f5ch26 - CHAPTER 26 CAPACITANCE 711 CHAPTER 26 Answer to...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: CHAPTER 26 CAPACITANCE 711 CHAPTER 26 Answer to Checkpoint Questions 1. 2. 3. 4. 5. 6. (a) same; (b) same (a) decreases; (b) increases; (c) decreases (a) V , q=2; (b) V=2, q (a) q0 = q1 + q34 ; (b) equal (C3 and C4 are in series) (a) same; (b) { (d) increase; (e) same (same potential dierence across same plate separation) (a) same; (b) decrease; (c) increase Answer to Questions 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. (a) increase; (b) same (a) series; (b) parallel; (c) parallel (a) parallel; (b) series (a) no; (b) yes; (c) all tie (a) C=3; (b) 3C ; (c) parallel parallel, C1 alone, C2 alone, series (a) equal; (b) less (a) same; (b) same; (c) more; (d) more (a) { (d) less (a) decreases; (b) decreases; (c) increases (a) 2; (b) 3; (c) 1 (a) less; (b) more; (c) equal; (d) more Increase plate separation d, but also plate area A, keeping A=d constant a, 2; b, 1; c, 3 B and C tie, then A 712 CHAPTER 26 CAPACITANCE 16. (a) increases; (b) increases; (c) decreases; (d) decreases; (e) same, increases, increases, increases Solutions to Exercises & Problems The minimum charge measurable is 1E qmin = CVmin = (50 pF)(0:15 V) = 7:5 pC : 2E (a) C = qV = 70 pC = 3:5 pF : 20 V (b) The capacitance is independent of q, i.e. it is still 3:5 pF. (c) q pC V = C = 200 pF = 57 V : 3:5 3E Charge ows until the potential dierence across the capacitor is the same as the emf of the battery. The charge on the capacitor is then q = C E and this is the same as the total charge that has passed through the battery. Thus q = (25 10 6 F)(120 V) = 3:0 10 3 C. 4E 2 F [0 ] = m = VCm = ( N mCC) m = NCm2 : = 5E (a) The capacitance of a parallel plate capacitor is given by C = 0 A=d, where A is the area of each plate and d is the plate separation. Since the plates are circular the plate area is A = R2 , where R is the radius of a plate. Thus 2 12 2 2 :2 C = 0 R = (8:85 10 1:3F/m)(8m 10 m) = 1:4 10 d 10 3 10 F = 140 pF : CHAPTER 26 CAPACITANCE 713 (b) The charge on the positive plate is given by q = CV , where V is the potential dierence across the plates. Thus q = (1:4 10 10 F)(120 V) = 1:7 10 8 C = 17 nC. Use C = A0 =d. Thus 6E A0 = (1:00 m2 )(8:85 10 d= C 1:00 F 12 C2=N m2 ) = 8:85 10 12 m : Since d is much less than the size of an atom ( 10 structed. (a) Use Eq. 26-17: 7E 10 m), this capacitor cannot be con- :0 C = 40 b ab a = (8:99 109 (40m2mm)(38::0 mm) 38:0 mm) = 84:5 pF : N =C2 )(40 0 mm (b) Let the area required be A. Then C = 0 A=(b a), or 5 38:0 A = C (b a) = (84:(8:pF)(40:0 mm 2=N m2mm) = 191 cm2 : 12 C 85 10 ) 0 The charge accumulated is given by (0:25 m)(15 103 V) = 4:1 10 7 C : q = CV = 40 RV = 8:99 109 N m2 =C2 8E (a) Use q = CV = 0 AV=d to solve for A: 9E 10 :0 A = Cd = (10(8:85 10F)(1C2N10 2 )m) = 1:1 10 3 m2 : 12 = m 12 3 0 (b) Now C0 = C d 1:0 mm d0 = (10 pF) 0:9 mm = 11 pF : 714 CHAPTER 26 CAPACITANCE (c) The new potential dierence is V 0 = q=C 0 = CV=C 0 . Thus V = V 0 V = (10 pF)(12 V) 12 V = 1:2 V : 11 pF In a microphone, mechanical pressure applied to the aluminum foil as a result of sound can cause the capacitance of the foil to change, thereby inducing a variable V in response to the sound signal. 