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Physics solutions

# Physics solutions - Yale University Department of Physics...

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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 ˆ x = 9 × 10 9 Nm 2 /C 2 (5 × 10- 9 C )(5 × 10- 9 C ) 3 2 × 10- 4 m 2 ˆ x = 25 × 10- 5 N ˆ x (2 pts) (7) 2 and, ~ F 2 on 1 =- k | q 1 || q 2 | r 2 2 (cos θ ˆ x + sin θ ˆ 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 θ ˆ x + sin θ ˆ y ) =- 18 × 10- 5 N (cos θ ˆ x + sin θ ˆ y ) (2 pts) (8) The angle, θ , is given by in the inverse tan of the ratio between the two sides, tan- 1 4 3 = 53 . 13 o → sin θ = 0 . 8 (1 pts) → cos θ = 0 . 6 (1 pts) (9) Alternatively, since we know all three sides of the triangle (it is a 3-4-5 right triangle), we can directly compute the sine and cosine of the angle using SOH-CAH-TOA. So, sin θ = 4 / 5 = 0 . 8, and cos θ = 3 / 5 = 0 . 6. ~ F net on 1 = 25 × 10- 5 N ˆ x- 18 × 10- 5 (0 . 6ˆ x + 0 . 8ˆ y ) N = (14 . 2 N ˆ x- 14 . 4 N ˆ 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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