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# solution37 - cherubin(mcc2645 – Homework 2 Electric Field...

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Unformatted text preview: cherubin (mcc2645) – Homework 2 Electric Field and Potential – Catala – (12054) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points An electric field does 10 J of work on a . 0008 C charge. What is the voltage change? Correct answer: 12500 V. Explanation: Let : W = 10 J and q = 0 . 0008 C . Work is W = qV V = W q = 10 J . 0008 C = 12500 V . 002 (part 2 of 2) 10.0 points The same electric field does 20 J of work on a . 0016 C charge. What is the voltage change? Correct answer: 12500 V. Explanation: Let : W = 20 J and q = 0 . 0016 C . The voltage change is V = W q = 20 J . 0016 C = 12500 V . 003 10.0 points Two charges are located along the x-axis. One has a charge of 6 . 4 μ C, and the second has a charge of- 3 . 4 μ C. The Coulomb constant is k e = 8 . 98755 × 10 9 N m 2 / C 2 . The acceleration of gravity is 9 . 81 m / s 2 . If the electrical potential energy associated with the pair of charges is- 4 . 3 × 10- 2 J, what is the distance between the charges? Correct answer: 4 . 54936 m. Explanation: Let : q 1 = 6 . 4 μ C , q 2 =- 3 . 4 μ C , U electric =- 4 . 3 × 10- 2 J , and k e = 8 . 99 × 10 9 N · m 2 / C 2 . U electric = k e q 1 q 2 r r = k e q 1 q 2 U electric = (8 . 99 × 10 9 N · m 2 / C 2 ) × (6 . 4 × 10- 6 C)(- 3 . 4 × 10- 6 C)- 4 . 3 × 10- 2 J = 4 . 54936 m . 004 10.0 points A charge moves a distance of 2 . 3 cm in the direction of a uniform electric field having a magnitude of 222 N / C. The electrical potential energy of the charge decreases by 122 . 642 × 10- 19 J as it moves. Find the magnitude of the charge on the moving particle. (Hint: The electrical poten- tial energy depends on the distance moved in the direction of the field.) Correct answer: 2 . 40192 × 10- 18 C. Explanation: Let : Δ d = 2 . 3 cm , E = 222 N / C and Δ U electric =- 122 . 642 × 10- 19 J . cherubin (mcc2645) – Homework 2 Electric Field and Potential – Catala – (12054) 2 Δ U electric =- q E Δ d q =- Δ U electric E Δ d =-- 122 . 642 × 10- 19 J (222 N / C)(0 . 023 m) = 2 . 40192 × 10- 18 C . 005 10.0 points The Coulomb constant is 8 . 98755 × 10 9 N m 2 / C 2 and the acceleration of gravity is 9 . 81 m / s 2 ....
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solution37 - cherubin(mcc2645 – Homework 2 Electric Field...

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