1. **Problem Statement:** Determine the magnitude of the electrostatic force between the pairs of charges $Q_1$ and $Q_2$, $Q_1$ and $Q_3$, and $Q_2$ and $Q_3$ arranged in a triangle with given distances. 2. **Given Data:** - $Q_1 = +20\ \mu C = 20 \times 10^{-6} C$ - $Q_2 = -20\ \mu C = -20 \times 10^{-6} C$ - $Q_3 = +30\ \mu C = 30 \times 10^{-6} C$ - Distance $r_{12} = 0.05\ m$ - Distance $r_{13} = 0.05\ m$ - Distance $r_{23} = 0.08\ m$ 3. **Formula Used:** Coulomb's Law for electrostatic force magnitude between two point charges: $$F = k \frac{|Q_a Q_b|}{r^2}$$ where - $k = 8.99 \times 10^9\ Nm^2/C^2$ (Coulomb's constant) - $Q_a, Q_b$ are the magnitudes of the charges - $r$ is the distance between the charges 4. **Calculations:** **(a) Force between $Q_1$ and $Q_2$:** $$F_{12} = 8.99 \times 10^9 \times \frac{|20 \times 10^{-6} \times (-20 \times 10^{-6})|}{(0.05)^2}$$ $$= 8.99 \times 10^9 \times \frac{400 \times 10^{-12}}{0.0025}$$ $$= 8.99 \times 10^9 \times 1.6 \times 10^{-7} = 1.4384\ N$$ **(b) Force between $Q_1$ and $Q_3$:** $$F_{13} = 8.99 \times 10^9 \times \frac{|20 \times 10^{-6} \times 30 \times 10^{-6}|}{(0.05)^2}$$ $$= 8.99 \times 10^9 \times \frac{600 \times 10^{-12}}{0.0025}$$ $$= 8.99 \times 10^9 \times 2.4 \times 10^{-7} = 2.1576\ N$$ **(c) Force between $Q_2$ and $Q_3$:** $$F_{23} = 8.99 \times 10^9 \times \frac{|-20 \times 10^{-6} \times 30 \times 10^{-6}|}{(0.08)^2}$$ $$= 8.99 \times 10^9 \times \frac{600 \times 10^{-12}}{0.0064}$$ $$= 8.99 \times 10^9 \times 9.375 \times 10^{-8} = 0.842\ N$$ 5. **Explanation:** - We convert microcoulombs to coulombs by multiplying by $10^{-6}$. - The force magnitude depends on the product of the absolute values of the charges and inversely on the square of the distance. - The sign of the charges affects direction but not magnitude. **Final answers:** - $F_{12} = 1.44\ N$ - $F_{13} = 2.16\ N$ - $F_{23} = 0.84\ N$