ch21-p021

# ch21-p021 - procedure is a maximum (and will be presented...

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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 q 1 or q 2 on q 3 . Let e = +1.60 × 10 19 C, then q 1 = q 2 = +2 e and q 3 = 4.0 e and we have F net = 2 F cos θ = 2(2 e )(4 e ) 4 πε o ( x 2 + d 2 ) x x 2 + d 2 = 4 e 2 x πε o ( x 2 + d 2 ) 3/2 . (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
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Unformatted text preview: 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 F net = 0. (d) The value of the net force at x = d / 2 is F net = 4.9 10 26 N. 21. If is the angle between the force and the x-axis, then cos = x x 2 + d 2 ....
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