Worked example: Three charges at the corners of an equilateral triangle of side 0.1 m: q1 = +2 μC, q2 = +2 μC, q3 = −3 μC. Find the system's potential energy. Step 1: List the three pairs: (1,2): k·q1·q2/r = (8.988 × 10⁹)(2 × 10⁻⁶)(2 × 10⁻⁶)/0.1 = +0.360 J. Step 2: (1,3): (8.988 × 10⁹)(2 × 10⁻⁶)(−3 × 10⁻⁶)/0.1 = −0.539 J. Step 3: (2,3): same as (1,3) = −0.539 J. Step 4: Sum: U = 0.360 − 0.539 − 0.539 = −0.718 J. Negative total: the system is bound — you would need to supply +0.718 J to pull all three apart to infinity. Step 5 (check via assembly): bring q1 (0 work), q2 (work = q2·k·q1/r = +0.360 J), q3 into V of q1+q2: V = k(2−... compute: V at corner 3 = k·q1/r + k·q2/r = 2 × 179.8 = 359.5... wait: k·2μC/0.1 = 179.8 V each, so V = 359.5 V; work = q3·V = (−3 × 10⁻⁶)(359.5) = −1.079 J. Total = 0.360 − 1.079 = −0.718 J. Matches.