PH202-5D Final Comprehensive Exam (August 10, 2007)

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1 NAME SCORE PH202-5D Final Comprehensive Exam (August 0, 2007) You may not open the textbook nor notebook. A letter size information may be used. A calculator may be used. However, mathematics or physics formula programmed in a calculator may not be used. Write down, reasoning, calculation and answer in the blank space after each problem. Use the backside of the sheet if necessary. Answers without reasonable explanation nor mathematical derivation will receive no point. Partial credit may be given to correct reasoning and mathematical procedures even if your final answers are wrong. Don t forget units in the final answers! Important Physical Constants: Electric charge of electron: e = C Mass of electron: m e = kg Mass of proton: m p = kg Permittivity of free space: ǫ 0 = 4πk = C 2 /(N m 2 ) where k = N m 2 /C 2 Permeability of free space: µ 0 = 4π 0 7 T m/a Speed of light in vacuum: c= m/s

2 . [5 pts.] Figure shows three point charges fixed in place. The charge at the coordinate origin has a value of q = +8.00µC; the other two have identical charge of q 2 = q 3 = 4.00µC. Draw electric field lines the figure below. You must draw a sufficient number of lines so that the electric field inside and outside of the triangle is clearly displayed. (At least 8 lines.) (b) Determine the net force (magnitude and direction) exerted on q by the other two charges. (c) What is the electrostatic potential energy of the whole system assuming that the eletrostatic potential energy is zero when the charges are infinitely far apart? (b) Let F 2 and F 3 be force exerted by q 2 and q 3, respectively. Using Coulomb s law, F 2 = F 3 = kq q 2 r 2 = ( ) ( ) ( ) (.30) 2 = 0.7N There x and y components are F 2x = 0.70cos(23 ) = 0.56N, F 3x = 0.70cos(23 ) = 0.56N, F 2y = 0.70sin(23 ) = 0.066N; F 3y = 0.70sin(23 ) = 0.066N; Therfore, the components of the net force are given by F x = F 2x + F3x = = 0.32N, F y = F 2y + F 3y = = 0.00N ANS: magnitude=0.32 N; direction=+x (0 ) (c) ANS: J U 3 = U 2 = kq q 2 = ( ) ( ) ( ) = 0.22J r 2.3 U 23 = kq 2q 3 = ( ) ( ) ( ) r sin(23 = 0.42J ) U NET = U 2 + U 3 + U 3 = = 0.30J 2

3 2. [5 pts.] Consider a parallel plate capacitor with a capacitance of 2.5 µf. Initially, there is no net electric charge on each plates. When a certain amount of electric charge is transfered from one plate to the other, the potential difference between two plates becomes 2.0 V. How many electrons are transfered between the plates? (b) What is the minimum energy needed to transfer that amount of charge? (c) A proton is released from the positively charged plate and travels to the negatively charged plate. What is its kinetic energy when it arrives at the negatively charged plate? Q = CV = ( ) 2.0 = C ANS:. 0 3 N = Q e = = (b) E = 2 CV 2 = 0.5 ( ) (2.0) 2 = J ANS: J (c) Using the law of mechanical energy conservation, ANS: J 2 mv2 = ev = ( ) 2.0 = J 3

4 3. [4 pts.] A metal rod of length 5 cm is sliding on two parallel metal tracks to the right with a velocity of.2 m/s. The tracks are connected at one end so that they and the rod form a closed circuit as shown in Figure. The rod has a resistance 0 Ω, and the tracks have negligible resistance. A uniform magnetic field of 2.0 T perpendicular to the plane of this circuit (into the page) is applied. What is the magnitude and direction (up or down) of the induced current through the rod? (b) What is the magnitude and the direction (left or right) of the force exerted on the rod by the magnetic field? emf = Φ t = BL x = BLv = = 0.36V t I = emf R = 0.36 = 0.036A 0. Since the magnetic flux into the page is increasing, the direction of the indueced magnetic field should out of page (Lenz s law). Therfore, the induced current in the rod is upward (RHL). ANS: A, up (b) Using RHL, the direction is left. ANS: 0.0 N, left F = ILB = = 0.0N 4

5 4. [4 pts] An object (the arrow in Figure) is placed 5.0 cm in front of a spherical concave mirror whose radius is 20.0 cm. Find the location of the image using a mirror equation. (b) If the height of the object is 6.00 cm, what is the image height? (c) Construct the image in Figure below by drawing. Leave rays and other lines used to construct the image in the figure so that the instructor knows how you find the image. (At least three rays should be used.) [Make it sure that the drawing is consistent with the calculation.] f = R/2 = 0.0cm and d o = 5.0cm. Hence, d i = 30.0cm. ANS: 30.0 cm in front of the mirror d i = f d o = = 30.0 (b) ANS: -2.0 cm h i = d i h o = = 2.0 d o 5.0 (c) 5

6 5. [5 pts.] A 2-V battery and four 6-Ω light bulbs are connected as shown in Figure. How much current is drawn from the battery? (b) How much power is consumed by the whole system? (c) Determine the power dissipated in the light bulb L 2. R 23 = 8.0Ω R eq = 48/5 = 9.6Ω ANS:.25 A (b) = + = R 23 R R = 8 R 23 = R + R 23 = = 24 = + = R eq R 23 R = 5 48 I = V R eq = =.25A P = IV =.25 2 = 5W ANS: 5 W (c) I 4 = V R 4 = 2 6 = 0.75A I = I I 4 = = 0.50A V = I R = = 8.0V ANS:.0 W V 2 = V V = = 4.0V P 2 = V 2 2 = 4.02 R 2 6 =.0W 6

7 6. [4 pts.] Two long, straight wires carry the identical currents of I =I 2 =2.0 A as shown in Figure. A 4.0-µC point charge is moving at 30 m/s along the wires in direction B shown in Figure. The distance between the wires are 6.0 cm and the distance between wire and the charge is 4.0 cm. Answers the following questions. The direction should be specified by a letter (U, D, F, B, L, R) shown in the figure. Find the magnitude and direction of the magnetic field at the position of the charge. (b) Determine the magnitude and direction of the force exerted on the charge by the magnetic field B is upward at the position of the charge B = µ 0I = 4π = T 2πr 2π0.04 B 2 = µ 0I 2 = 4π = T 2πr 2 2π0.02 B 2 is downward at the position of the charge. Hence the net magnetic field is B B 2 = T. ANS: T, direction = D (b) F = qvb = = N ANS: N, direction=l 7

8 7. [2 pts.] A 375-nm thick soap film is surrounded on both sides by air as shown in Figure. The soap film has an index of reflecetion n =.33. When sunlight strikes the film nearly perpendicularly, two colors look bright in the reflected light due to constructive interference. Noting that sunlight consists of visible light whose wavelengths extend from 380 nm to 750 nm. Which wavelengths in this range become bright? 2t = ( N + ) λ 2 n λ = 2tn N + 2 = N + 2 = 998 N + 2 where N is non-negative integer. For N = 0,,2,3,..., λ = 996,665,399,285,... Hence N = and N = 2 correspond to the visible light. ANS: 399 nm and 665 nm 8

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