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Timestamp: 2019-04-22 00:13:50+00:00

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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.
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.
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.
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.
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).
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.
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.
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.
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.

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