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# Sol-Ch22 - CHAPTER 22 E xercises 1(a F = 90(6/16 10/4 i =...

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Exercises 1. (a) F =90(6/16 – 10/4)i = –191i N (b) F = (9 ×   10 9 )[10/4 – 15/36](10–8)i =188i N; 2. (a) 67.5i – 80 j     µN; (b) F x = –F 32 cos θ – F 34 ; F y = F 32 sin θ .F = –84.8i +13 j   µN 3. (a) F x = 0, F y = F 32 sin60o, so F = 208 j   N; (b) F x = (F 23 – F 21 )cos60o, F y = –(F 21 + F 23 ) sin60o F = 80i – 277 j   N 4. (a) F x = kQ 2 [2/16 + 0.8(4/25)] × 10 4 =0.364 mN F y = kQ 2 [6/9 – 0.6(4.25)]   ×   10 4 = 0.822 mN (b) F x = kQ 2 [6/16 + 0.8(3/25)]   ×   10 4 = 0.678 mN F y = kQ 2 [–6/9 + 0.6(3/25)] × 10 4 = –0.857 mN 5. (a) 27/x 2 = 3/(1 – x) 2 , leads to x = 0.75 m; (b) 27/(1 + d) 2 = 3/d 2 , leads to d = 0.5 m or x = 1.5 m 6. kQ 2 = GmM, so Q = (GmM/k) 1/2 = 5.72 × 10 –4 m 7 r 2 = kQ 2 /F leads to r = 1.52 × 10 –14 m 8. (a) kqQ/r 2 = 4.61 × 10 3 N; (b) a = F/m = 6.88 c 1029 m/s 2 9. (a) ke 2 /r 2 = 4.21 × 10 –8 N; (b) ke 2 /r 2 = 2.90 × 10 –9 N 10. (a) kqQ(–0.222i – 0.250 j   ) or F = 0.334 kqQ at 48.4 ° below –x axis ; (b) Need F = 0.334 kgQ opposite to (a). Thus r = 2.73 m at 48.4 ° below the –x axis, or , (–1.82 m, –2.04 m) 11. By symmetry q 1 = q 2 and they must be negative. Consider force on the 2 µC on the left.

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