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Unformatted text preview: 20. If is the angle between the force and the xaxis, then cos = x . x + d2
2 We note that, due to the symmetry in the problem, there is no y component to the net force on the third particle. Thus, F represents the magnitude of force exerted by q1 or q2 on q3. Let e = +1.60 1019 C, then q1 = q2 = +2e and q3 = 4.0e and we have Fnet = 2F cos = 2(2e)(4e) 4o (x2 + d2) 4e2 x x = . o (x2 + d2 )3/2 x2 + d2 (a) To find where the force is at an extremum, we can set the derivative of this expression equal to zero and solve for x, but it is good in any case to graph the function for a fuller understanding of its behavior and as a quick way to see whether an extremum point is a maximum or a miminum. In this way, we find that the value coming from the derivative procedure is a maximum (and will be presented in part (b)) and that the minimum is found at the lower limit of the interval. Thus, the net force is found to be zero at x = 0, which is the smallest value of the net force in the interval 5.0 m x 0. (b) The maximum is found to be at x = d/ 2 or roughly 12 cm. (c) The value of the net force at x = 0 is Fnet = 0. (d) The value of the net force at x = d/ 2 is Fnet = 4.9 1026 N. ...
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This note was uploaded on 09/06/2009 for the course PHYSICS 14358 taught by Professor Bolton during the Spring '09 term at Kansas State University.
 Spring '09
 Bolton
 Force

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