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QUIZ-I PHYS-102 SOLUTION SPRING-08 Time allowed: 50 min. Notes: 1. For Prob.1 please circle the answer of your choice for each part. 2. For problems 2 and 3, your solutions must have adequate details to get full credit. ________________________________________________________________________ Name (Please PRINT) _________________________________ ________________________________________________________________________ r r q kQ Q FE = qE ; F = 12 2 ; 1/40 = k = 9 x109 Nm 2C !2 , " E = # E.dA = en ; r !o 0 = 8.85x10-12 C2/Nm2 _______________________________________________________________________ 1a. The E-field at the center of a charged conducting shell of radius R carrying a charge q is measured to be zero. The magnitude of the E-field at a point R/2 away from the center would be: [a] zero [b]4 kq/R2 N/C [c] 2 kq/R2 N/C 1b. A thin, uniformly charged ring lies in the x-y plane with its center at the origin as shown. The net E-field at point P on the z-axis will point: z [a] somewhere in the x-y plane. [b] along the positive z-axis . [c] E-field at P is zero because of the symmetry of the charge distribution. 1c. Two point charges are situated symmetrically about the origin. The E-field measured at the origin is zero. At point P, the E-field points in the positive x-direction. From this one can conclude that: [a] both charges are positive and equal in magnitude. [b] both charges are negative and equal in magnitude. [c] the charges are of opposite polarity but of equal magnitude. P + + x + + o + + + + + + + y y E P q1 0 q2 x 1d. A test charge of +3 C is at a point P where an electric external field is directed to the right and has a magnitude of 4 106 N/C. If the test charge is replaced with another charge of -3 C, the external electric field at P: [a] is unaffected, [b] reverses direction, or [c] changes in a way that cannot be determined. 2. A solid sphere of radius R = 50.0 cm has a total positive charge of 25.0 C uniformly distributed throughout its volume. Using Gauss' Law, calculate the magnitude of the electric field at: [a] 20.0 cm from the center of the sphere. E4r2 = 25.0*10-6 r3/0R3 E = 25.0*10-6 r/4 0R3 = 25.0*10-6 (0.2)/4 0(0.5)3 = 3.6*105 N/C " E.dA = q r r enc . / !o [b] 80.0 cm from the center of the sphere. E4r2 = 25.0*10-6 /0 E = 25.0*10-6 /4 0r2 = 25.0*10-6 /4 0(0.8)2 = 3.5*105 N/C 3. Three charges are placed at the three vertices of a right-angle triangle as shown in the diagram below. [a] Determine the force exerted on Q3 by charge Q1. Label this force F31 and express it in the usual i, j notation and show it on the diagram. Y Q2 = 4.0 C = tan-1(4/3) = 53.10 F31 = - 9x109x8.0x6.25x10-12/(3.0x10-2)2i = - 500i N 4.0 cm 5.0cm O Q 1= - 8.0 C 3.0 cm ! F31 " FT X F32 [b] Determine the force exerted on Q3 by charge Q2. Label this force F32, show it on the diagram and express it in the usual i, j notation. F32 = 9x109x4.0x6.25x10-12/(5x10-2)2[ cos53.1o i sin53.1oj]= 54i - 72j N [c] Determine the magnitude and direction of the net force exerted on Q3. FT = F32 + F32 = (54 500) i 72j N = - 446 i 72j N Magnitude , FT = [(446)2 + (72)2 ]1/2 = 451.8 ; = tan-1(72/446) = 9.2o
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Path: Drexel >> PHYS >> 102 Spring, 2008
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Path: Drexel >> PHYS >> 102 Spring, 2008
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Path: Drexel >> PHYS >> 102 Spring, 2008
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Path: Drexel >> PHYS >> 102 Spring, 2008
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Path: Drexel >> COOP >> 101 Spring, 2008
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Path: American University of Sharjah >> MECHANICAL >> 223,224,22 Spring, 2008
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Path: American University of Sharjah >> MECHANICAL >> 223,224,22 Spring, 2008
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Path: American University of Sharjah >> MECHANICAL >> 223,224,22 Spring, 2008
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Path: American University of Sharjah >> MCE >> 240 Spring, 2008
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Path: American University of Sharjah >> MCE >> 240 Spring, 2008
Description: American University of Sharjah College of Engineering Mechanical Engineering Department Spring 2008 Due Date: Sunday, April 22nd, 2008. Statics MCE 220 Assignment # 14 Solve the following problems from the textbook (Engineering Mechanics: Statics ...
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Path: American University of Sharjah >> MCE >> 240 Spring, 2008
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Path: American University of Sharjah >> MCE >> 240 Spring, 2008
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Path: American University of Sharjah >> MCE >> 240 Spring, 2008
Description: American University of Sharjah College of Engineering Mechanical Engineering Department Spring 2008 Due Date: Thursday, April 3rd, 2008. Statics MCE 220-01 Assignment # 11 Solve the following problems from the textbook (Engineering Mechanics: Stat...
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Path: American University of Sharjah >> MECHANICAL >> 223,224,22 Spring, 2008
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Path: American University of Sharjah >> MCE >> 240 Spring, 2008
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Path: American University of Sharjah >> MECHANICAL >> 223,224,22 Spring, 2008
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Path: American University of Sharjah >> MECHANICAL >> 223,224,22 Spring, 2008
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Path: American University of Sharjah >> MECHANICAL >> 223,224,22 Spring, 2008
Description: CHAPTER 16 KEY EQUATIONS Equation Number Key Equation 16.1 Pin = 3 rmsI rms cos( ) V 16.2 16.3 Pout = Toutm m = nm = 16.6 16.7 16.8 16.9 16.11 16.12 16.13 16.14 16.15 16.16 16.17 16.18 16.20 16.21 16.22 16.23 16.27 16.28 16.34 Pout 100% Pin 2 ...
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Path: American University of Sharjah >> MECHANICAL >> 223,224,22 Spring, 2008
Description: CHAPTER 17 KEY EQUATIONS Equation Number Key Equation 17.13 s = P 2 120f 17.14 ns = 17.16 17.17 17.23 17.24 17.25 P s - m ns - nm s = = s ns slip = s Pdev = 3 1-s Pr = 3Rr (I r )2 s Rr (I r )2 17.26 17.27 Ps = 3Rs I s2 Pin = 3I sVs cos( ...
