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Three identical charges +Q are fixed at the vertices of an equilateral triangle of side a. A particle of charge -q and mass m sits on the axis through the centroid, perpendicular to the triangle's plane, at distance x from the centre. What is the magnitude of the net force on this charge?

1. (1/(4*pi*eps0)) * (9*sqrt(3)*Q*q*x)/(3x² + a²)^(3/2)
2. (1/(4*pi*eps0)) * (27*sqrt(3)*Q*q*x)/(3x² + a²)^(3/2)
3. (1/(4*pi*eps0)) * (2*sqrt(2)*Q*q*x)/(2x² + a²)^(3/2)
4. (1/(4*pi*eps0)) * (4*sqrt(2)*Q*q*x)/(2x² + a²)^(3/2)

Correct answer: (1/(4*pi*eps0)) * (9*sqrt(3)*Q*q*x)/(3x² + a²)^(3/2)

Solution

Centroid-to-vertex distance = a/sqrt(3), so each source charge is r = sqrt(x² + a²/3) away. The force from one charge has axial component F1*(x/r). The three perpendicular components cancel by symmetry. Net = 3 * kQq * x / r³ = 3kQqx / (x² + a²/3)^(3/2) = 9*sqrt(3)*kQqx / (3x² + a²)^(3/2).

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