problem07_37

# University Physics with Modern Physics with Mastering Physics (11th Edition)

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7.37: a) . 6 12 7 13 r b r a r U r F - = - = b) Setting 0 = r F and solving for r gives . ) 2 ( 6 / 1 min b a r = This is the minimum of potential energy, so the equilibrium is stable. c) . 4 2 4 ) ) / 2 (( ) ) / 2 (( ) ( 2 2 2 2 6 6 / 1 12 6 / 1 6 min 12 min min a b a b a ab b a b b a a r b r a r U - = - = - = - = To separate the particles means to remove them to zero potential energy, and requires the negative of this, or . 4 2 0 a b E = d) The expressions for 0 E and min r in terms of a and b are . 2 4 6 min 2 0 b a r a b E = = Multiplying the first by the second and solving for b gives 6 min 0 2 r E b = , and substituting this into the first and solving for
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Unformatted text preview: a gives 12 min r E a = . Using the given numbers, . m J 10 41 . 6 ) m 10 13 . 1 )( J 10 54 . 1 ( 2 m J 10 68 . 6 ) m 10 13 . 1 )( J 10 54 . 1 ( 6 78 6 10 18 12 138 12 10 18 ⋅ × = × × = ⋅ × = × × =------b a (Note: the numerical value for a might not be within the range of standard calculators, and the powers of ten may have to be handled seperately.)...
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