( N m 2 /C 2 )( C)( C) J

Electrical Energy and Capacitance Practice 8A, p. 669 Chapter 8. PE electric = 6.3 0 9 J q = q = q p + q n = ().60 0 9 C + ()(0) = 3.0 0 9 C kcqq (8.99 0 9 N /C )(3.0 0 9 C) r = = P Ee lectric 6.3 0 9 J r =.46 0 9. q = 6.4 µc = 6.4 0 6 C q = 3. µc = 3. 0 6 C PE electric = 4. 0 J kcqq r = = P Ee lectric r = 4.5 (8.99 0 9 N /C )(6.4 0 6 C)( 3. 0 6 C) 4. 0 J 3. N = 0 3 q e =.60 0 9 PE electric = 7. 0 J q = Nq e = (0 3 )(.60 0 9 C) =.60 0 6 C q = Nq e = ( 0 3 )(.60 0 9 C) =.60 0 6 C kcqq (8.99 0 9 N /C )(.60 0 6 C)(.60 0 6 C) r = = P Ee lectric 7. 0 J r = 0.3 4. d =.0 c =.0 0 E = 5 N/C PE electric = 6.9 0 9 J q = PE ( 6.9 0 9 J) = =.60 0 9 C Ed (5 N/C)(.0 0 ) Section Review, p. 669 5. 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 =.0 0 6 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 = 6.0 0 4 J PE = qed = (35 C)(.0 0 6 N/C)(.0 0 3 ) = 7.0 0 0 J Section One Pupil s Edition Ch. 8

Practice 8B, p. 673. r =.0 c q =.60 0 9 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 (8.99 0 9 N /C )(.60 0 9 C) = r.0 0 =.4 0 7 = k Cq (8.99 0 9 N /C )(5.0 0 9 C) = =60 r (0.350 /) = k Cq (8.99 0 9 N /C )( 3.0 0 9 C) = = 50 r (0.350 /) tot = + = 60 50 = 0 r = diag onal (.0 ) + (. 0 ) 4. 0 = = + 4.0 8. 0 r = =.4 = k Cq (8.99 0 9 N /C )(5.0 0 6 C) = r.4 = k Cq (8.99 0 9 N /C )(3.0 0 6 C) = r.4 3 = k Cq 3 (8.99 0 9 N /C )(3.0 0 6 C) = r.4 =3. 0 4 =.9 0 4 =.9 0 4 4 = k Cq 4 (8.99 0 9 N /C )( 5.0 0 6 C) = = 3. 0 4 r.4 tot = + + 3 + 4 tot = (3. 0 4 ) + (.9 0 4 ) + (.9 0 4 ) + ( 3. 0 4 ) tot = 3.8 0 4 Section Review, p. 675. d = 0.060 c E = 3.0 0 6 / 5. E = 8.0 0 4 / d = 0.50 q =.60 0 9 C 6. E =.0 0 6 / d =.60 k = E d = (3.0 0 6 /)(0.060 0 ) =.8 0 3 =.8 0 3 a. = E d = (8.0 0 4 /)(0.50 ) = 4.0 0 4 b. PE = qed = (.60 0 9 C)(8.0 0 4 /)(0.50 ) PE = 6.4 0 5 J = E d = (.0 0 6 /)(.60 0 3 ) =.6 0 9 =.6 0 9 Ch. 8 Holt Physics Solution Manual

