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30. (a) The two charges are q = αQ (where α is a pure number presumably less than 1 and greater than zero) and Q − q =(1 − α ) Q . Thus, Eq. 22-4 gives F = 1 4 πε 0 ( αQ )((1 − α ) Q ) d 2 = Q 2 α (1 − α ) 4 πε 0 d 2 . (b) The graph below, of F versus α , has been scaled so that the maximum is 1. In actuality, the maximum value of the force is F max = Q 2 / 16 πε 0 d 2 . 0 1 Force 1 alpha (c) It is clear that α = 1 2 gives the maximum value of F . (d) Seeking the half-height points on the graph is diﬃcult without grid lines or some of the special
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