# Sol-Ch25 - CHAPTER 25 E xercises 1(a qV = 1.87 1 0 2 8 e...

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Exercises 1. (a) qΔV = 1.87 × 10 28 eV; (b) 5 × 10 7 s = 1.58 y 2. (a) (80)(3600) = 2.88 × 10 5 C; (b) qV = 3.46 × 10 6 J 3. W ext = q(V f – V i ), thus 4 × 10 –7 – (–5 nC) (–20 – Vi). Find V i = 60 V 4. (a) V B – V A = –E Δs = –(–180k)(10 –1 k) = +18 V (b) ΔV = Ed, so d = ΔV/E = 27/180 = 0.15 m 5. V B – V A = – E x dx – E y dy = +6 V 6. V(x) – V(0) = – E x dx. (a) –a ln(x/x o ); (b) (e –Bx –1 )A/B 7. 1/2 mv 2 = eV, V = 2.847 × 10 –12 v 2 (a) 3.10 × 10 –7 V; (b) 3.57 × 10 –4 V; (c) 2.56 × 10 3 V 8. 1/2 mv 2 = eV, V = 5.22 × 10 –9 v 2 (a) 5.68 × 10 –4 V; (b) 0.655 V; (c) 4.70 MV 9. v = (2eV/m) 1/2 ; (a)2.05 × 10 6 m/s; (b) 4.80 × 10 4 m/s 10. V = Ed = 3 kV 11. W ext = qΔ = (–2 µC)(–10V) = 2 × 10 –5 J 12. (a) V B – V A = –E ds = –(600 V/m)(–0.04 m) = +24 V (b) U B – U A = q(V B – V A ) = (–3 µC)(24 V) = –7.2 × 10 –5 J 13. V = Ed = (F/q)d = 150 V 14. V = mv 2 /2q : (a) 9.33 × 10 6 V; (b)1.19 × 10 7 V 15. V = Ed: (a) 216 V; (b) 52 kV CHAPTER 25

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16. (a) E = ΔV/d = 4 kV/m; (b) W = ΔK = –qΔV = +1.92 × 10 –17 J; (c) ΔV = +120 V; (d) ΔU = –1.92 × 10 –17 J 17. W = ΔK + ΔV = 1/2 mv 2 + q(–6000) = 1.6 +0.09 = 1.69 J 18. (a) U = ke 2 /r = 2.3 × 10 –13 J; (b) ΔK = –ΔU, mv 2 = –ke 2 (1/4 –1)10 –15 , so v = 1.02 ×  10 7 m/s 19. ΔK = –ΔU = k(48e)(44e)/r = 6.95 × 10 –11 J 20. (a) V = k(10 –4 )(4/4(2) 1/2 + 2/4 – 3/4) = 4.11 × 10 5 V (b) U = qV = –0.823 J (c) Six terms lead to U = k10 –10 [–1/2 –7/2(2) 1/2 ] = –2.68 J 21. V = (9 × 10 3 ) (0.6 + 2.2 – 3.6 + 4.8)/(0.0707) = 5.09 × 10 5 V W = ΔU = qΔV = (–5 µ)(5.09 × 10 5 V) = – 2.55 J 22. (a) ΔV = –2kQ/3 = –3 ×  10 4 V; (b) ΔK = –qΔV =, so 1/2 mv 2 = 2kqQ/3, thus v = 20 m/s. 23. (a) ΔV = 2kQ(1/3 – 1/5) = 12 kV;
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Sol-Ch25 - CHAPTER 25 E xercises 1(a qV = 1.87 1 0 2 8 e...

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