( N m 2 /C 2 )( C)( C) J
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- Ἰσμήνη Ζάρκος
- 6 χρόνια πριν
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1 Electrical Energy and Capacitance Practice 8A, p. 669 Chapter 8. PE electric = J q = q = q p + q n = () C + ()(0) = C kcqq ( N /C )( C) r = = P Ee lectric J r = q = 6.4 µc = C q = 3. µc = C PE electric = 4. 0 J kcqq r = = P Ee lectric r = 4.5 ( N /C )( C)( C) 4. 0 J 3. N = 0 3 q e = PE electric = 7. 0 J q = Nq e = (0 3 )( C) = C q = Nq e = ( 0 3 )( C) = C kcqq ( N /C )( C)( C) r = = P Ee lectric 7. 0 J r = d =.0 c =.0 0 E = 5 N/C PE electric = J q = PE ( J) = = C Ed (5 N/C)(.0 0 ) Section Review, p E = 50 N/C, in the positive x direction q = C q oves fro the origin to (0.0 c, 50.0 c). 6. q = 35 C d =.0 k E = N/C The displaceent in the direction of the field (d) is 0.0 c. PE = qed = ( 0 6 C)(50 N/C)(0.0 0 ) PE = J PE = qed = (35 C)( N/C)( ) = J Section One Pupil s Edition Ch. 8
2 Practice 8B, p r =.0 c q = C. q = 5.0 nc q = 3.0 nc r = 35.0 c 3. q = 5.0 C q = 3.0 C q 3 = 3.0 C q 4 = 5.0 C Each charge is at the corner of a.0.0 square. = k Cq ( N /C )( C) = r.0 0 = = k Cq ( N /C )( C) = =60 r (0.350 /) = k Cq ( N /C )( C) = = 50 r (0.350 /) tot = + = = 0 r = diag onal (.0 ) + (. 0 ) 4. 0 = = r = =.4 = k Cq ( N /C )( C) = r.4 = k Cq ( N /C )( C) = r.4 3 = k Cq 3 ( N /C )( C) = r.4 = = = = k Cq 4 ( N /C )( C) = = r.4 tot = tot = ( ) + ( ) + ( ) + ( ) tot = Section Review, p d = c E = / 5. E = / d = 0.50 q = C 6. E = / d =.60 k = E d = ( /)( ) = = a. = E d = ( /)(0.50 ) = b. PE = qed = ( C)( /)(0.50 ) PE = J = E d = ( /)( ) = = Ch. 8 Holt Physics Solution Manual
3 Practice 8C, p. 68. C = 4.00 F =.0 =.50 a. Q = C = ( F)(.0 ) = C b. PE = C( ) = (0.5)( F)(.50 ) = J. Q = 6.0 C =.5 Q a. C = = C = F =.50 b. PE = C( ) = (0.5)( F)(.50 ) = J 3. C =.00 pf Q = 8.0 pc a. = Q C = C.00 0 = F 9.00 =.5 b. Q = C = (.00 0 F)(.5 ) = C 4. C =.00 F d =.00 A = C d (.00 F)( ) = = e C /N Section Review, p. 68. A =.0 c d =.0 a. C = e 0A ( C /N )( ) = = d F = 6.0 b. Q = C = ( F)(6.0 ) = C 3. C =.35 pf =.0 4. d = A = E = N/C 4. q = C at the origin q = C PE = J PE = C( ) = (0.5)(.35 0 F)(.0 ) PE = J a. C = e 0A ( C /N )( ) = =. 0 8 F d b. = E d Q = C = C( E d) = (. 0 8 F)( N/C)(800.0 ) = 8 C Q =±8 C Chapter Review and Assess, pp PE = PE f PE i PE i = 0 J because r i = r f = k Cqq ( N /C )( C)( C) = =0.300 = 30.0 c PEf J Section One Pupil s Edition Ch. 8 3