10E You want to nd the radius of the combined spheres, then use C = 40 R to nd the capacitance. When the drops combine the volume is doubled. It is then V = 2(4=3)R3 . The new radius R0 is given by 4 (R0 )3 = 2 4 R3 ; 3 3 so R0 = 21=3 R : The new capacitance is C 0 = 40 R0 = 40 (21=3 R) = 21=3 C = 1:26C . In Eq. 26-14, let b = a + d with d=a 1, we have 11P L aL C = 20 ln(L ) = 20 ln(1 + d=a) 20 L = 0 (2d ) 0dA ; b=a d=a where A = 2aL is roughly the surface area of either of the cylinders. Here the approximation ln(1 + x) x for jxj 1 was used. Since b a d and a b, from Eq. 26-17 12P ab a C = 40 b ab a = 0 (4d ) 0 (4d ) = 0dA : 2 13P dC = d A = 0 dA 0 A dx = 0 A 1 dA 1 dx 2 dT dT 0 x x x A dT x dT 1 dA 1 dxdT x dT = C A dT x dT : CHAPTER 26 CAPACITANCE 715 (b) Set dC=dT = 0 to obtain So x = 2Al = 2(23 10 6 = C ) = 4:6 10 5 = C : 14E 1 dA = = 2 = : Al x A dT Al The equivalent capacitance is given by Ceq = q=V , where q is the total charge on all the capacitors and V is the potential dierence across any one of them. For N identical capacitors in parallel Ceq = NC , where C is the capacitance of one of them. Thus NC = q=V and 1 00 N = VqC = (110 V)(1::00 C 10 6 F) = 9090 : 15E F)(5 2 Ceq = C3 + CC1 CC = 4:00 F + (10::00F + 5::00 F) = 7:33 F : + 10 00 F 1 2 16P ( F + 00 F)(4 Ceq = CC1 + C2 )C3 = (10::00F + 55::00 F + 4::00 F) = 3:16 F : 10 00 F 1 + C2 + C3 The charge that passes through meter A is 17E q = Ceq V = 3CV = 3(25:0 F)(4200 V) = 0:315 C : 18E (a) 00 F)(4 2 Ceq = CC1 CC = (6::00 F + 4::00 F) = 2:40 F : 6 00 F 1+ 2 (b) q = Ceq V = (2:40 F)(200 V) = 4:80 104 C: (c) V1 = q=C1 = 4:80 104 C=2:40 F = 120 V, and V2 = V V1 = 200 V 120 V = 80 V: 19E (a) Now Ceq = C1 + C2 = 6:00 F + 4:00 F = 10:0 F: 716 CHAPTER 26 CAPACITANCE (b) q1 = C1 V = (6:00 F)(200 V) = 1:20 10 3 C; q2 = C2 V = (4:00 F)(200 V) = 8:00 10 4 C: (c) V1 = V2 = 200 V: Let x be the separation of the plates in the lower capacitor. Then the plate separation in the upper capacitor is a b x. The capacitance of the lower capacitor is C` = 0 A=x and the capacitance of the upper capacitor is Cu = 0 A=(a b x), where A is the plate area. Since the two capacitors are in series the equivalent capacitance is determined from 1 = 1 + 1 = x +a b x = a b: Ceq C` Cu 0 A 0 A 0 A Thus the equivalent capacitance is given by Ceq = 0 A=(a b) and is independent of x. (a) The equivalent capacitance of the three capacitors connected in parallel is Ceq = 3C = 30 A=d = 0 A=(d=3). Thus the required spacing is d=3. (b) Now Ceq = C=3 = 0 A=3d, so the spacing should be 3d. (a) The equivalent capacitance is Ceq = C1 C2 =(C1 + C2 ): Thus the charge q on each capacitor is 22P 21P 20P C V q = Ceq V = C 1 C2C = (2:0 2:F)(8:0 8F)(300 V) = 4:8 10 4 C : 0 F + :0 F 1+ 2 The potential dierences are: V1 = q=C1 = 4:8 10 4 C=2:0 F = 240 V, V2 = V V1 = 300 V 240 V = 60 V: 0 0 (b) Now we have q1 =C1 = q2 =C2 = V 0 (V 0 being the new potential dierence across each 0 + q2 = 2q. Solve for q1 , q2 and V : 0 0 0 capacitor) and q1 8 > q0 = 2C1 q = 2(2:0 F)(4:8 10 4 C) = 1:9 10 4 C ; > 1 > > > C1 + C2 2:0 F + 8:0 F < 0 = 2q q1 = 7:7 10 4 C ; > q2 0 > > 0 q1 1:92 10 4 C >V = = > : C 2:0 F = 96 V : 1 (c) Now the capacitors will simply discharge themselves, leaving q1 = q2 = 0 and V1 = V2 = 0. CHAPTER 26 CAPACITANCE 717 For maximum capacitance the two groups of plates must face each other with maximum area. In this case the whole capacitor consists of (n 1) identical single capacitors connected in parallel. Each capacitor has surface area A and plate separation d so its capacitance is given by C0 = 0 A=d. Thus the total capacitance of the combination is 1) C = (n 1)C0 = (n d 0 A : 23P (a) and (b) The original potential dierence V1 across C1 is (3 16 V1 = CCeq V = 10::0 FF)(100 V) = 21:1 V : + 5:00 F 1 + C2 Thus V1 = 100 V 21:1 V = 79 V and q1 = C1 V1 = (10:0 F)(79 V) = 7:9 10 4 C: (a) Put ve such capacitors in series. Obviously the equivalent capacitance is 2:0 F=5 = 0:40 F, and with each capacitor taking a 200-V potential dierence the equivalent capacitor can withstand 1000 V. (b) As one possibility, you can take three identical arrays of capacitors, each array being a ve-capacitor combination described in part (a) above, and hook up the arrays in parallel. The equivalent capacitance is now Ceq = 3(0:40 F) = 1:2 F; and with each capacitor taking a 200-V potential dierence the equivalent capacitor can withstand 1000 V. 25P 24P (a) The potential dierence across C1 is V1 = 10 V. Thus q1 = C1 V1 = (10 F)(10 V) = 1:0 10 4 V. (b) Let C = 10 F. Consider rst the three-capacitor combination consisting of C2 and its two closest neighbors, each of capacitance C . The equivalent capacitance of this combination is Ceq = C + CC2 C = 1:5C : +C The voltage drop across this combination is then 2 26P CV 2 1 V = C CVC = C + 115C = 5 V1 : + eq : 718 CHAPTER 26 CAPACITANCE Since this voltage dierence is divided equally between C2 and the one connected in series with it, the voltage dierenc across C2 satis es V2 = V=2 = V1 =5. Thus 10 V q2 = C2 V2 = (10 F) 5 = 2:0 10 5 V : The charge initially carried by the 100-pF capacitor is q1 = C1 V1 = (100 pF)(50 V) = 5:0 10 9 C. When connected with the second capacitor, the charge on the rst one 0 reduces to q1 = C1 V10 = (100 pF)(35 V) = 3:5 10 9 C, which means that the second one 0 now has a charge of q2 = q1 q1 = 1:5 10 9 C: Thus its capacitance is C2 = q2 =V2 = 1:5 10 9 C=35 V = 43 pF: (a) Firstly, the equivalent capacitance of C2 and the the other 4:0-F capacitor connected in series with it is given by 4:0 F=2 = 2:0 F. This combination is then connected in parallel with each of the two 2:0-F capacitors, resulting in an equivalent capacitance C = 2:0 F + 2:0 F + 2:0 F = 6:0 F. This four-capacitor combination is then connected in series with another combination, which consists of the two 3:0-F capacitors connected in parallel, whose equivalent capacitance in turn is C 0 = 3:0 F + 3:0 F = 6:0 F. So nally the equivalent capacitance of the entire circuit is given by 0 F)(6 Ceq = CCCC 0 = (6::00F + 6::00F) = 3:0 F : + 6 F (b) Let V = 20 V be the potential dierence supplied by the battery. Then q = Ceq V = (3:0 F)(20 V) = 6:0 10 5 C. (c) The potential dierence across C1 is given by V1 = CCVC 0 = 6(6:0 FF)(20 V) = 10 V ; + :0 + 6:0 F and the charge carried by C1 is q1 = C1 V1 = (3:0 F)(10 V) = 3:0 10 5 C. (d) The potential dierence across C2 is given by V2 = V V1 = 20 V 10 V = 10 V. Thus the charge carried by C2 is q2 = C2 V2 = (2:0 F)(10 V) = 2:0 10 5 C. (e) Since this voltage dierence V2 is divided equally between C3 and the other 4:0-F capacitors connected in series with it, the voltage dierenc across C3 is given by V3 = V2 =2 = 10 V=2 = 5:0 V. Thus q3 = C3 V3 = (4:0 F)(5:0 V) = 2:0 10 5 C: 29P 28P 27P (a) After the switches are closed the potential dierences across the capacitors are the same and the two capacitors are in parallel. The potential dierence from a to b is given by Vab = Q=Ceq , where Q is the net charge on the combination and Ceq is the equivalent capacitance. CHAPTER 26 CAPACITANCE 719 The equivalent capacitance is Ceq = C1 + C2 = 4:0 10 6 F. The total charge on the combination is the net charge on either pair of connected plates. The charge on capacitor 1 is q1 = C1 V = (1:0 10 6 F)(100 V) = 1:0 10 4 C and the charge on capacitor 2 is q2 = C2 V = (3:0 10 6 F)(100 V) = 3:0 10 4 C ; so the net charge on the combination is 3:0 10 4 C 1:0 10 4 C = 2:0 10 4 C. The 2:0 10 4 C = 50 V : Vab = 4:0 10 6 F (b) The charge on capacitor 1 is now q1 = C1 Vab = (1:0 10 6 F)(50 V) = 5:0 10 5 C. (c) The charge on capacitor 2 is now q2 = C2 Vab = (3:0 10 6 F)(50 V) = 1:5 10 4 C. 30P potential dierence is The charges on capacitors 2 and 3 are the same, so these capacitors may be replaced by an equivalent capacitance determined from 1 = 1 + 1 = C2 + C3 : Ceq C2 C3 C2 C3 Thus Ceq = C2 C3 =(C2 + C3 ). The charge on the equivalent capacitor is the same as the charge