Sol-Ch24 - CHAPTER 24 E xercises 1 2 3 = E Acos60 = 1 0.2...

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Exercises 1. Φ = EAcos60 ° = 10.2 N.m 2 /C 2. Φ = EAcos53 ° = 0.867 N.m 2 /C 3. Projected area is πR 2 , so Φ = πR 2 E 4. Φ = E A = (70i + 90k N/C) (1441 × 10 –4 k m 2 ) = 1.30 N.m 2 /C 5. Φ = (q 1 + q 2 )/ ϵ 0 = –2.26 × 10 5 N.m 2 /C (inward) 6. Q = ϵ 0 Φ = ϵ 0 (6)(3 ×  10 4 ) = 1.59 µC 7. (a)Q/ ϵ 0 = 6.78 × 10 6 N.m 2 /C; (b) 1.13 × 10 6 N.m 2 /C; (c) No for (a), yes for (b) 8. zero for “near” faces; (1/8)(1/3)q/ ϵ 0 = q/24 ϵ 0 for “far” faces 9. (a) Q = 4πR 2 σ = 8.04 ×  10 – 12 C, E = kQ/R 2 = 11.3 N/C; (b) E = kQ/r 2 = 7.23 N/C 10. (a) Inside : E = 1.44 × 10 5 /r 2 ; Outside : E = 0.72 × 10 5 /r 2 (b) Inner : –16 µC; Outer : +8 µC 11. E = kQ/R 2 = k(4πR 2 σ)/R 2 = 4πkσ = σ/ ϵ 0 12. (a)E = kQ/R 2 , so Q = ER 2 /k = –3.56 × 10 –11 C (b) No. A spherically symmetric distribution 13. (a) Zero : (b) σ/ ϵ 0 14. (a) 2σ/ ϵ 0 ; (b) σ/ ϵ 0 15. (a) σ/ ϵ 0 ; (b) Zero CHAPTER 24
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16. (a) (a + bL)L 2 – aL 2 = bL 3 ; (b) bL 3 = Q/ ϵ 0 so Q = bL
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Sol-Ch24 - CHAPTER 24 E xercises 1 2 3 = E Acos60 = 1 0.2...

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