4 5. r i = 55 c PE =. 0 8 J q = C q = C PE i = k Cqq ( N /C )( C)( C) = r i (55 0 ) PE i = J PE f = PE + PE i = (. 0 8 J) + ( J) =. 0 8 J r f = k Cqq ( N /C )( C)( C) = PEf. 0 8 J r f =.. E = N/C d =.5 c = E d = ( N/C)(.5 0 ) = = F = N q = 56.0 C d = 0.0 c = E d = F d ( N)(0.00 ) = = 54 q C 4. q =+8.0 C q = 8.0 C q 3 = C r,p = 0.35 r,p = 0.0 = k Cq ( N /C )( C) = r, P 0.35 = k Cq ( N /C )( C) = r, P 0.0 (r,p ) + (r,p ) = (r 3,P ) r 3,P = (r, P ) + ( r,p ) = (0.3 5 ) + ( 0. 0 ) 3 = k Cq3 ( N /C )( 0 6 C) = r3, P (0.3 5 ) + ( 0. 0 ) =. 0 5 = =.0 C = 6.0 pf 7. C = 0.0 F = 6500 ( N /C )( 0 6 C) 3 = ( N /C )( 0 6 C) 3 = = tot = = (. 0 5 ) + ( ) + ( ) tot = Q = C = (6.0 0 F)(.0 ) = 7. 0 C Q = ±7. 0 C a. Q = C = ( F)(6500 ) = C b. PE = C( ) = (0.5)( F)(6500 ) = 4. J Ch. 8 4 Holt Physics Solution Manual
5 8. C = 5 F C = 5.0 F = 0 9. = E = 00.0 N/C PE = C ( ) = (0.5)(5 0 6 F)(0 ) = 0.8 J PE = C ( ) = (0.5)( F)(0 ) = J PE tot = PE + PE = 0.8 J + (3.6 0 J) = 0. J k C rq = =r E kcq r r = = 0 = E 0. 0 N/C q = r (600.0 )(3.000 ) = kc N /C = C 30. d = 3.0 E = N/C Q =.0 C C = e 0A Q = d = Q E d e 0 AE d = Qd Q A = e 0E A = pr r = A p = e p = ( C) r = 0. 0 E Q ( C /N )( N/C)(p) 3. q = 8.0 C q =.0 C q 3 = 4.0 C r, = 3.0 c r,3 = (3.0 c ) + ( 6. 0 c ) 3. = d = 0.30 c PE, = k Cqq ( N /C )( C)( C) = =4.8 J r, PE,3 = k Cqq 3 ( N /C )( C)( C) = r, 3 ( ) + ( ) ( N /C )( C)( C) PE,3 = ( ) + ( ) ( N /C )( C)( C) PE,3 = =4.3 J PE,tot = PE, + PE,3 = 4.8 J J = 9. J E = = d = / Section One Pupil s Edition Ch. 8 5
6 33. A = 5.00 c d =.00 Q = pc a. = Q C C = e 0A d Q Qd ( C)( ) = = = = ε e 0A ( C /N )( ) 0 da 90.4 b. E = 90.4 = d = / 34. A = 75 c d = Q = pc a. C = e 0A ( C /N )( ) = = d ( ) b. = Q C = C = 0.9 F F 35. KE = J e = kg p = kg a. v e = K b. v p = K e E = ( 9 p E = ( ) ( ) ( J 3 k ) = /s g J 7 k ) g = /s 36. = v i = 0 /s q = C p = kg a. KE f = PE = q = ( C)(5 700 ) = J b. v f = K E p f = ( ) ( J 7 k ) g =. 0 6 /s 37. = 0 v i = 0 /s p = kg q = C 38. d = 5.33 = q = C d = ( ) =.43 KE f = PE = q p v f = q v f = q = p ()( C)(0 ) = /s kg a. E = = d = / b. F = qe = ( C)( /) = N F = N c. PE = qed = ( C)( /)( ) PE = J Ch. 8 6 Holt Physics Solution Manual
7 39. q = C q = C q 3 = C r, = r,3 = 4.0 c r,3 =.0 c r + (0.00 ) = (0.040 ) r = ( ) ( ) = k C r q = ( N /C )( C) ( ) ( ) ( N /C )( C) = ( ) ( ) ( N /C )( C) = = r = r 3 = = 0.00 = k Cq ( N /C )( C) = = 4500 r = k Cq 3 ( N /C )( C) = = 4500 r tot = = (00 ) + ( 4500 ) + ( 4500 ) = q = C at the origin q = C at x =.00, y = 0.00 For the location between the two charges, tot = + = 0 = = k Cq P k = C q (.00 P) k C k = Cq Pq (.0 0 P) Pq = (.00 P)(q ) Pq = (.00 )(q ) Pq P(q q ) = (.00 )(q ) P = (.0 0 )( q ) (.00 )( C) = =0.545 q q ( C) ( C) P is to the right of the origin, at x = For the location to the left of the y-axis, = k Cq P k = C q (.00 + P) k C k = Cq Pq ( P) Pq = (.00 + P)(q ) Pq = (.00 )(q ) + Pq P(q + q ) = (.00 )(q ) P = (. 