on either of the two capacitors in the combination and the potential dierence across the equivalent capacitor is given by q2 =Ceq . The potential dierence across capacitor 1 is q1 =C1 , where q1 is the charge on this capacitor. The potential dierence across the combination of capacitors 2 and 3 must be the same as the potential dierence across capacitor 1, so q1 =C1 = q2 =Ceq . Now some of the charge originally on capacitor 1 ows to the combination of 2 and 3. If q0 is the original charge, conservation of charge yields q1 + q2 = q0 = C1 V0 , where V0 is the original potential dierence across capacitor 1. Solve the two equations q1 =C1 = q2 =Ceq and q1 + q2 = C1 V0 for q1 and q2 . The second equation yields q2 = C1 V0 q1 and, when this is substituted into the rst, the result is q1 = C1 V0 q1 : C1 Ceq Solve for q1 . You should get 2V C2 C 2 (C2 3 ) q1 = C C1+ 0C = C C 1 V0 = C C 1+ C + C+ V0 C : eq 1 1 2 1 C3 C2 3 2 3 C2 + C3 + C1 720 CHAPTER 26 CAPACITANCE The charges on capacitors 2 and 3 are CC C 2( 2 3 q2 = q3 = C1 V0 q1 = C1 V0 C C 1+CC + C+)V0 C = C C C1C 2C 3 V0 C C : 1 2 1 C3 C2 3 1 2+ 1 3+ 2 3 (a) 31P C V F)(3 0 q1 = q3 = C 1 C3C = (1:010 F :+ F)(12 V) = 9:0 C ; + : 3:0 F C V F)(4 0 q2 = q4 = C 2 C4C = (2:020 F :+ F)(12 V) = 16 C : + : 4:0 F 2 4 1 3 (b) Now the voltage dierence V1 across C1 and C2 is C (3:0 + :0 V) V1 = C + C3 + C4 + C V = 1:0 F + 2:F F4+ 3:F)(12+ 4:0 F = 8:4 V : 0 0 F 1 2 + C3 4 Thus q1 = C1 V1 = (1:0 F)(8:4 V) = 8:4 C, q2 = C2 V1 = (2:0 F)(8:4 V) = 17 C, q3 = C3 (V V1 ) = (3:0 F)(12 V 8:4 V) = 11 C, and q4 = C4 (V V1 ) = (4:0 F)(12 V 8:4 V) = 14 C. In the rst case the two capacitors are eectively connected in series so the output potential dierence is Vout = CVin =2C = Vin =2 = 50:0 V: In the second case the lower diode acts as a wire so Vout = 0. Let v = 1:00 m3 . The energy stored is 1 U = uv = 2 0 E 2 v 1 = 2 (8:85 10 12 C2=N m2 )(150 V/m)2 (1:00 m3 ) = 9:96 10 8 J : 34E 33E 32P (a) (b) 1 1 U = 2 CV 2 = 2 (61:0 10 3 F)(10:0 103 V)2 = 3:05 106 J : U = (3:05 106 J)=(3:6 106 J= kW h) = 0:847 kW h : CHAPTER 26 CAPACITANCE 721 The energy stored by a capacitor is given by U = 1 CV 2 , where V is the potential dierence 2 across its plates. You must convert the given value of the energy to joules. Since a joule is a wattsecond, simply multiply by (103 W= kW)(3600 s= h) to obtain 10 kWh = 3:6 107 J. Thus 7 C = 2U = 2(3:6 10 2J) = 72 F : V2 (1000 V) (a) 36E 35E (b) No, because we don't know the volume of the space inside the capacitor where the electric eld is present. Use U = 1 CV 2 . As V is increased by V the energy stored in the capacitor increases 2 correspondingly from U to U + U : U + U = 1 C (V + V )2 . Thus (1 + V=V )2 = 2 1 + U=U , or r V = 1 + U 1 = p1 + 10% 1 = 4:9% : 37E 1 1 U = 2 CV 2 = 2 (130 10 12 F)(56:0 V)2 = 2:04 10 7 J : V U (a) 38E m C = 0dA = (8:85 10 1C =N10 3)(40 10 m ) = 3:5 10 :0 m (b) q = CV = (35 pF)(600 V) = 2:1 10 8 C = 21 nC. (c) U = 1 CV 2 = 1 (35 pF)(21 nC)2 = 6:3 10 6 J = 6:3 J. 2 2 (d) E = V=d = 600 V=1:0 10 3 m = 6:0 105 V/m: 12 2 2 4 2 11 F = 35 pF : (e) : 10 J u = U = (40 10 643m2 )(1:0 10 3 m) = 1:6 J/m3 : v 6 39E The total energy is the sum of the energies stored in the individual capacitors. Since they are connected in parallel the potential dierence V across the capacitors is the same and the total energy is 1 1 U = 2 (C1 + C2 )V 2 = 2 (2:0 10 6 F + 4:0 10 6 F)(300 V)2 = 0:27 J : 722 CHAPTER 26 CAPACITANCE 2 1 1 e 2 u = 2 0 E 2 = 2 0 4 r2 = 32e r4 : 2 0 0 (b) From the expression above u / r 4 . So for r ! 0 u ! 