exam1S - Physics 21 Fall, 2007 Solution, Hour Exam #1 The...

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Physics 21 Fall, 2007 Solution, Hour Exam #1 The graders for the problems were: 1 Belony, 2 McGeehan, 3 Sweeney, 4 Kanofsky, 5 Lyu For questions about the grading, see the grader by Oct. 19. Problem 1. 2 3 4 I 1 I 3 I 2 10.0 V 8.0 V a b loop 1 loop 2 loop 3 (a) (10 pts.) Write the loop and junction equations needed to determine the currents I 1 , I 2 ,and I 3 in the circuit shown. Indicate clearly the loop used to determine each loop equation. The junction and loop equations are: junction: I 2 = I 1 + I 3 loop 1: 8 2 I 1 3 I 2 =0 loop 2: 10 3 I 2 4 I 3 loop 3: 8 2 I 1 10 + 4 I 3 (b) (6 pts.) Determine the currents I 1 , I 2 I 3 (including the correct sign) by explicit solution of the equations. You must show your work. We can use any two loop equations; the di±erence be- tween the loop 1 and 2 equations is the loop 3 equation. Let’s use 1 and 2. First rearrange them and then elim- inate I 3 using the junction equation. loop 1: 2 I 1 +3 I 2 =8 loop 2: 3 I 2 +4 I 3 =10 3 I 2 +4( I 2 I 1 )=10 loop 2 ± : 4 I 1 +7 I 2 Multiply loop 1 by 2 and add to loop 2 ± : loop 1 ± : 4 I 1 +6 I 2 =16 loop 2 ± : 4 I 1 I 2 add: 13 I 2 =26 I 2 =2A Now back substitute to get all the currents: I 1 =1A ,I 2 3 (c) (4 pts.) What is the potential di±erence V b V a between the points marked a and b on the diagram? Use the calculated values of the currents and show how you get your answer. The potential di±erence can be calculated by any path. There are three ways to get the answer: V b V a =4 I 3 +8=10+2 I 1 3 I 2 =12V . Problem 2. The thin rod shown in the diagram is extends along the x axis from x =0to x = a . A total charge Q is spread uniformly along the rod. x y P = (0, y ,0) +a x' dQ r - r ' (a) (3 pts.) Write an expression for the linear charge density λ on the rod. λ = Q/a . (b) (4 pts.) Give the components of the vector ( r r ± )from the charge dQ shownontherodat x = x ± to the point P on the y axis. The ²eld point r is y ˆ j , and the charge point r ± is x ± ˆ i . Then r r ± = x ± ˆ i + y ˆ j . (c) (4 pts.) Determine the electric ²eld d E at the point P due to the element dQ of charge at the point x ± on the rod. Express d E in terms of Q , a , x ± y . Start with the general expression (from equation sheet) d E (at r )= 1 4 π± 0 dQ ( r r ± ) | r r ± | 3 , where dQ = λdx ± =( Q/a ) dx ± and | r r ± | = p ( x ± ) 2 + y 2 .
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exam1S - Physics 21 Fall, 2007 Solution, Hour Exam #1 The...

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