Practice 8C, p. 68. C = 4.00 F =.0 =.50 a. Q = C = (4.00 0 6 F)(.0 ) = 4.80 0 5 C b. PE = C( ) = (0.5)(4.00 0 6 F)(.50 ) = 4.50 0 6 J. Q = 6.0 C =.5 Q a. C = = 6.0 0 6 C =.5 4.8 0 6 F =.50 b. PE = C( ) = (0.5)(4.8 0 6 F)(.50 ) = 5.4 0 6 J 3. C =.00 pf Q = 8.0 pc a. = Q C = 8.0 0 C.00 0 = F 9.00 =.5 b. Q = C = (.00 0 F)(.5 ) = 5.0 0 C 4. C =.00 F d =.00 A = C d (.00 F)(.00 0 3 ) = =.3 0 8 e 0 8.85 0 C /N Section Review, p. 68. A =.0 c d =.0 a. C = e 0A (8.85 0 C /N )(.0 0 4 ) = = d.0 0 3 8.8 0 3 F = 6.0 b. Q = C = (8.8 0 3 F)(6.0 ) = 5.3 0 C 3. C =.35 pf =.0 4. d = 800.0 A =.00 0 6 E =.0 0 6 N/C 4. q = 9.00 0 9 C at the origin q = 3.00 0 9 C PE = 8.09 0 J PE = C( ) = (0.5)(.35 0 F)(.0 ) PE = 9.7 0 J a. C = e 0A (8.85 0 C /N )(.00 0 6 ) = =. 0 8 F d 800.0 b. = E d Q = C = C( E d) = (. 0 8 F)(.0 0 6 N/C)(800.0 ) = 8 C Q =±8 C Chapter Review and Assess, pp. 683 687 PE = PE f PE i PE i = 0 J because r i = r f = k Cqq (8.99 0 9 N /C )(9.00 0 9 C)(3.00 0 9 C) = =0.300 = 30.0 c PEf 8.09 0 7 J Section One Pupil s Edition Ch. 8 3

5. r i = 55 c PE =. 0 8 J q =.60 0 9 C q =.60 0 9 C PE i = k Cqq (8.99 0 9 N /C )(.60 0 9 C)(.60 0 9 C) = r i (55 0 ) PE i = 4. 0 8 J PE f = PE + PE i = (. 0 8 J) + ( 4. 0 8 J) =. 0 8 J r f = k Cqq (8.99 0 9 N /C )(.60 0 9 C)(.60 0 9 C) = PEf. 0 8 J r f =.. E =.7 0 6 N/C d =.5 c = E d = (.7 0 6 N/C)(.5 0 ) =.6 0 4 =.6 0 4 3. F = 4.30 0 N q = 56.0 C d = 0.0 c = E d = F d (4.30 0 N)(0.00 ) = = 54 q 56.0 0 6 C 4. q =+8.0 C q = 8.0 C q 3 = C r,p = 0.35 r,p = 0.0 = k Cq (8.99 0 9 N /C )(8.0 0 6 C) = r, P 0.35 = k Cq (8.99 0 9 N /C )( 8.0 0 6 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 (8.99 0 9 N /C )( 0 6 C) = r3, P (0.3 5 ) + ( 0. 0 ) =. 0 5 = 3.6 0 5 6. =.0 C = 6.0 pf 7. C = 0.0 F = 6500 (8.99 0 9 N /C )( 0 6 C) 3 = 0. + 0.0 40 (8.99 0 9 N /C )( 0 6 C) 3 = =.7 0 5 0. 6 tot = + + 3 = (. 0 5 ) + ( 3.6 0 5 ) + (.7 0 5 ) tot = 4. 0 5 Q = C = (6.0 0 F)(.0 ) = 7. 0 C Q = ±7. 0 C a. Q = C = (0.0 0 6 F)(6500 ) =.3 0 3 C b. PE = C( ) = (0.5)(0.0 0 6 F)(6500 ) = 4. J Ch. 8 4 Holt Physics Solution Manual

8. C = 5 F C = 5.0 F = 0 9. = 600.0 E = 00.0 N/C PE = C ( ) = (0.5)(5 0 6 F)(0 ) = 0.8 J PE = C ( ) = (0.5)(5.0 0 6 F)(0 ) = 3.6 0 J PE tot = PE + PE = 0.8 J + (3.6 0 J) = 0. J k C rq = =r E kcq r r = 600.0 = 0 = 3.000 E 0. 0 N/C q = r (600.0 )(3.000 ) = kc 8.99 0 9 N /C =.00 0 7 C 30. d = 3.0 E = 3.0 0 6 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 = (.0 0 6 C) r = 0. 0 E Q (8.85 0 C /N )(3.0 0 6 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 (8.99 0 9 N /C )(8.0 0 6 C)(.0 0 6 C) = =4.8 J r, 3.0 0 PE,3 = k Cqq 3 (8.99 0 9 N /C )(8.0 0 6 C)(4.0 0 6 C) = r, 3 (0.0 30 ) + ( 0. 06 0 ) (8.99 0 9 N /C )(8.0 0 6 C)(4.0 0 6 C) PE,3 = (9.0 0 4 ) + ( 3. 6 0 3 ) (8.99 0 9 N /C )(8.0 0 6 C)(4.0 0 6 C) PE,3 = =4.3 J 4. 5 0 3 PE,tot = PE, + PE,3 = 4.8 J + 4.3 J = 9. J E = = d 0.30 0 = 4.0 0 3 / Section One Pupil s Edition Ch. 8 5