1. (a) 40E Use E = q=40 R2 = V=R. Thus 1 1 u = 2 0 E 2 = 2 0 V R 42P 41P 2 1 = 2 (8:85 10 12 C2=N m2 ) 8000 V 2 3 0:050 m = 0:11 J=m : The total energy stored in the capacitor bank is 1 1 U = 2 Ctotal V 2 = 2 (2000)(5:00 10 6 F)(50; 000 V)2 = 1:3 107 J : Thus the cost is (1:3 107 J)(3:0 cent= kW h) = 10 cents : 3:6 106 J= kW h (a) In the rst case U = q2 =2C ; and in the second case U = 2(q=2)2 =2C = q2 =4C: So the energy is now 4:0 J=2 = 2:0 J: (b) It becomes the thermal energy generated in the wire connecting the capacitors during the discharging process. (a) 44P 43P (b) 2 4 2 U1 = 2q = (4:8:010 6C) = 5:8 10 2 J ; C1 2(2 10 F) 2 4 2 U2 = 2q = (4:8:010 6C) = 1:4 10 2 J : C2 2(8 10 F) 02 4 2 0 = q1 = (1:9 10 C) = 9:2 10 3 J ; U1 6 00 00 (c) U1 = U2 = 0. 2(2:0 10 F) 02 4 2 2 U20 = 2qC = (7:7:010 6C) = 3:7 10 2 J : 10 F) 2 2(8 2C1 CHAPTER 26 CAPACITANCE 723 0 0 00 00 Note that U1 + U2 > U1 + U2 > U1 + U2 , since the system partially discharges as it goes from situation (a) to (b), and completely discharges in (c). (a) and (b). The voltage dierence across C1 and C2 is given by 00 F)(100 V1 = V2 = C +C3 V+ C = 10:0 (4:+ 500 F +V):00 F = 21:1 V : F : 4 1 C2 3 Also, V3 = V V1 = V V2 = 100 V 21:1 V = 78:9 V: Thus q1 = C1 V1 = (10:0 F)(21:1 V) = 2:11 10 4 C, q2 = C2 V2 = (5:00 F)(21:1 V) = 1:05 10 4 C, and q3 = q1 + q2 = 2:11 10 4 C + 1:05 10 4 C = 3:16 10 4 C: (c) U1 = 1 C1 V12 = 1 (10:0 F)(21:1 V)2 = 2:22 10 3 J, U2 = 1 C2 V22 = 1 (5:00 F) 2 2 2 2 (21:1 V)2 = 1:11 10 3 J, and U3 = 1 C3 V32 = 1 (4:00 F)(78:9 V)2 = 1:25 10 2 J. 2 2 46P 45P (a) Let q be the charge on the positive plate. Since the capacitance of a parallel-plate capacitor is given by 0 A=d, the charge is q = CV = 0 AV=d. After the plates are pulled apart their separation is 2d and the potential dierence is V 0 . Then q = 0 AV 0 =2d and V 0 = 2d q = 2d 0dA V = 2V : A A 0 0 (b) The initial energy stored in the capacitor is 1 Ui = 2 CV 2 = 0 AV 2d and the nal energy stored is 1 0A 1 0A Uf = 2 2d (V 0 )2 = 2 2d 4V 2 = 0 AV : d 2 2 This is twice the initial energy. (c) The work done to pull the plates apart is the dierence in the energy: W = Uf Ui = 0 AV 2 =2d. (a) 47P q3 = C3 V = (4:00 F)(100 V) = 4:00 10 4 mC ; C V q1 = q2 = C 1 C2C = (10:0 :F)(5:005:F)(100 V) = 3:33 10 4 C : 10 0 F + 00 F 1+ 2 724 CHAPTER 26 CAPACITANCE (b) V1 = q1 =C1 = 3:33 10 4 C=10:0 F = 33:3 V, V2 = V V1 = 100 V 33:3 V = 66:7 V, and V3 = V = 100 V: (c) Use Ui = 1 Ci Vi2 , where i = 1; 2; 3. The answers are U1 = 5:6 mJ, U1 = 11 mJ, and 2 U1 = 20 mJ. You rst need to nd an expression for the energy stored in a cylinder of radius R and length L, whose surface lies between the inner and outer cylinders of the capacitor (a < R < b). The energy density at any point is given by u = 1 0 E 2 , where E is the magnitude of the 2 electric eld at that point. If q is the charge on the surface of the inner cylinder then the magnitude of the electric eld at a point a distance r from the cylinder axis is given by E = 2q Lr 0 (see Eq. 26-12) and the energy density at that point is given by 2 1 u = 2 0 E 2 = 82 q L2 r2 : 0 48P The energy in the cylinder is the volume integral UR = udV : q2 2rdr = q2 Z R dr = q2 ln R : UR = 82 L2 r2 40 L2 a r 40 L2 a 0 a To nd an expression for the total energy stored in the capacitor, replace R with b: q2 ln b : Ub = 4 L2 a 0 You want the ratio UR =Ub to be 1=2, so 1 b ln R = 2 ln a a p p p p or, since 1 ln(b=a) = ln( b=a), ln(R=a) = ln( b=a). This means R=a = b=a or R = ab. 2 R 49P Z Now dV = 2rLdr, so Z The charge is held constant while the plates are being separated, so write the expression for the stored energy as U = q2 =2C , where q is the charge and C is the capacitance. The CHAPTER 26 CAPACITANCE 725 If the plate separation increases by dx the energy increases by dU = (q2 =20 A)dx. Suppose the agent pulling the plate apart exerts force F . Then the agent does work Fdx and if the plates begin and end at rest