This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 345 right triangle), we can directly compute the sine and cosine of the angle using SOHCAHTOA. 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 . ×...
View
Full
Document
This note was uploaded on 03/15/2011 for the course PHYS 181 taught by Professor Stephenirons during the Spring '08 term at Yale.
 Spring '08
 STEPHENIRONS
 Physics, Charge

Click to edit the document details