spring_06_exam4_sol_phy2049

spring_06_exam4_sol_phy2049 - PHY2049 Spring 2006 Exam 4...

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PHY2049 Spring 2006 Exam 4 Solutions 1 Prof. Darin Acosta Prof. Greg Stewart May 1, 2006 Exam 4 (Final) Solutions 1. Four charges are arranged into a square with side length a=1 cm as shown in the figure. The charges (clockwise from top left) are q 1 = 10 nC, q 2 = -10 nC, q 3 = -10 nC, and q 4 = 10 nC (1nC = 10 -9 C). What is the magnitude of the electric field at the center of the square in N/C? a. 5.1 × 10 6 b. 7.2 × 10 6 c. 1.3 × 10 6 d. 2.5 × 10 6 e. 0 Solution: By symmetry the field in the vertical direction will cancel out, and the field will point in the +x horizontal direction. Magnitude of the field from one charge is: 2 2 6 2 where 2 2 cos45 for one charge 22 N 4 5.1 10 all charges 2C x x qa EK r r q a q a == ⇒= D 2. A charge of 10 -6 C is distributed uniformly across a large circular sheet with a radius of 1m that is nonconducting. What is the magnitude of the electric field in N/C at a point 0.5 cm above the center of the circle? a. 1.8 × 10 4 b. 3.6 × 10 4 c. 3.6 × 10 8 d. 3.2 × 10 -7 e. 5.6 × 10 4 Solution: A 1m sheet compared to a position 0.5 cm above it makes the sheet essentially infinite. By using Gauss’ Law, we can determine the electric field to be constant and given by 0 2 E σ ε = . The charge density is 6 7 2 10 3.18 10 C q Ar π = × . So the field is 1.8 × 10 4 N/C.
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PHY2049 Spring 2006 Exam 4 Solutions 2 3. Two electrons are fixed 2~cm apart. A third electron is shot from infinity and comes to a rest midway between the two. What was its initial velocity? a. 320 m/s b. 80 m/s c. 2.9 × 10 -7 m/s d. 1.0 × 10 5 m/s e. 0 Solution: First find the electric potential at the point between the 2 electrons assuming V=0 at infinity: /2 ee VK K dd =+ The potential energy, then, for an electron ( q= -e ) to be at rest there, is: 2 4 e Uq V K d == . Thus, an electron must have a kinetic energy equal to U initially: 2 2 2 1 4 2 8 320 m/s K e Em v K d Ke v md ⇒= = 4. An air filled parallel plate capacitor has a constant charge q=10 -4 C on one of its plates and the voltage across the capacitor is 100 V. If the plates are pulled apart to 3 times their original separation, what potential difference in Volts now exists between the plates? a. 300 b. 100 c. 33 d. 900 e. 11 Solution: The capacitance of a parallel plate capacitor is 0 A C d ε = , so if the plates are pulled apart 3 times further, the capacitance is 3 times smaller. Then since qC V = , the voltage must increase 3 times to 300 V if the charge is to remain constant. 5. Two identical light bulbs (with identical resistances) are connected in series to a constant voltage source. If the light bulbs are instead connected in parallel to the constant voltage source, what is the ratio of the brightness of a light bulb in the parallel configuration to its former brightness when connected in series? a. 4 b. 0.25 c. 16 d. 2 e. 1 Solution: Let the light bulbs have resistance R and let the voltage be V from the source. The current through the circuit with the resistors in series is iV R = . The brightness of one light is related to the power dissipated by one resistor, which is 2 2 4 V PiR R .
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PHY2049 Spring 2006 Exam 4 Solutions 3
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spring_06_exam4_sol_phy2049 - PHY2049 Spring 2006 Exam 4...

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