1. **Problem:** Determine the magnitude of the electrostatic force between the pairs of charges Q₁ and Q₂, Q₁ and Q₃, and Q₂ and Q₃ arranged in a triangle. 2. **Formula:** Coulomb's Law states the electrostatic force between two point charges is $$F = k \frac{|Q_1 Q_2|}{r^2}$$ where $k = 8.99 \times 10^9 \text{ Nm}^2/\text{C}^2$, $Q_1$ and $Q_2$ are the charges, and $r$ is the distance between them. 3. **Given:** - $Q_1 = +20 \mu C = 20 \times 10^{-6} C$ - $Q_2 = -20 \mu C = -20 \times 10^{-6} C$ - $Q_3 = +30 \mu C = 30 \times 10^{-6} C$ - $r_{12} = 0.05 m$ - $r_{13} = 0.05 m$ - $r_{23} = 0.08 m$ 4. **Calculate each force:** - Between $Q_1$ and $Q_2$: $$F_{12} = 8.99 \times 10^9 \times \frac{|20 \times 10^{-6} \times (-20 \times 10^{-6})|}{(0.05)^2}$$ $$= 8.99 \times 10^9 \times \frac{400 \times 10^{-12}}{0.0025} = 8.99 \times 10^9 \times 1.6 \times 10^{-7} = 1.4384 \times 10^3 \text{ N}$$ - Between $Q_1$ and $Q_3$: $$F_{13} = 8.99 \times 10^9 \times \frac{|20 \times 10^{-6} \times 30 \times 10^{-6}|}{(0.05)^2}$$ $$= 8.99 \times 10^9 \times \frac{600 \times 10^{-12}}{0.0025} = 8.99 \times 10^9 \times 2.4 \times 10^{-7} = 2.1576 \times 10^3 \text{ N}$$ - Between $Q_2$ and $Q_3$: $$F_{23} = 8.99 \times 10^9 \times \frac{|(-20 \times 10^{-6}) \times 30 \times 10^{-6}|}{(0.08)^2}$$ $$= 8.99 \times 10^9 \times \frac{600 \times 10^{-12}}{0.0064} = 8.99 \times 10^9 \times 9.375 \times 10^{-8} = 842.8 \text{ N}$$ 5. **Final answers:** - $F_{12} = 1438.4$ N - $F_{13} = 2157.6$ N - $F_{23} = 842.8$ N All forces are magnitudes; directions depend on charge signs.