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**Unformatted text preview: **tan(10 ) cos(10 ) 1.00 10 N 0.30 10 N 0.41 10 kg 4.1g tan(10 ) (9.8 N/kg)tan10 F F mg F F m g---ﾴ-= ﾴ = ﾴ-ﾴ-ﾴ = = = ﾴ = ﾴ ﾴ 25.75. Model: The charges are point charges. Visualize: Solve: The forces on the -1.0 nC charge lie along the line connecting the pairs of charges. Since the 10 nC charge is positive, 10 F r points as shown. The angle formed by the dashed lines where they meet is 90 ﾴ , so q F r must point towards q in order for 10 q F F F = + r r r to hold. Therefore 10 cos(30 ) F F = ﾴ The distance between the 10 nC and -1.0 nC charges is 5.0sin30 cm 2.5 cm. r = ﾴ = Thus 9 9 9 2 2 4 10 2 2 (10 10 C)(1.0 10 C) (9.0 10 N m /C ) 1.44 10 N (2.5 10 m) F----ﾴ ﾴ = ﾴ = ﾴ ﾴ Thus, 4 4 1.44 10 N 1.7 10 N cos(30 ) F--ﾴ = = ﾴ ﾴ...

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