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Unformatted text preview: klarin (sjk772) 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, 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 . 062 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 . 4844 m. Explanation: Let : n = 2 10 13 , q e = 1 . 6 10 19 C , U electric = . 062 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 . 062 J = 1 . 4844 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 2 2 k ma 2. bardbl vectorv bardbl = q radicalBigg 3 6 k ma 3. bardbl vectorv bardbl = q radicalBigg 6 5 k ma 4. bardbl vectorv bardbl = q radicalBigg 6 2 k ma correct 5. bardbl vectorv bardbl = q radicalBigg 3 2 k ma 6. bardbl vectorv bardbl = q radicalBigg 6 6 k ma 7. bardbl vectorv bardbl = q radicalBigg 6 3 k ma 8. bardbl vectorv bardbl = q radicalBigg 3 5 k ma 9. bardbl vectorv bardbl = q radicalBigg 2 5 k ma 10. bardbl vectorv bardbl = q radicalBigg 3 3 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 klarin (sjk772) homework 09 Turner (56705) 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 82 with a uniform electric field in the horizontal plane (shown in the figure)....
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 Fall '10
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