Resolução - Halliday - Volume 3 - Eletricidade E Magnetismo - P22 032

Transcript

32. (a) Using Coulomb’s law, we obtain
2 kq q1 q2 = 2 = F = 4πε0 r2 r 8.99 × 109 N·m2 C2  (1.00 m)2 (1.00 C)2 = 8.99 × 109 N . (b) If r = 1000 m, then
 2 8.99 × 109 N·m (1.00 C)2 C2 q1 q 2 kq 2 F = = = = 8.99 × 103 N . 4πε0 r2 r2 (1.00 × 103 m)2