this must equal the increase in stored energy. Thus capacitance of a parallel-plate capacitor is given by C = 0 A=x, where A is the plate area and x is the plate separation, so 2 U = 2q x : A 0 q2 Fdx = 2 A dx 0 and The net force on a plate is zero so this must also be the magnitude of the force one plate exerts on the other. The force can also be computed as the product of the charge q on one plate and the electric eld E1 due to the charge on the other plate. Recall that the eld produced by a uniform plane surface of charge is E1 = q=20 A. Thus F = q2 =20 A. According to the result of 49P the force on either capacitor plate is given by F = q2 =20 A, where q is the charge on one plate and A is the area of a plate. The electric eld between the plates is E = q=0 A, so q = 0 AE and 50P q2 : F = 2 A 0 E F = 02A A = 1 0 AE 2 : 0 2 2 2 2 The force per unit area of plate is Note that the eld E that enters this equation is the total eld, due to charges on both plates. According to the result of 50P the electrostatic force acting on a small area A is Fe = 1 0 E 2 A. The electric eld at the surface is E = q=40 R2 , where q is the charge on the 2 bubble. Thus q 2 2 1 Fe = 2 0 4 R2 A = 32q2A 4 : R 0 0 51P F = 1 E2 : A 20 726 CHAPTER 26 CAPACITANCE This force is outward. The force of the gas inside is the product of the pressure inside and 3 the area: Fg = p(V0 =V ) A. Since V0 = (4=3)R0 and V = (4=3)R3 , R3 Fg = p R0 A : 3 Solve for q2 . You should get This force is outward. The force of the air outside is Fa = p A. This force is inward. Since the bubble surface is in equilibrium, the sum of the forces must vanish: Fe + Fg Fa = 0. This means 3 q2 + p R0 p = 0 : 322 R4 R3 0 q2 = 322 0 R4 p 3 R0 = 322 pR(R3 R3 ) : 1 R3 0 0 If the original capacitance is given by C = 0 A=d, then the new capacitance is C 0 = 0 A=2d. Thus C 0 =C = =2 or = 2C 0 =C = 2(2:6 pF=1:3 pF) = 4:0. The capacitance with the dielectric in place is given by C = C0 , where C0 is the capacitance before the dielectric is inserted. The energy stored is given by U = 1 CV 2 = 1 C0 V 2 , 2 2 so 6 = C2U 2 = (7:4 2(7:4 10 J) V)2 = 4:7 : V 10 12 F)(652 According to Table 26{1 you should use pyrex. 54E 53E 52E 0 Use C = 0 A=d / =d: To maximize C we need to choose the material with the greatest value of =d. It follows that the mica sheet should be chosen. (a) Use C = 0 A=d to solve for d: 12 C2= m2 2 d = 0 A = (8:85 1050 10N12 F)(0:35 m ) = 6:2 10 2 m : C 55E (b) Use C / . The new capacitance is C 0 = C (=air ) = (50 pf)(5:6=1:0) = 280 pF: CHAPTER 26 CAPACITANCE 727 The capacitance of a cylindrical capacitor is given by C = C0 = 20 L ; ln(b=a) where C0 is the capacitance without the dielectric, is the dielectric constant, L is the length, a is the inner radius, and b is the outer radius. The capacitance per unit length of the cable is C = 20 = 2(2:6)(8:85 10 12 F/m) = 8:1 10 11 F/m = 81 pF/m : L ln(b=a) ln [(0:60 mm)=(0:10 mm)] 57P 56E The capacitance is given by C = C0 = 0 A=d, where C0 is the capacitance without the dielectric, is the dielectric constant, A is the plate area, and d is the plate separation. The electric eld between the plates is given by E = V=d, where V is the potential dierence between the plates. Thus d = V=E and C = 0 AE=V . Solve for A: CV A = E : 0 For the area to be a minimum, the electric eld must be the greatest it can be without breakdown occurring. That is, 10 8 F)(4:0 103 V) A = 2:8(8:(7:010 12 F/m)(18 106 V/m) = 0:63 m2 : 85 58P (a) Use Eq. 26-14: : m) C = 20 ln(L ) = 2(8:99 109 N(4m7)(0:215ln(3:8 cm=3:6 cm) = 0:73 nF : 2 =C ) b=a (b) The breakdown potential is (14 kV/mm)(3:8 cm 3:6 cm) = 28 kV: 2 (a) Since u = 1 0 E 2 , we select the material with the greatest value of Emax , when 2 Emax is