Charges −q, +q, −q at corners of equilateral triangle (side a). Find net force on +q at centroid O.

Distance from centroid to vertex: r = a/√3

Force magnitude from each corner on +q at O:

F = kq²/r² = kq²/(a²/3) = 3kq²/a²

Directions (place +q corner at top, −q corners at bottom-left and bottom-right):

From +q at top: repels downward (0, −F)

From −q at bottom-L: attracts toward B-L (−F√3/2, −F/2)

From −q at bottom-R: attracts toward B-R (+F√3/2, −F/2)

Sum of components:

Ex = 0 − F√3/2 + F√3/2 = 0

Ey = −F − F/2 − F/2 = −2F

F_net = 2F = 6kq²/a²

Answer: F_net = 6kq²/a², directed away from the +q corner (toward midpoint of opposite side)

Symmetry only guarantees x-cancellation here. The y-components all add never assume the force is zero without resolving components.