ch21-p034 - 34. Let d be the vertical distance from the...

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From symmetry, we see that there is no net force in the vertical direction on q 2 = – e sitting at a distance R to the left of the coordinate origin. We note that the net x force caused by q 3 and q 4 on the y axis will have a magnitude equal to 3 22 2 00 0 2c o s o s o s 44 ( / c o s ) 4 qe qe qe rR R θ πε == . Consequently, to achieve a zero net force along the x axis, the above expression must equal the magnitude of the repulsive force exerted on q 2 by q 1 = – e . Thus, 32 3 o s 2 c o s qe e e q RR π επ ε =⇒ = . Below we plot q/e as a function of the angle (in degrees): The graph suggests that q/e < 5 for < 60º, roughly. We can be more precise by solving the above equation. The requirement that q 5 e leads to 31 / 3 1 5c o s 2cos (10) e e ≤⇒ which yields 62.34º. The problem asks for “physically possible values,” and it is reasonable to suppose that only positive-integer-multiple values of e are allowed for q . If we let q = ne , for n = 1 … 5, then N will be found by taking the inverse cosine of the cube root of (1/2 n ). 34. Let d be the vertical distance from the coordinate origin to
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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ch21-p034 - 34. Let d be the vertical distance from the...

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