Worked example: Three charges sit at the corners of an equilateral triangle of side 0.1 m: qA = +2 μC at the bottom-left, qB = +2 μC at the bottom-right, qC = −2 μC at the top. Find the net force on qC. Step 1: Force on qC due to qA: magnitude F = k·|qA·qC|/r² = (8.988 × 10⁹)(2 × 10⁻⁶)(2 × 10⁻⁶)/0.01 = 3.60 N, attractive, pointing from C down-left toward A (60° below the negative x-axis... precisely, along the side CA). Step 2: By symmetry, force due to qB has the same magnitude 3.60 N, pointing down-right toward B. Step 3: Resolve: each makes 60° with the vertical. Horizontal components (±3.60·sin60°) cancel; vertical components add: Fy = −2 × 3.60·cos60° = −3.60 N (downward). Step 4: Net force on qC is 3.60 N straight down. The x-components cancelled by symmetry — superposition plus symmetry halves the work.