E 2 5 k q 2 r 2 2 5 1 899 3 10 9 n m 2 c 2 21 600 3

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E 2 5 k 0 q 2 0 r 2 2 5 1 8.99 3 10 9 N # m 2 / C 2 21 6.00 3 10 2 9 C 2 1 0.400 m 2 2 5 337 N / C E 1 5 k 0 q 1 0 r 1 2 5 1 8.99 3 10 9 N # m 2 / C 2 21 4.00 3 10 2 9 C 2 1 1.20 m 2 2 5 25.0 N / C 2 x E x 5 E 1 x 1 E 2 x 5 2 899 N / C 1 1 2 150 N / C 2 5 2 1049 N / C. E 2 5 k 0 q 2 0 r 2 2 5 1 8.99 3 10 9 N # m 2 / C 2 21 6.00 3 10 2 9 C 2 1 0.600 m 2 2 5 150 N / C E 1 5 k 0 q 1 0 r 1 2 5 1 8.99 3 10 9 N # m 2 / C 2 21 4.00 3 10 2 9 C 2 1 0.200 m 2 2 5 899 N / C q 1 q 2 E 1 E 1 E 2 E 2 E 1 E 2 x y c a b 0.20m 0.20m 0.80m 1.20m 2 1 E S E 5 k 0 q 0 r 2 . 1 1.60 3 10 2 19 C 21 8.74 3 10 3 N / C 2 5 1.40 3 10 2 15 N, F 5 qE 5 E 5 8.74 3 10 3 N / C, 6.54 3 10 3 N / C, E 5 E 1 1 E 2 5 E 2 5 k 0 q 2 0 r B 2 2 5 1 8.99 3 10 9 N # m 2 / C 2 2 12.5 3 10 2 9 C 1 0.350 m 2 2 5 9.17 3 10 2 N / C E 1 5 k 0 q 1 0 r B 1 2 5 1 8.99 3 10 9 N # m 2 / C 2 2 6.25 3 10 2 9 C 1 0.100 m 2 2 5 5.619 3 10 3 N / C E 2 2 E 1 5 8.74 3 10 3 N / C, E 5 E 2 5 k 0 q 2 0 r A 2 2 5 1 8.99 3 10 9 N # m 2 / C 2 2 12.5 3 10 2 9 C 1 0.100 m 2 2 5 1.124 3 10 4 N / C E 1 5 k 0 q 1 0 r A 1 2 5 1 8.99 3 10 9 N # m 2 / C 2 2 6.25 3 10 2 9 C 1 0.150 m 2 2 5 2.50 3 10 3 N / C 17-12 Chapter 17
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17.42. Set Up: For a point charge, is toward a negative charge and away from a positive charge. The two charges and their fields at each point are shown in Figures 17.42a-d. Figure 17.42 Solve: (a) and are in opposite directions, so the resultant electric field is zero. (b) The resultant electric field has magnitude 2660 N/C and is in the direction. (c) and makes an angle counterclockwise from the axis. (d) has magnitude 1380 N/C and is in the direction. 1 y E S E y 5 E 1 y 1 E 2 y 5 2 E 1 sin u 5 2 1 863 N / C 2 sin 53.1° 5 1380 N / C. E x 5 E 1 x 1 E 2 x 5 0. E 1 5 E 2 5 1 8.99 3 10 9 N # m 2 / C 2 2 6.00 3 10 2 9 C 1 0.250 m 2 2 5 863 N / C. u 5 53.1°. E 5 " E x 2 1 E y 2 5 526 N / C. 1 x 360° 2 75.7° 5 284° E S f 5 75.7° tan f 5 P E y E x P 5 3.92. E y 5 E 1 y 1 E 2 y 5 2 337 N / C 2 173 N / C 5 2 510 N / C. E x 5 E 1 x 1 E 2 x 5 0 1 130 N / C 5 1 130 N / C. E 2 y 5 2 E 2 sin u 5 2 173 N / C. E 2 x 5 E 2 cos u 5 130 N / C. u 5 53.1°. E 2 5 1 8.99 3 10 9 N # m 2 / C 2 2 6.00 3 10 2 9 C 1 0.500 m 2 2 5 216 N / C. E 1 y 5 2 337 N / C. E 1 x 5 0, E 1 5 1 8.99 3 10 9 N # m 2 / C 2 2 6.00 3 10 2 9 C 1 0.400 m 2 2 5 337 N / C. 1 x E y 5 0. E x 5 E 1 x 1 E 2 x 5 E 1 1 E 2 5 2660 N / C. E 2 5 1 8.99 3 10 9 N # m 2 / C 2 2 6.00 3 10 2 9 C 1 0.450 m 2 2 5 266 N / C E 1 5 1 8.99 3 10 9 N # m 2 / C 2 2 6.00 3 10 2 9 C 1 0.150 m 2 2 5 2397 N / C E x 5 E y 5 E 5 0. E S 1 5 E S 2 E 1 5 E 2 x y E 1 E 2 q 1 q 2 0.15 m 0.15 m (a) E 2 E 1 0.15 m 0.15 m 0.15 m y x q 2 q 1 (b) (c) (d) 0.15 m x y E 1 x E 2 y E 1 E 2 q 1 q 2 0.15 m 0.50 m 0.40 m u u q 2 q 1 y x E 2 y E 2 x E 2 E 1 y E 1 x E 1 0.20 m 0.15 m 0.15 m 0.25 m u u u u E S E 5 k 0 q 0 r 2 . Electric Charge and Electric Field 17-13
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17.43. Set Up: For a point charge, is toward a negative charge and away from a positive charge. Let and For the net electric field to be zero, and must have equal magnitudes and opposite directions. Figure 17.43 Solve: The two charges and the directions of their electric fields in three regions are shown in Figure 17.43a. Only in region II are the two electric fields in opposite directions. Consider a point a distance x from so a distance from gives and is the positive solution. The electric field is zero at a point between the two charges, 0.24 m from the 0.500 nC charge.
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