ch21-p022 - 22. We note that the problem is examining the...

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Unformatted text preview: 22. We note that the problem is examining the force on charge A, so that the respective distances (involved in the Coulomb force expressions) between B and A, and between C and A, do not change as particle B is moved along its circular path. We focus on the endpoints (θ = 0º and 180º) of each graph, since they represent cases where the forces (on A) due to B and C are either parallel or antiparallel (yielding maximum or minimum force magnitudes, respectively). We note, too, that since Coulomb’s law is inversely proportional to r² then the (if, say, the charges were all the same) force due to C would be one-fourth as big as that due to B (since C is twice as far away from A). The charges, it turns out, are not the same, so there is also a factor of the charge ratio ξ (the charge of C divided by the charge of B), as well as the aforementioned ¼ factor. That is, the force exerted by C is, by Coulomb’s law equal to ±¼ξ multiplied by the force exerted by B. (a) The maximum force is 2F0 and occurs when θ = 180º (B is to the left of A, while C is the right of A). We choose the minus sign and write 2 F0 = (1 − ¼ξ) F0 ⇒ ξ=–4. One way to think of the minus sign choice is cos(180º) = –1. This is certainly consistent with the minimum force ratio (zero) at θ = 0º since that would also imply 0 = 1 + ¼ξ ⇒ ξ=–4. (b) The ratio of maximum to minimum forces is 1.25/0.75 = 5/3 in this case, which implies 1 + ¼ξ 5 = ⇒ ξ = 16 . 3 1 − ¼ξ Of course, this could also be figured as illustrated in part (a), looking at the maximum force ratio by itself and solving, or looking at the minimum force ratio (¾) at θ = 180º and solving for ξ. ...
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