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A total charge Q is split into two parts Q1 and Q2 (so Q1 + Q2 = Q) placed a fixed distance R apart. For what division is the Coulomb force of repulsion between them maximum?

1. Q1 = Q/2, Q2 = Q/2
2. Q2 = Q/4, Q1 = 3Q/4
3. Q2 = Q/3, Q1 = 2Q/3
4. Q2 = Q/5, Q1 = 4Q/5

Correct answer: Q1 = Q/2, Q2 = Q/2

Solution

With R fixed, F is proportional to the product Q1*Q2 under the constraint Q1 + Q2 = Q. The product of two numbers with a fixed sum is maximized when the numbers are equal. Hence Q1 = Q2 = Q/2 gives maximum repulsive force.

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