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Unformatted text preview: howard (cah3459) – homework 09 – Turner – (56705) 1 This printout should have 11 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Initially, both metal spheres are neutral. In a charging process, 1 × 10 13 electrons are removed from one metal sphere and placed on a second sphere. Then the electrical poten tial energy associated with the two spheres is found to be . 061 J . The Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 and the charge on an electron is 1 . 6 × 10 19 C . What is the distance between the two spheres? Correct answer: 0 . 377183 m. Explanation: Let : n = 1 × 10 13 , q e = 1 . 6 × 10 19 C , U electric = . 061 J , and k e = 8 . 98755 × 10 9 N · m 2 / C 2 . q 1 = nq e q 2 = nq e U electric = k e q 1 q 2 r r = k e q 1 q 2 U electric = k e [ nq e ] 2 U electric = (8 . 98755 × 10 9 N · m 2 / C 2 ) · [(1 × 10 13 ) (1 . 6 × 10 19 C)] 2 . 061 J = . 377183 m . 002 10.0 points Four charges are fixed at the corners of a square centered at the origin as follows: q at ( a, + a ); 2 q at (+ a, + a ); 3 q at (+ a, a ); and 6 q at ( a, a ). A fifth charge + q with mass m is placed at the origin and released from rest. Find the speed when it is a great distance from the origin, where the potential energy of the fifth charge due to the four point charges is negligible. 1. $v = q 6 √ 3 k ma 2. $v = q 6 √ 5 k ma 3. $v = q 3 √ 5 k ma 4. $v = q 3 √ 2 k ma 5. $v = q 2 √ 2 k ma 6. $v = q 3 √ 3 k ma 7. $v = q 6 √ 2 k ma correct 8. $v = q 2 √ 5 k ma 9. $v = q 6 √ 6 k ma 10. $v = q 3 √ 6 k ma Explanation: x a +6 q a 3 q +2 q + q + q, m √ 2 a The initial energy of the charge is E i = K i + U i = U i howard (cah3459) – homework 09 – Turner – (56705) 2 = q k q √ 2 a + 2 k q √ 2 a + ( 3 q ) k √ 2 a + 6 k q √ 2 a = 6 k q 2 √ 2 a . The final energy is E f = 1 2 mv 2 . From energy conversation, we have E i = E f 6 k q 2 √ 2 a = 1 2 mv 2 v = q 6 √ 2 k ma . 003 10.0 points A charged particle is connected to a string that is is tied to the pivot point P . The particle, string, and pivot point all lie on a horizontal table (consequently the figure below is viewed from above the table). The particle is initially released from rest when the string makes an angle 37 ◦ with a uniform electric field in the horizontal plane (shown in the figure). 37 ◦ 2 . 2 m 462 V / m P . 01 kg 3 μ C initial release ω parallel P Determine the speed of the particle when the string is parallel to the electric field. Correct answer: 0 . 350429 m / s. Explanation: Let : L = 2 . 2 m , q = 3 μ C = 3 × 10 6 C , m = 0 . 01 kg , and θ = 37 ◦ ....
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This note was uploaded on 10/11/2010 for the course PHY 56715 taught by Professor Turner during the Spring '10 term at University of Texas.
 Spring '10
 Turner

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