its dielectric strength. Thus we choose strontium titanate, with the corresponding minimum volume U Vmin = UU = 2E 2 = (310)(8:85 10 2(250=kJ)m2 )(8 kV/mm)2 = 2:85 m3 : 12 C2 N max 0 max 59P 728 CHAPTER 26 CAPACITANCE 2 0 (b) Solve 0 from U = 1 0 0 Emax Vmin : 2 0 = V 02UE 2 = (8:85 10 0 min max 60P 12 C2=N m2 )(0:0870 m3 )(8 kV/mm)2 2(250 kJ) = 1:01 104 : (a) Label the capacitor whose dielectric constant is + as capacitor 1, and the other one as capacitor 2. For parallel connection Ceq = C1 + C2 = 0 A(d+ ) + 0 A(d ) = 20dA : (b) Since C1 > C2 we need to evaluate q1 , the charge on capacitor 1: 1 Q 1 + : C1 Q = (0 A=d)( + )Q q1 = C + C ( A=d)( + ) + ( A=d)( ) = 2 1 2 0 0 (a) The length d is eectively shortened by b so C 0 = 0 A=(d b). (b) U = q2 =2C = C 0 = 0 A=(d b) = d : U 0 q2 =2C 0 C A=d d b 0 61P (c) The work done is W = U = U 0 q2 1 U = 2 C0 1 = q2 (d b d) = q2 b : C 20 A 20 A Since W < 0 the slab is sucked in. (a) C 0 = 0 A=(d b), the same as part (a) in 61P. (b) U = 1 CV 2 = C = 0 A=d = d b : 2 0 1 0 2 C 0 A=(d b) U d (c) The work done is 2C 62P V 0 W = U = U 0 1 U = 2 (C 0 C )V 2 = 02A d 1 b 1 V 2 = 0 AbV 2 : d 2d(d b) Since W > 0 the slab must be pushed in. CHAPTER 26 CAPACITANCE 729 The capacitor can be viewed as two capacitors C1 and C2 in parallel, each with surface area A=2 and plate separation d, lled with dielectric materials with dielectric constans 1 and 2 , respectively. Thus 0 (A=2)1 + 0 (A=2)2 = 0 A 1 + 2 : C = C1 + C2 = d d d 2 Assume there is charge q on one plate and charge q on the other. Calculate the electric eld at points between the plates and use the result to nd an expression for the potential dierence V between the plates, in terms of q. The capacitance is C = q=V . The electric eld in the upper half of the region between the plates is where A is the plate area. The electric eld in the lower half is Take d=2 to be the thickness of each dielectric. Since the eld is uniform in each region the potential dierence between the plates is E1 d + E2 d = qd 1 + 1 = qd 1 + 2 ; V= 2 2 20 A 1 2 20 A 1 2 so q 0 C = V = 2dA 1 2 : + 1 2 64P 63P q E1 = A ; 1 0 q E2 = A : 2 0 Notice that this expression is exactly the same as the expression for the equivalent capacitance of two capacitors in series, one with dielectric constant 1 and the other with dielectric constant 2 . Each has plate area A and plate separation d=2. Also notice that if 1 = 2 the expression reduces to C = 1 0 A=d, the correct result for a parallel plate capacitor with plate area A, plate separation d, and dielectric constant 1 . 65P Let C1 = 0 (A=2)1 =2d = 0 A1 =4d, C2 = 0 (A=2)2 =d = 0 A2 =2d, and C3 = 0 A3 =2d. Note that C2 and C3 are eectively connected in series, while C1 is eectively connected in parallel with the C2 -C3 combination. Thus A )( 2 2)( 3 C = C1 + CC2 CC = 04d 1 + (0 A=d=2+= =3 =2) 2 3 2 2 + 23 0A 2 3 = 4d 1 + + : 2 3 730 CHAPTER 26 CAPACITANCE 66E (a) The electric eld in the region between the plates is given by E = V=d, where V is the potential dierence between the plates and d is the plate separation. The capacitance is given by C = 0 A=d, where A is the plate area and is the dielectric constant, so d = 0 A=C and 12 F) V E = C = 5:4(8:85 (50 V)(100 10 10 4 m2 ) = 1:0 104 V/m : 10 12 F/m)(100 0A (b) The free charge on the plates is qf = CV = (100 10 12 F)(50 V) = 5:0 10 9 C. (c) The electric eld is produced by both the free and induced charge. Since the eld of a large uniform layer of charge is q=20 A, the eld between the plates is q q E = 2 fA + 2 fA 2qiA 2qiA ; 0 0 0 0 where the rst term is due to the positive free charge on one plate, the second is due to the negative free charge on the other plate, the third is due to the positive induced charge on one dielectric