1.58  10 3 C 6 and V2  Q / C 2  (1.58  10 3 C) (4.0 μF)  395 V  V3  660 V  395 V  265 V. b) Disconnecting them from the voltage source and reconnecting them to themselves we must have equal potential difference, and the sum of their charges must be the sum of the original charges: Q1  C1V and Q2  C2V  2Q  Q1  Q2  (C1  C2 ) V 2Q 2(1.58  10 3 C) V    316 V. C1  C2 10.0  10 6 F  Q1  (4.00  10 6 F)(316 V)  1.26  103 C.  Q2  (6.00  10 6 F)(316 V)  1.90  10 3 C. 24.63: a) Reducing the furthest right leg yields C   1 6.9  F  1 6.9  F  1  6.9  F 1  2.3 μF  C1 / 3. It combines in parallel with a C 2  C  4.6 μF  2.3 μF  6.9 μF  C1. So the next reduction is the same as the first: C  2.3 μF  C1 / 3. And the next is the same as the second, leaving 3 C1 ’s in series so Ceq  2.3 μF  C1 / 3. b) For the three capacitors nearest points a and b: QC1  C eqV  (2.3  10 6 F)(420 V)  9.7  10 4 C and QC 2  C2V2  (4.6  10 6 F) (420 V) 3  6.44  10 4 C. c) Vcd  1  420 V   46.7 V, since by symmetry the total voltage drop over the 3 3 equivalent capacitance of the part of the circuit from the junctions between a, c and d, b is 420 V, and the equivalent capacitance is that of three equal capacitors C1 in series. 3 Vcd is the voltage over just one of those capacitors, i.e., 1 3 of 420 V. 3 24.64: (a) Cequiv  C1  C2  C3  60 μF Q  CV  (60 μ F) (120 V)  7200 μC 1 1 1 1 (b)    C equiv C1 C 2 C 3 Cequiv  5.45 μF Q  CV  (5.45 μF)(120 V)  654 μC
19. 24.65: a) Q is constant. with the dielectric: V  Q C  Q ( KC0 ) without the dielectric: V0  Q C0 V0 / V  K , so K  (45.0 V)/(11.5 V)  3.91 b) Let C0  ε0 A d be the capacitance with only air between the plates. With the dielectric filling one-third of the space between the plates, the capacitor is equivalent to C1 and C 2 in parallel, where C1 has A1  A / 3 and C 2 has A2  2 A / 3 C1  K C0 3 , C2  2 C0 3; Ceq  C1  C2  (C0 3) ( K  2) Q Q 3   3   3  V    K  2   V0  K  2   (45.0 V)  5.91   22.8 V    Ceq C0       24.66: a) This situation is analagous to having two capacitors C1 in series, each with separation 1 2 (d  a ). Therefore C   1 C1  C1 1  1  1 C1  2 1 ε0 A 2 (d  a) 2  ε0 A d a . ε0 A ε A d d b) C   0  C0 . d a d d a d a c) As a  0, C  C0 . And as a  d , C  . 24.67: a) One can think of “infinity” as a giant conductor with V  0. b) C  V  (Q / 4πε 0 R )  4πε0 R, where we’ve chosen V  0 at infinity. Q Q c) Cearth  4πε0 Rearth  4πε0 (6.4  106 m)  7.1  10 4 F. Larger than, but comparable to the capacitance of a typical capacitor in a circuit.
20. 1 24.68: a) r  R : u  ε0 E 2  0. 2 2 1  Q  Q2 b) r  R : u  ε0 E  ε0  1 2   . 2 2  4πε0 r 2  32π 2 ε0 r 4     Q 2 dr Q2 c) U   udV  4π  r udr  8πε0  r 2 8πε0 R  2  R R 2 1 Q d) This energy is equal to 2 4 πε 0 R which is just the energy required to assemble all the charge into a spherical distribution. (Note, being aware of double counting gives the factor of 1 2 in front of the familiar potential energy formula for a charge Q a distance R from another charge Q.) 2 Q2 e) From Equation (24.9): U  QC  8πε 0 R from part (c)  C  4πε0 R, as in 2 Problem (24.67). 2 1 1  kQr  kQ 2 r 2 24.69: a) r  R:u  ε0 E 2  ε0  3   2 2  R  8πR 6  2 1 1  kQ  kQ 2 b) r  R : u  ε0 E 2  ε0  2   . 2 2  r  8πr 4 R R kQ 2 4 kQ 2 c) r  R : U   udV  4π  r 2udr  2 R6  r dr  . 0 0 10 R   kQ 2 dr kQ 2 3kQ 2 r  R : U   udV  4π  r 2udr   r 2 2R  U  . R 2 R 5R 2 1 1  λ  λ2 24.70:  a) u  ε0 E 2  ε0   2 2  2πε r  2 2  0  8π ε0 r . rb Lλ 2 dr U λ2 b) U   udV  2πL  urdr     ln(rb / ra ). 4πε0 ra r L 4πε0 c) Using Equation (24.9): Q2 Q2 λ2 L U  ln (rb / ra )  ln (rb / ra )  U of part (b). 2C 4 0 L 4 0