P22_027 - to removing the ion. The forces of the eight...

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27. (a) Every cesium ion at a corner of the cube exerts a force of the same magnitude on the chlorine ion at the cube center. Each force is a force of attraction and is directed toward the cesium ion that exerts it, along the body diagonal of the cube. We can pair every cesium ion with another, diametrically positioned at the opposite corner of the cube. Since the two ions in such a pair exert forces that have the same magnitude but are oppositely directed, the two forces sum to zero and, since every cesium ion can be paired in this way, the total force on the chlorine ion is zero. (b) Rather than remove a cesium ion, we superpose charge e at the position of one cesium ion. This neutralizes the ion, and as far as the electrical force on the chlorine ion is concerned, it is equivalent
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Unformatted text preview: to removing the ion. The forces of the eight cesium ions at the cube corners sum to zero, so the only force on the chlorine ion is the force of the added charge. The length of a body diagonal of a cube is 3 a , where a is the length of a cube edge. Thus, the distance from the center of the cube to a corner is d = ( 3 / 2) a . The force has magnitude F = k e 2 d 2 = ke 2 (3 / 4) a 2 = (8 . 99 10 9 N m 2 / C 2 )(1 . 60 10 19 C) 2 (3 / 4)(0 . 40 10 9 m) 2 = 1 . 9 10 9 N . Since both the added charge and the chlorine ion are negative, the force is one of repulsion. The chlorine ion is pushed away from the site of the missing cesium ion....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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