Electric Potential Energy-solutions

Electric Potential Energy-solutions - gilvin (jg47854) - 7:...

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R + Q 2 R + 2 2 R = k e 2 Q 2 gilvin (jg47854) - 7: Electric Potential Energy - meyers - (21235) This print-out should have 19 questions. The potentials of the 5 cases are Multiple-choice questions may continue on 1 the next column or page - find all choices before answering. Thios homwework is due Tuesday, February 22, at midnight Tucson time. 001 (part 1 of 2) 10.0 points Pictured below are 5 different configurations of point charges: +Q +Q 1) b b R +Q −Q 2) b b R −Q −Q 3) b b R +Q 4) b +Q +Q +Q 5) b b b R R Which of the following choices has the high- est potential energy? 1. Configuration (3) and Configuration (5) 2. Configuration (1) and Configuration (5) 3. Configuration (1) 4. Configuration (1) and Configuration (2) 5. Configuration (2) and Configuration (3) 6. Configuration (5) correct 7. Configuration (1) and Configuration (3) 8. Configuration (2) 9. Configuration (3) 10. Configuration (4) Explanation: U 1 = k e Q 2 R U 2 = −k e Q 2 R U 3 = k e Q 2 R U 4 = 0 U 5 = k e Q 2 R 002 (part 2 of 2) 10.0 points Which of the following choices has the lowest potential energy? 1. Configuration (4) 2. Configuration (1) and Configuration (5) 3. Configuration (1) and Configuration (2) 4. Configuration (2) correct 5. Configuration (3) 6. Configuration (2) and Configuration (3) 7. Configuration (1) 8. Configuration (5) 9. Configuration (1) and Configuration (3) 10. Configuration (3) and Configuration (5) Explanation: Only U 2 has a negative potential. 003 10.0 points Two alpha particles (helium nuclei), each con- sisting of two protons and two neutrons, have an electrical potential energy of 6.5 × 10 −19 J . Given: k e = 8.98755 × 10 9 N m 2 /C 2 , q p = 1.6021 × 10 −19 C , and g = 9.8 m/s 2 . What is the distance between these parti- cles at this time? Correct answer: 1.4196 × 10 −9 m. Explanation:
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gilvin (jg47854) - 7: Electric Potential Energy - meyers - (21235) 2 Let: U electric = 6.5 × 10 −19 J , 005 10.0 points k e = 8.98755 × 10 9 N m 2 /C 2 , q p = 1.6021 × 10 −19 C , q α = 2 q p = 3.2042 × 10 −19 C , and q n = 0 C . q 1 = q 2 = 2 q p + 2 q n = 2 (1.6021 × 10 −19 C) + 2 (0 C) = 3.2042 × 10 −19 C U electric = k e q 1 q 2 r r = k e q 1 q 2 U electric = k e q 1 U electric = (8.99 × 10 9 N · m 2 /C 2 ) × (3.2042 × 10 19 C) 2 6.5 × 10 −19 J = 1.4196 × 10 −9 m . 004 10.0 points A charge moves a distance of 1.6 cm in the direction of a uniform electric field having a magnitude of 225 N/C. The electrical potential energy of the charge decreases by 90.5043 × 10 −19 J as it moves. Find the magnitude of the charge on the
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This note was uploaded on 04/29/2011 for the course PHYS 202 taught by Professor Doc during the Spring '11 term at North Texas.

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Electric Potential Energy-solutions - gilvin (jg47854) - 7:...

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