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Challenge Problem Session 1 Solution on General Physics

Challenge Problem Session 1 Solution on General Physics -...

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Solution to Challenge Problem Set 1 (CP1) July 11, 2013 25.74. Model: The charged spheres are point charges. Visualize: The figure shows the free-body diagram of the forces on the sphere with the negative charge that is shown in Figure CP25.74. The force E F is due to the electric field. The force e F is the attractive force between the positive and the negative spheres. The tension in the string and the gravitational force are the remaining two forces on the spheres. Solve: The two electrical forces are calculated as follows: 9 5 2 E (100 10 C)(1.00 10 N/C) 1.00 10 N F q E - - = = = 9 2 2 9 2 5 2 1 2 e 2 2 2 (9.0 10 N m /C )(100 10 C) 9.0 10 N m /C K q q F r r r - - = = = From the geometry of Figure CP25.74, 5 2 3 e 2 9.0 10 N m /C 2 sin(10 ) 2(0.50 m)sin(10 ) 0.174 3.0 10 N (0.174 m) r L E F - - = = = = = From the free-body diagram, e E cos10 sin10 T mg T F F ᄡ = ᄡ + = Rearranging and dividing the two equations gives E e 2 2 2 E e sin(10 ) tan(10 )
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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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