33. A = 5.00 c d =.00 Q = 400.0 pc a. = Q C C = e 0A d Q Qd (400.0 0 C)(.00 0 3 ) = = = = ε e 0A (8.85 0 C /N )(5.00 0 4 ) 0 da 90.4 b. E = 90.4 = d.00 0 3 = 9.04 0 4 / 34. A = 75 c d = 0.0400 Q = 500.0 pc a. C = e 0A (8.85 0 C /N )(75 0 4 ) = = d (0.0400 0 3 ) b. = Q C = 50 0.0 0 C 3.87 0 9 = 0.9 F 3.87 0 9 F 35. KE =.00 0 9 J e = 9.09 0 3 kg p =.673 0 7 kg a. v e = K b. v p = K e E = ( 9 p E = ( ) (. 0. 0 9 ) (. 0.6 7 3 0 0 0 9 J 3 k ) = 4.69 0 5 /s g 0 0 0 9 J 7 k ) g =.09 0 4 /s 36. = 5 700 v i = 0 /s q =.60 0 9 C p =.673 0 7 kg a. KE f = PE = q = (.60 0 9 C)(5 700 ) = 4. 0 5 J b. v f = K E p f = ( ) ( 4..6 7 3 0 0 5 J 7 k ) g =. 0 6 /s 37. = 0 v i = 0 /s p =.673 0 7 kg q =.60 0 9 C 38. d = 5.33 = 600.0 q =.60 0 9 C d = (5.33.90 ) =.43 KE f = PE = q p v f = q v f = q = p ()(.60 0 9 C)(0 ) =.5 0 5 /s.673 0 7 kg a. E = 600. 0 = d 5.33 0 3 =.3 0 5 / b. F = qe = (.60 0 9 C)(.3 0 5 /) =.8 0 4 N F =.8 0 4 N c. PE = qed = (.60 0 9 C)(.3 0 5 /)(.43 0 3 ) PE = 4.39 0 7 J Ch. 8 6 Holt Physics Solution Manual

39. q = 5.0 0 9 C q = 5.0 0 9 C q 3 = 5.0 0 9 C r, = r,3 = 4.0 c r,3 =.0 c r + (0.00 ) = (0.040 ) r = (0.0 40 ) ( 0. 0 0 ) = k C r q = (8.99 0 9 N /C )(5.0 0 9 C) (0.0 40 ) ( 0. 0 0 ) (8.99 0 9 N /C )(5.0 0 9 C) = (.6 0 3 ) (. 0 0 4 ) (8.99 0 9 N /C )(5.0 0 9 C) = =00. 5 0 3 r = r 3 = 0.0 0 = 0.00 = k Cq (8.99 0 9 N /C )( 5.0 0 9 C) = = 4500 r 0.00 3 = k Cq 3 (8.99 0 9 N /C )( 5.0 0 9 C) = = 4500 r 3 0.00 tot = + + 3 = (00 ) + ( 4500 ) + ( 4500 ) = 7800 40. q = 3.00 0 9 C at the origin q = 8.00 0 9 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 )( 3.00 0 9 C) = =0.545 q q ( 3.00 0 9 C) (8.00 0 9 C) P is 0.545 to the right of the origin, at x = 0.545. For the location to the left of the y-axis, = 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 = (. 00 ) (q ) (.00 )( 3.00 0 9 C) = =.0 q + q ( 3.00 0 9 C) + (8.00 0 9 C) P is.0 to the left of the origin, at x =.0. Section One Pupil s Edition Ch. 8 7

