problem21_72

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

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21.72: a) Free body diagram as in 21.71. Each charge still feels equal and opposite electric forces. b) . N 0285 . 0 20 sin so N 0834 . 0 20 cos 1 2 2 1 r q kq T F mg T e = = ° = = ° = (Note: m.) 342 . 0 m)sin20 500 . 0 ( 2 1 = ° = r c) From part (b), . C 10 71 . 3 2 13 2 1 - × = q q d) The charges on the spheres are made equal by connecting them with a wire, but we still have 2 2 2 0 4 1 e N 0453 . 0 tan r Q πε θ mg F = = = where . 2 2 1 q q Q + = But the separation 2 r is known: m) 500 . 0 ( 2 2 = r m. 500 . 0 30 sin = ° Hence: 2 2 0 2 4 2 1 r F πε Q e q q = = + . C 10 12 . 1 6 -
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Unformatted text preview: × = This equation, along with that from part (b), gives us two equations in 1 q and C 10 24 . 2 . 6 2 1 2-× = + q q q and . C 10 70 . 3 2 13 2 1-× = q q By elimination, substitution and after solving the resulting quadratic equation, we find: C 10 06 . 2 6 1-× = q and C 10 80 . 1 7 2-× = q ....
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• NoProfessor
• Equations, Quadratic equation, Elementary algebra

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