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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
onefourth 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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 Spring '08
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 Physics, Charge, Force

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