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PHY132 Sp04 Final Solutions

# University Physics with Modern Physics with Mastering Physics (11th Edition)

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PHY 132 Final Exam May 13, 2004 Version Put FULL NAME, ID#, and EXAM VERSION on the front cover of the BLUE BOOKLET! To avoid problems in grading: do all problems in order , write legibly , and show all work for partial credit ! Note: make sketches where appropriate and show your axis conventions. In the problems, directions must be given as “ up ”, “ down ”, “ left ”, “ right ”, “ out of- “ and “ into the paper ”, “ CW ” (clockwise), or “ CCW ”. Include units, watch the number of significant digits, and check results for consistency. This exam lasts 2.5 hr. Success! 1. The two metal plates of a parallel-plate capacitor have been charged with charges + Q and –Q respectively, where Q= 6.00 μC. The plates have area A= 0.600 m 2 and separation distance d= 0.885 mm. The volume be- tween the plates is filled with air ( K= 1.00). ( 50 points ) a. Use Gauss’ Law to prove that the electric field E between the plates equals σ / ε 0 , where σ Q / A . Solution: (10 points) draw a “pill box” through one of the plates: the surface charge density inside the box is σ Q / A . The box has area A’ and the field between the plates is E . Because of the symmetry, the electric field OUTSIDE the plates must be zero, and in between the field must be uniform and pointing from + Q to – Q . Gauss’ Law then gives: EA’ = σ A’/ ε 0 Î E = σ / ε 0 b. Calculate the voltage difference between the two plates. Solution: V = 0 d E · d x = Ed = σ d/ ε 0 = Qd/ ( A ε 0 ) = 1000 V (10 points) c. Calculate the capacitance of the parallel-plate system. Solution: C = Q / V = 6.00 nF (B : 5.00 nF) (10 points) d. An electron ( m e = 9.11×10 –31 kg) is let go from rest near the plate with charge –Q ; calculate the speed of the electron when it arrives at the second plate. Solution: a = F/m = eE/m = eV/ ( md ); v 2 = 2 ad = 2 eV/m Î v =1.87×10 7 m/s (10 points) e. Calculate the energy stored in the capacitor. Solution: U = ½ QV = 3.00×10 –3 J (B : 2.50×10 –3 J) (10 points) 2. Consider the circuit shown in the Figure. ( 40 points ) a. Calculate the currents through R 1 , R 2 , and R 3 . Solution: I 2 = I 1 + I 3 (30 points) 8 – 5 I 1 – 10 I 2 = 0 Î 8 – 15 I 1 – 10 I 3 = 0 Î 16 – 30 I 1 – 20 I 3 = 0 12 – 10 I 3 – 10 I 2 = 0 Î 12 – 20 I 3 – 10 I 1 = 0 Î 4 – 30 I 1 + 10 I 1 = 0 Î I 1 = 4/20 = 0.20 A Î I 3 = 10/20 = 0.50 A Î I 2 = 0.70 A (B: 8 – 10 I 1 – 10 I 2 = 0 Î 8 – 20 I 1 – 10 I 3 = 0 Î 4 – 10 I 1 – 5 I 3 = 0 20 Î 16 – 17 I – 7 I 3 – 10 I 2 = 0 Î 20 – 17 I 3 – 10 I 1 = 0 A b. tween voltages at the points A and B ; beware of the sign! 3 + 5 I 3 = 0 Î I 3 = 12/16 = 0.75 Î I 1 = 1/4/10 = 0.025 A Î I 2 = 0.775 A) Calculate the voltage difference V A V B be Solution: V A V B = 3 I 1 – 6 I 3 = –2.4 V (B: 6 I 1 – 5 I 3 =–3.6 V) (10 points) R 1 =2 + 8 V 3 R 2 =10 A Í I 2 I 1 Î B + − 12 V I 3 Î 6 R 3 =4

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a B ext 3. Consider the rectangular current loop with dimensions a =4.00 cm, b =15.0 cm as shown, and lying in the plane of the paper. The current I= 4.00 A in the loop is in the clockwise direction. The uniform external magnetic field B ext = 2.00 T, directed as shown. ( 40 points ) b a. Calculate the magnetic forces (magnitude and direction!) on each of the four sides
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PHY132 Sp04 Final Solutions - PHY 132 Final Exam Version A...

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