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Unformatted text preview: marout (ssm979) – homework 09 – Turner – (56725) 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, 2 × 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: 1 . 50873 m. Explanation: Let : n = 2 × 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 ) · [(2 × 10 13 ) (1 . 6 × 10 − 19 C)] 2 − . 061 J = 1 . 50873 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. bardbl vectorv bardbl = q radicalBigg 3 √ 6 k ma 2. bardbl vectorv bardbl = q radicalBigg 3 √ 5 k ma 3. bardbl vectorv bardbl = q radicalBigg 2 √ 2 k ma 4. bardbl vectorv bardbl = q radicalBigg 6 √ 5 k ma 5. bardbl vectorv bardbl = q radicalBigg 3 √ 2 k ma 6. bardbl vectorv bardbl = q radicalBigg 3 √ 3 k ma 7. bardbl vectorv bardbl = q radicalBigg 6 √ 3 k ma 8. bardbl vectorv bardbl = q radicalBigg 6 √ 2 k ma correct 9. bardbl vectorv bardbl = q radicalBigg 6 √ 6 k ma 10. bardbl vectorv bardbl = q radicalBigg 2 √ 5 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 marout (ssm979) – homework 09 – Turner – (56725) 2 = q parenleftbigg k q √ 2 a + 2 k q √ 2 a + ( − 3 q ) k √ 2 a + 6 k q √ 2 a parenrightbigg = 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 radicalBigg 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 31 ◦ with a uniform electric field in the horizontal plane (shown in the figure)....
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 Fall '10
 TURNER,J
 Charge, Electric Potential, Potential Energy, Electric charge

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