# ass9 - SOLUTIONS ASSIGNMENT 9 plus Problem 23.25 23.10...

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SOLUTIONS, ASSIGNMENT 9 plus Problem 23.25 23.10. I DENTIFY : The work done on the alpha particle is equal to the difference in its potential energy when it is moved from the midpoint of the square to the midpoint of one of the sides. S ET U P : We apply the formula . a b a b W U U = - In this case, a is the center of the square and b is the midpoint of one of the sides. Therefore center side center side . W U U = - There are 4 electrons, so the potential energy at the center of the square is 4 times the potential energy of a single alpha-electron pair. At the center of the square, the alpha particle is a distance r 1 = 50 nm from each electron. At the midpoint of the side, the alpha is a distance r 2 = 5.00 nm from the two nearest electrons and a distance r 2 = 125 nm from the two most distant electrons. Using the formula for the potential energy (relative to infinity) of two point charges, 0 0 (1/ 4 )( / ), U qq r π = P the total work is center side center side W U U = - = 0 1 0 2 0 3 1 1 1 4 2 2 4 4 4 e e e q q q q q q r r r α - + P P P Substituting q e = e and q = 2 e and simplifying gives 2 center side 0 1 2 3 1 2 1 1 4 4 W e r r r = - - + P E XECUTE : Substituting the numerical values into the equation for the work gives ( 29 2 19 21 2 1 1 4 1.60 10 C 6.08 10 J 5.00 nm 50 m 125 nm W - - = - × - + = × ??? E VALUATE : Since the work is positive, the system has more potential energy with the alpha particle at the center of the square than it does with it at the midpoint of a side. 23.14.

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## This note was uploaded on 03/11/2010 for the course PHYSICS 1523 taught by Professor Dfs during the Fall '10 term at Dallas.

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ass9 - SOLUTIONS ASSIGNMENT 9 plus Problem 23.25 23.10...

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