Six identical positive charges q are placed at the six vertices A, B, C, D, E and F of a regular hexagon of circumradius R. The magnitude of the net electrostatic force on the charge at vertex A is F. First, the charge at A is moved to the centre; let the force magnitude on it there be F₁. Then the charge at E is also moved to the centre, so that the charge present at the centre becomes 2q; let the force magnitude on this central charge due to the charges remaining at the vertices be F₂. Which ordered pair (F₁,F₂) is correct?
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Question

Six identical positive charges q are placed at the six vertices A, B, C, D, E and F of a regular hexagon of circumradius R. The magnitude of the net electrostatic force on the charge at vertex A is F. First, the charge at A is moved to the centre; let the force magnitude on it there be F₁. Then the charge at E is also moved to the centre, so that the charge present at the centre becomes 2q; let the force magnitude on this central charge due to the charges remaining at the vertices be F₂. Which ordered pair (F₁,F₂) is correct?

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G1 Physics · Accepted Answer

For a regular hexagon, the force on the original charge at A is radial. Contributions from the two adjacent charges give √3(kq²/R²), the two next-nearest charges give (1/3)(kq²/R²), and the opposite charge gives (1/4)(kq²/R²). Hence F = (√3+7/12)kq²/R² = [(12√3+7)/12](kq²/R²). After moving A to the centre, the five remaining vertex charges produce the same field magnitude as the missing A-vertex charge would have produced, so F₁=kq²/R²=12F/(12√3+7). After E is also removed from the vertices, th…