685_Physics ProblemsTechnical Physics

# 685_Physics ProblemsTechnical Physics - 25 Chapter 23...

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Chapter 23 25 P23.66 (a) The distance from each corner to the center of the square is LLL 22 2 F H G I K J + F H G I K J = . The distance from each positive charge to Q is then z L 2 2 2 + . Each positive charge exerts a force directed +q +q +q +q z –Q L /2 L /2 x FIG. P23.66 along the line joining q and Q , of magnitude kQq zL e 2 + . The line of force makes an angle with the z -axis whose cosine is z 2 + The four charges together exert forces whose x and y components add to zero, while the z -components add to Fk =− + 4 2 32 kQqz e ej ± (b) For >> , the magnitude of this force is F L L zm a z e e z F H G I K J = 4 2 42 2 3 a f Therefore, the object’s vertical acceleration is of the form az z ω 2 with 2 33 128 == af mL mL e e . Since the acceleration of the object is always oppositely directed to its excursion from equilibrium and in magnitude proportional to it, the object will execute simple harmonic motion with a period given by
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## This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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