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Unformatted text preview: PHY2049 Spring 2010 Profs. P. Avery, A. Rinzler, S. Hershfield Final Exam Solution 1. A proton traveling along the x axis (toward increasing x ) has a speed of 1 . × 10 5 m/s. At time t = 0 it enters a region of space in which there is an electric field of 100.0 N/C ˆ j . At a time t = 14 μ s later the proton’s velocity makes an angle with respect to the + x axis that is: Answer: +53 ◦ Solution: The force on the proton is | e | (100 N/C ) in the y-direction so the acceleration is a = | e | (100 N/C ) /m p in the positive y-direction. The x-component of the velocity is constant at v x = 1 . × 10 5 m/s, and the y-component of the velocity is v y = at . The angle the velocity makes with the x-axis is is tan − 1 ( v y /v x ). 2. In the Figure 12.0 V is applied to the network of capacitors where the capacitance values are in μ F. The charge on the positive plate of the upper 10 μ F capacitor (in μ C) is: Answer: 37.5 Solution: The effective capacitance of the 10 μF and 12 μF capacitors in parallel is 22 μF , and the effective capacitance of the entire network is 6 . 875 μF with charge (6 . 875 μF )(12 V ) = 82 . 5 μC . Because capacitors in series have the same charge, this is also the charge on the effective 22 μF capacitor resulting from the 10 μF and 12 μF capacitors in parallel. The voltage across this effective capacitor is Q/C = 82 . 5 μC/ 22 μF = 3 . 75 V . Since capacitors in parallel have the same voltage, this is also the voltage across the upper 10 μF capacitor. The charge on that capacitor is CV = (10 μF )(3 . 75 V ) = 37 . 5 μC . 3. The resistor network in the diagram has 8V across its end terminals labeled A and B. The current through, and voltage drop across, the 6kΩ resistor are, respectively: Answer: 1mA, 6V Solution: The effective resistance of the top segment is 8 k Ω so the current through it is 8 V/ 8 k Ω = 1 mA . This is also the current through the 6 k Ω resistor, which has a voltage drop of IR = (1 mA )(6 k Ω) = 6 mV . 4. At time t = 0 an ideal 12.00 V power supply is connected to an uncharged series RC circuit in which the resistor is 120Ω and the capacitor is 47.0 pF. The charge accumulated on the positive plate of the capacitor at a time t = 10 RC is closest to, Answer: 564 pC Solution: The charge on the capacitor as a function of time is (12 V )(47 × 10 − 12 F )(1 − e − t/ ( RC ) ). At t = 10 RC the exponential factor is very small, e − t/ ( RC ) ≪ 1, i.e. the capacitor is very nearly charged to its final value of (12 V )(47 × 10 − 12 F ) = 5 . 64 × 10 − 12 C . 5. An object is reflected in a concave spherical mirror. The object distance p is less than the focal length f ( p < f ). The image can be (the correct answer contains only true statements from the following list of possibilities): A) real, B) virtual, C) upright,...
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This note was uploaded on 10/26/2011 for the course PHY 2049 taught by Professor Any during the Fall '08 term at University of Florida.
- Fall '08