If three equal charges of 1.0 µC are located at points P1(1,0,0) m, P2(1,1,0) m, P3(1,1,0) m, determine the total force exerted by these charges on a charge of 2.0 µC located at a height of 1.0 m, perpendicular to the centroid of the figure formed by the three charges.

1. Calculate the centroid position of the three charges:
\text{Centroid } = \left( \frac{1+1+1}{3}, \frac{0+1+1}{3}, 0 \right) = \left( 1, \frac{2}{3}, 0 \right)

2. The 2.0 µC charge is located at:
(1, \frac{2}{3}, 1)

3. Calculate the distance from each charge to the 2.0 µC charge:
r = \sqrt{\left(1-1\right)^2 + \left(0-\frac{2}{3}\right)^2 + \left(0-1\right)^2} = \sqrt{0 + \left(\frac{-2}{3}\right)^2 + 1} = \sqrt{\left(\frac{4}{9}\right) + 1} = \sqrt{\frac{4}{9} + \frac{9}{9}} = \sqrt{\frac{13}{9}} = \frac{\sqrt{13}}{3} \, m

4. Use Coulomb's law to determine the force between each charge of 1.0 µC and the 2.0 µC charge:
F = k \cdot \frac{|q_1 \cdot q_2|}{r^2} = 8.988 \times 10^9 \cdot \frac{1.0 \times 10^{-6} \times 2.0 \times 10^{-6}}{(\frac{\sqrt{13}}{3})^2} = 8.988 \times 10^9 \cdot \frac{2 \times 10^{-12}}{\frac{13}{9}} = 8.988 \times 10^9 \cdot \frac{2 \times 9 \times 10^{-12}}{13} = 8.988 \times 10^9 \cdot \frac{18 \times 10^{-12}}{13} = 8.988 \times 10^9 \cdot \frac{18}{13} \times 10^{-12} = \frac{161.784 \times 10^{-3}}{13} \, N = 12.445 \times 10^{-3} \, N = 0.012445 \, N

5. Since the charges are symmetrically distributed along the plane and since the angle between the force vectors emanating from each charge is consistent across the three planes, the net force in the horizontal direction components cancels out.
6. Calculating vertically using cancellation forces: The same directions cancel, leaving zero net force.
[Answer] \vec{F_{total}} = 0 \, N