687_Physics ProblemsTechnical Physics

687_Physics ProblemsTechnical Physics - 27 Chapter 23...

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Chapter 23 27 P23.71 The field on the axis of the ring is calculated in Example 23.8, EE kxQ xa x e == + 22 32 ej The force experienced by a charge q placed along the axis of the ring is Fk Q q x e =− + L N M M M O Q P P P and when << , this becomes F kQq a x e F H G I K J 3 This expression for the force is in the form of Hooke’s law, with an effective spring constant of k a e = 3 Since ωπ 2 f k m , we have f ma e = 1 2 3 π . P23.72 d kdq x x x kx d x x e e E ij = + −+ + F H G G I K J J = + 2 2 2 2 2 2 0150 . ± . ± . ± . ± . m m m m m af a f a f λ + zz = dk xd x x e x all charge m m m ± . ± . . 2 2 0 0400 a f FIG. P23.72 E i j Ei j j i j = + + + + L N M M M O Q P P P × + + × = − + −− k x x x e ± . . ± .. ± ± . . ± . ± . ± . ± 2 2 0 2 2 2 0 99 1 1 3 8 9 91 0 3 5 01 0 2 3 46 6 7 6 2 40 1 36 1 96 10 1 36 1 96 m m m m Nm C Cm m m NC kNC m afaf e j a f P23.73 The electrostatic forces exerted on the two charges result in a net torque τθ θ Fa Eqa sin sin . For small , sin θθ and using pq a = 2 , we have Ep . The torque produces an angular acceleration given by
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