4. d = 5.0 c = 550 v i,e = 0 /s v i,p = 0 /s p =.673 0 7 kg e = 9.09 0 3 kg q =.60 0 9 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 = (.673 0 7 kg) + (9.09 0 3 kg) =.674 0 7 kg t e = t p = ()(5.0 0 ) (9.09 0 3 kg)(.673 0 7 kg) (.674 0 7 kg)(.60 0 9 C)(550 ) t e = t p = 7. 0 9 s b. v e = a e t e = q (.60 0 ed 9 C)(550 )(7. 0 9 s) t e = = (9.09 0 3 kg)(5.0 0 ) q (.60 0 v p = a p t p = pd 9 C)(550 )(7. 0 9 s) t p = = (.673 0 7 kg)(5.0 0 ) x c. t p,tot = tot p d q x tot = d t p,tot = pd q t p,tot = 3. 0 7 s = ()(.673 0 7 kg)(5.0 0 ) (.60 0 9 C)(550 ).4 0 7 /s 7.6 0 3 /s 4. = 60.0 PE =.9 0 7 J q = PE =.9 0 7 J = 3.0 0 9 C 60.0 Ch. 8 8 Holt Physics Solution Manual

43. = 00.0 Q = 400.0 C Q C = = 400.0 0 6 C = 4.000 0 6 F 00.0 44. = 4.5 0 6 v i = 0 /s q =.60 0 9 C p =.673 0 7 kg a. KE f = PE = q = (4.5 0 6 )(.60 0 9 C) = b. v f = K E p f = ( ). 6 ( 7. 7 3 3 0 J 0 7 k ) g =.9 0 7 /s 7. 0 3 J 45. v f,positron = 9.0 0 7 /s positron = 9.09 0 3 kg q =.60 0 9 C proton =.673 0 7 kg KE = PE v = q = positron( vf,positron) = q =.3 0 4 v f,proton = p q roton v f,proton =. 0 6 /s (9.09 0 3 kg)(9.0 0 7 /s) ()(.60 0 9 C) = ()(.3 0 4 )(.60 0 9 C).673 0 7 kg 46. v f = (0.600)(3.00 0 8 /s) e = p = 9.09 0 3 kg q e =.60 0 9 C q p =.60 0 9 C a. PE = KE f q = v f e = evf = qe (9.09 0 3 kg)[(0.600)(3.00 0 8 /s)] ()(.60 0 9 C) e = 9. 0 4 47. = 00 q =.60 0 9 C e = 9.09 0 3 kg p =.673 0 7 kg 48. C = 3750 pf Q =.75 0 8 C d = 6.50 0 4 b. p = pvf (9.09 0 3 kg)[(0.600)(3.00 0 8 /s)] = qp ()(.60 0 9 C) p = 9. 0 4 a. KE f = PE v f = q v f,e = q e b. v f,p = p q ()(00 )(.60 0 9 C) = = 9.09 0 3 kg.8 0 7 /s ()(00 )(.60 0 = 9 C) = 6.5 0 5 /s.673 0 7 kg a. = Q 8 C =.75 0 C 3750 0 = 4.67 F b. E = 4.67 = 6.50 d 0 4 = 780 / Section One Pupil s Edition Ch. 8 9

49. r =.50 0 3 d =.40 0 4 = 0. a. C = e 0A = e 0pr (8.85 0 C /N )(p)(.50 0 3 /) = d d.40 0 4 C = 3.0 0 3 F b. Q = C = (3.0 0 3 F)(0. ) = 3.7 0 4 C c. PE electric = Q = (0.5)(3.7 0 4 C)(0. ) =. 0 5 J d =.40 0 4 d. = E d E = 0. = d.40 0 4 = 860 / d =.0 0 4.40 0 4 = 4.00 0 5 = E d = (860 /)(4.00 0 5 ) = 3.4 0 Q = (0.707)Q e. 3 = Q C Because the capacitance has not changed, 3 = (0.707)( ). 3 = (0.707)( ) = (0.707)(0. ) = 8.5 0 Ch. 8 0 Holt Physics Solution Manual