F x 5 0 f 1 x 5 0 f 2 x 5 2 f 3 x 2 1 0200 m 2 5

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F x 5 0. F 1 x 5 0, F 2 x 5 2 F 3 x " 2 1 0.200 m 2 5 0.2828 m. q 4 q 1 r 14 2 1.00 m C 1.35 3 10 2 3 N F 1 5 k 0 q 1 q 4 0 r 14 2 5 1 8.99 3 10 9 N # m 2 / C 2 2 1 3.00 3 10 2 9 C 21 1.00 3 10 2 6 C 2 1 0.1414 m 2 2 5 1.35 3 10 2 3 N. F S net 5 F S 1 . F S 2 1 F S 3 5 0 F S 3 F S 2 r 5 1 1 2 2 " 2 1 0.200 m 2 5 0.1414 m. q 4 q 1 r 14 2 1.00 m C 0.20 m 0.20 m q 1 F 2 F 3 F 1 q 2 q 3 q 4 0.20 m 0.20 m q 1 y x q 3 F 1 F 2 q 4 q 2 F 3 (a) (b) F 5 k 0 qq r 0 r 2 . 1 x 327° F S u 5 32.6°. tan u 5 P F y F x P 5 0.640 F 5 " F x 2 1 F y 2 5 8.01 3 10 2 5 N. F S 17-24 Chapter 17
17.71. Set Up: The ball is in equilibrium, so for it and The force diagram for the ball is given in Figure 17.71. is the force exerted by the electric field. Since the electric field is horizontal, is horizontal. Use the coordinates shown in the figure. The tension in the string has been replaced by its x and y components. Figure 17.71 Solve: gives and gives so q is negative and is to the right, so is to the left in the figure. Reflect: The larger the electric field E the greater the angle the string makes with the wall. 17.72. Set Up: is toward a negative charge and away from a positive charge. At the origin, due to the charge is in the direction, toward the charge. Solve: (a) so is away from Q so Q is positive. gives (b) so is toward Q so Q is negative. 17.73. Set Up: The gravity force (weight) has magnitude and is downward. The electric force is Solve: (a) To balance the weight the electric force must be upward. The electric field is downward, so for an upward force the charge q of the person must be negative. gives and (b) The repulsive force is immense and this is not a feasible means of flight. F 5 k 0 qq r 0 r 2 5 1 8.99 3 10 9 N # m 2 / C 2 2 1 3.9 C 2 2 1 100 m 2 2 5 1.4 3 10 7 N. 0 q 0 5 mg E 5 1 60 kg 21 9.80 m / s 2 2 150 N / C 5 3.9 C. mg 5 0 q 0 E w 5 F F S 5 q E S . w 5 mg 0 Q 0 5 E 2 r 2 k 5 1 76.2 N / C 21 0.600 m 2 2 8.99 3 10 9 N # m 2 / C 2 5 3.05 3 10 2 9 C. E S E 2 x 5 E x 2 E 1 x 5 2 45.0 N / C 2 31.2 N / C 5 2 76.2 N / C. E x 5 2 45.0 N / C, 0 Q 0 5 E 2 r 2 k 5 1 13.8 N / C 21 0.600 m 2 2 8.99 3 10 9 N # m 2 / C 2 5 5.53 3 10 2 10 C. E 2 5 k 0 Q 0 r 2 E S E 2 x 5 E x 2 E 1 x 5 1 45.0 N / C 2 31.2 N / C 5 13.8 N / C. E x 5 1 45.0 N / C, E x 5 E 1 x 1 E 2 x . E 1 x 5 1 31.2 N / C. E 1 5 1 8.99 3 10 9 N # m 2 / C 2 2 1 5.00 3 10 2 9 C 2 1 1.20 m 2 2 5 31.2 N / C. 1 x 2 5.00 nC E S 1 E S E 5 k 0 q 0 r 2 . E S F S E E 5 F E 0 q 0 5 3.78 3 10 2 2 N 1.11 3 10 2 6 C 5 3.41 3 10 4 N / C F E 5 0 q 0 E F E 5 1 mg cos u 2 sin u 5 mg tan u 5 1 12.3 3 10 2 3 kg 21 9.80 m / s 2 2 tan17.4° 5 3.78 3 10 2 2 N F E 2 T sin u 5 0. F E 2 T x 5 0. a F x 5 0 T 5 mg cos u . T cos u 2 mg 5 0 T y 2 mg 5 0. a F y 5 0 x y T T y T x F E u mg F S E F S 5 q E S . F E a F y 5 0. a F x 5 0 Electric Charge and Electric Field 17-25
17.74. Set Up: The force diagram for the charge is given in Figure 17.74. is the force exerted on the charge by the uniform electric field. The charge is negative and the field is to the right, so the force exerted by the field is to the left. is the force exerted by the other point charge. The two charges have opposite signs, so the force is attractive. Take the axis to be to the right, as shown in the figure.

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