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Unformatted text preview: Yale University Department of Physics Physics 181 University Physics, Spring 2011 Assignment #1 Solutions by Camille Avestruz Problem 1: Isolated conducting spheres. HRW 21.2. Figure 1: Problem 1 In (a), both spheres have charge q, Q 1 = Q 2 = q (1 pts) (1) In part (b), when the neutral charged sphere 3 comes into contact with sphere 1, the total charge between 1 and 3 is conserved and distributes equally, so, Q 1 + Q 3 = Q 1 + Q 3 = q + 0 and Q 1 = Q 3 Q 1 = Q 3 = q 2 (2 pts) (2) In part (c), when sphere 3 comes into contact with sphere 2, the total charge between 2 and 3 is conserved and distributes equally, so, Q 2 + Q 3 = Q 2 + Q 00 3 = q + q 2 and Q 2 = Q 00 3 Q 2 = Q 00 3 = 3 q 4 (2 pts) (3) The initial force between the two spheres was,  ~ F  = k q 2 r 2 (1 pts) (4) The final force between the two spheres is,  ~ F  = k 3 q 2 2 4 r 2 (2 pts) (5) F F = 3 8 = . 375 (2 pts) (6) Problem 2: Force vectors. Figure 2: Problem 2 The net force vector on q 1 is given by the sum of forces from q 2 and q 3 . ~ F 3 on 1 = k  q 1  q 3  r 2 1 x = 9 10 9 Nm 2 /C 2 (5 10 9 C )(5 10 9 C ) 3 2 10 4 m 2 x = 25 10 5 N x (2 pts) (7) 2 and, ~ F 2 on 1 = k  q 1  q 2  r 2 2 (cos x + sin y ) = 9 10 9 Nm 2 /C 2 (5 10 9 C )(10 10 9 C ) (3 2 + 4 2 ) 10 4 m 2 (cos x + sin y ) = 18 10 5 N (cos x + sin y ) (2 pts) (8) The angle, , is given by in the inverse tan of the ratio between the two sides, tan 1 4 3 = 53 . 13 o sin = 0 . 8 (1 pts) cos = 0 . 6 (1 pts) (9) Alternatively, since we know all three sides of the triangle (it is a 345 right triangle), we can directly compute the sine and cosine of the angle using SOHCAHTOA. So, sin = 4 / 5 = 0 . 8, and cos = 3 / 5 = 0 . 6. ~ F net on 1 = 25 10 5 N x 18 10 5 (0 . 6 x + 0 . 8 y ) N = (14 . 2 N x 14 . 4 N y ) 10 5 (2 pts) (10) The corresponding magnitude and angle of the net force on 1 is,  ~ F net on 1  = p 14 . 2 2 + 14 . 4 2 10 5 N = 2 ....
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