surface, and the fourth is due to the negative induced charge on the other dielectric surface. Note that the eld due to the induced charge is opposite the eld due to the free charge, so they tend to cancel. The induced charge is therefore qi = qf 0 AE = 5:0 10 9 C (8:85 10 = 4:1 10 9 C = 4:1 nC : 67E 12 F/m)(100 10 4 m2 )(1:0 104 V/m) (a) The electric eld E1 in the free space between the two plates is E1 = q=0 A while that inside the slab is E2 = E1 = = q=0 A. Thus V0 = E1 (d b) + E2 b = (q=0 A)(d b + b=), and the capacitance is (c) q C = V = (d0 A + b b) 0 12 C2=N m2 )(115 10 4 m2 )(2:61) (8 85 = (2::61)(1:10 0:780)(10 2 m) + (0:780 10 2 ) = 13:4 pF : 24 (b) q = CV = (13:4 10 12 F)(85:5 V) = 1:15 nC: : 10 9 C E1 = qA = (8:85 10 121C15 N m2 )(115 10 4 m2 ) = 1:13 104 N/C : 2= 0 E2 = E1 = 1:13 :10 N/C = 4:33 103 N/C : 2 61 4 (d) CHAPTER 26 CAPACITANCE 731 (a) Use C = C0 , where C0 is given by Eq. 26-17: 68P C = 40 b ab a : (b) q = CV = 40 V ab=(b a): (c) 0 = q 1 1 = 40 ( 1)V ab : q b a (a) Apply Gauss's law with dielectric: q=0 = EA, and solve for : 69P q = EA = (8:85 10 0 8:9 10 7 C 12 C2=N m2 )(1:4 10 6 V/m)(100 10 4 m2 ) = 7:2 : (b) The charge induced is q0 = q 1 1 = (8:9 10 7 C) 1 1 = 7:7 10 7 C : 7:2 (a) 70P (b) Use the result of Exercise 67, part (a): 12 C2=N 2 2 C0 = 0dA = (8:85 101:2 10 2m )(0:12 m ) = 89 pF : m 12 C2= 2 :12 2 C = (d0 A + b = (4:(8:85:2 10:40)(10 N2 m )(0(4:0m )(4:8)m) = 120 pF : b) 8)(1 0 m) + 10 3 (c) Before the insertion: q = C0 V (89 pF)(120 V) = 11 nC: Since the battery is disconnected, q will remain the same after the insertion of the slab. (d) E = q=0 A = 11 nC=(8:85 10 12 C2=N m2 )(0:12 m2 ) = 10 kV/m: (e) E 0 = E= = (10 kV/m)=4:8 = 2:1 kV/m: (f) V = E (d b)+E 0 b = (10 kV/m)(1:2 0:40)(10 2 m)+(2:1 kV/m)(0:4010 3 m) = 88 V. (g) The work done is q2 1 1 Wext = U = 2 C C 0 9 C)2 1 = (11 10 2 89 10 12 F 1 120 10 12 F = 1:7 10 7 J : 732 CHAPTER 26 CAPACITANCE (a) Since u = 1 0 E 2 , Eslab = Eair =slab , and U = uv, the fraction of energy stored in the 2 air gaps is 2 ( 1 frac = E 2 A(dEair A+d b)E 2 Ab = 1 + (E =E )2 [b=(d b)] b) slab slab slab slab air air 1 = 1 + (2:61)(1=2:61)2 [0:780=(1:24 0:780)] = 0:606 : 71P (b) The fraction of energy stored in the slab is 1 0:606 = 0:394. Assume the charge on one plate is +q and the charge on the other plate is q. Find an expression for the electric eld in each region, in terms of q, then use the result to nd an expression for the potential dierence V between the plates. The capacitance is C = q=V . The electric eld in the dielectric is Ed = q=0 A, where is the dielectric constant and A is the plate area. Outside the dielectric (but still between the capacitor plates) the eld is E = q=0 A. The eld is uniform in each region so the potential dierence across the plates is V = E b + E (d b) = qb + q(d b) = q b + (d b) : d 72P The capacitance is 0 A 0 A 0 A The result does not depend on where the dielectric is located between the plates; it might be touching one plate or it might have a vacuum gap on each side. For the capacitor of Sample Problem 27{10, = 2:61, A = 115 cm2 = 115 10 4 m2 , d = 1:24 cm = 1:24 10 2 m, and b = 0:78 cm = 0:78 10 2 m, so 12 F/m)(115 10 4 m2 2: C = 2:61(1:2461(8:852 10 0:78 10 2 m) + 0:78 )10 2 m 10 m = 1:34 10 11 F = 13:4 pF ; q 0A C = V = (d0bA + b = d ( 1) : ) b in agreement with the result found in the sample problem. If b = 0 and = 1, then the expression derived above yields C = 0 A=d, the correct expression for a parallel-plate capacitor with no dielectric. If b = d then the derived expression yields C = 0 A=d, the correct expression for a parallel-plate capacitor completely lled with a dielectric. ...
View Full Document

Ask a homework question - tutors are online