PHY2054_S10_exam1_solutions - PHY2054 Solutions Exam I 1-3...

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February 12, 2010 PHY2054 Solutions Exam I 1-3. Particle 1 has a mass of 1.0 g and charge 6.0 μ C, and particle 2 has a mass of 2.0 g and charge (3.0| 5.0| 8) μ C. While particle 2 is held in place, particle 1 is released from rest at a distance of 2.0 mm from particle 2. After a long time, what is the speed of particle 1 [in m/s]? Answer: 400|520|660 Energy conservation determines that E f = ½ mv 2 = E i = kq 1 q 2 /d. Solve for v to get the answers. 4-6. Four point charges are placed at the four corners of a 3.0 cm × 3.0 cm square, in two different configurations as shown in the figure. The magnitude Q of the charges is (3.0|2.0|1.0) μ C. Which configuration has higher electrical potential energy and what is its value [in J]? Answer: Configuration 1 (-3.8|-1.7|-0.42) The potential energy , where the sum is over all pairs of charges (q 1 , q 2 ) that are located a distance d apart. You can tell the answer for the highest energy configuration is 1 since here the like charges are closer by and like charges always make positive contribution to the potential energy, increasing it. The terms involving near pairs cancel in the U sum. Only the diagonals survive and lead to the answer. 7-9. Three point charges ( Q 1 = - 5 . 0 μ C, Q 2 = 4 . 0 μ C, and Q 3 = 6 . 0 μ C) are located at the corners of an equilateral triangle, as shown. The electric potential due to these charges is (4.0| 8.0| 16) V at the center of the triangle. If Q 2 is increased to 8.0 μ C and Q 3 is decreased to 4.0 μ C, what will be the electric potential at the center [in V]? Answer: ( 5.6| 11.2| 22) The potential is given by the sum over all the charges which happen to be located at the same distance from the center of the triangle. Thus the sum is over only the charges (alternatively, the value given for one set of charges determines the size of the triangle). Therefore V 2 being the potential with new charges and V 1 the potential for original charges, 10-12. In the electric circuit shown, capacitors 1-4 have the same capacitance of 2.0 μ F each, and the amount of charge stored in capacitor 2 is 3.0 μ C. What is the total energy stored in the four capacitors [in μ J]?
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Answer: (27| 48| 32) Since E = ½ CV 2 , we need the equivalent capacitance as well as the potential difference. The equivalent capacitance across the series part C = [1/2 +1/2+1/2] -1 = 2/3 μ F, which in parallel with 2 μ F gives C = 8/3 μ F. The charge across capacitance 2 is 3
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PHY2054_S10_exam1_solutions - PHY2054 Solutions Exam I 1-3...

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