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(C) Let the total number of charges be $N$. We divide these into two groups of $q$ and $N-q$ charges, where each charge is $Q$. The force between the

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$\frac{N^2}{4(N-1)}$

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(C) Let the total number of charges be $N$. We divide these into two groups of $q$ and $N-q$ charges, where each charge is $Q$. The force between the two groups is $F = k \cdot (qQ) \cdot ((N-q)Q) / r^2 = (kQ^2/r^2) \cdot q(N-q)$.To maximize the force, we need to maximize $q(N-q)$. This occurs when $q = N/2$, giving $q(N-q) = N^2/4$.To minimize the force, we place the minimum possible number of charges in one group, which is $q = 1$. This gives $q(N-q) = 1(N-1) = N-1$.The ratio of maximum force to minimum force is $(N^2/4) / (N-1) = N^2 / (4(N-1))$.Thus, the correct option is $C$.

$N$ point charges are distributed into two groups and separated by a fixed distance. Then the ratio of maximum to minimum forces between the two groups is ($N$ is even and greater than $2$).

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