00 ) (q ) (.00 )( C) = =.0 q + q ( C) + ( C) P is.0 to the left of the origin, at x =.0. Section One Pupil s Edition Ch. 8 7
8 4. d = 5.0 c = 550 v i,e = 0 /s v i,p = 0 /s p = kg e = kg q = C F qe q a. a = = = d x = a t t e = x a e = q x e e d t p = q x p t e = t p q x e p d e d = x e e = x p p x e + x p = d x e = d x p (d x p ) e = x p p q x p d e x p e = x p p d e = x p p + x p e p d de x p = p + e d t e = t p = q x p p d e p + e p d = = ( d p e q e + ) q p p + e = ( kg) + ( kg) = kg t e = t p = ()(5.0 0 ) ( kg)( kg) ( kg)( C)(550 ) t e = t p = s b. v e = a e t e = q (.60 0 ed 9 C)(550 )( s) t e = = ( kg)(5.0 0 ) q (.60 0 v p = a p t p = pd 9 C)(550 )( s) t p = = ( kg)(5.0 0 ) x c. t p,tot = tot p d q x tot = d t p,tot = pd q t p,tot = s = ()( kg)(5.0 0 ) ( C)(550 ) /s /s 4. = 60.0 PE = J q = PE = J = C 60.0 Ch. 8 8 Holt Physics Solution Manual
9 43. = 00.0 Q = C Q C = = C = F = v i = 0 /s q = C p = kg a. KE f = PE = q = ( )( C) = b. v f = K E p f = ( ). 6 ( J 0 7 k ) g = /s J 45. v f,positron = /s positron = kg q = C proton = kg KE = PE v = q = positron( vf,positron) = q = v f,proton = p q roton v f,proton =. 0 6 /s ( kg)( /s) ()( C) = ()( )( C) kg 46. v f = (0.600)( /s) e = p = kg q e = C q p = C a. PE = KE f q = v f e = evf = qe ( kg)[(0.600)( /s)] ()( C) e = = 00 q = C e = kg p = kg 48. C = 3750 pf Q = C d = b. p = pvf ( kg)[(0.600)( /s)] = qp ()( C) p = a. KE f = PE v f = q v f,e = q e b. v f,p = p q ()(00 )( C) = = kg /s ()(00 )(.60 0 = 9 C) = /s kg a. = Q 8 C =.75 0 C = 4.67 F b. E = 4.67 = 6.50 d 0 4 = 780 / Section One Pupil s Edition Ch. 8 9
10 49. r = d = = 0. a. C = e 0A = e 0pr ( C /N )(p)( /) = d d C = F b. Q = C = ( F)(0. ) = C c. PE electric = Q = (0.5)( C)(0. ) =. 0 5 J d = d. = E d E = 0. = d = 860 / d = = = E d = (860 /)( ) = Q = (0.707)Q e. 3 = Q C Because the capacitance has not changed, 3 = (0.707)( ). 3 = (0.707)( ) = (0.707)(0. ) = Ch. 8 0 Holt Physics Solution Manual
10.0 C N = = = electrons C/electron C/electron. ( N m 2 /C 2 )( C) 2 (0.050 m) 2.
Electric Forces and Fields Section Review, p. 633 Givens Chapter 17 3. q 10.0 C q 10.0 C N 6.5 10 19 electrons 1.60 10 19 C/electron 1.60 10 19 C/electron Practice 17A, p. 636 1. q 1 8.0 C q 8.0 C r 5.0
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