1. **Problem 4:** Find values of $x$ where the magnitude of the electrostatic force on particle 3 from particles 1 and 2 is minimum and maximum, and the corresponding magnitudes. 2. Given: $q_1 = q_2 = +3.20 \times 10^{-19}$ C, $q_3 = +6.40 \times 10^{-19}$ C, $d=0.17$ m, particle 3 moves from $x=0$ to $x=5$ m. 3. The force between charges is given by Coulomb's law: $$F = k_e \frac{|q_a q_b|}{r^2}$$ where $k_e = 8.99 \times 10^9$ Nm$^2$/C$^2$. 4. Distance from particle 3 at $(x,0)$ to each particle 1 and 2 at $(0,\pm d)$ is: $$r = \sqrt{x^2 + d^2}$$ 5. Forces from particles 1 and 2 on 3 have equal magnitude due to symmetry, directions along lines from (0,d) and (0,-d) to (x,0). 6. Each force magnitude: $$F = k_e \frac{q_1 q_3}{(x^2 + d^2)}$$ 7. Components along y-axis cancel; x-components add. The x-component of each force is: $$F_x = F \frac{x}{\sqrt{x^2 + d^2}} = k_e \frac{q_1 q_3 x}{(x^2 + d^2)^{3/2}}$$ 8. Total force magnitude on particle 3 is twice this: $$F_{total} = 2 F_x = 2 k_e \frac{q_1 q_3 x}{(x^2 + d^2)^{3/2}}$$ 9. To find extrema, differentiate $F_{total}$ w.r.t. $x$: $$\frac{dF}{dx} = 2 k_e q_1 q_3 \frac{(x^2 + d^2)^{3/2} - x \cdot \frac{3}{2} (x^2 + d^2)^{1/2} 2x}{(x^2 + d^2)^3}$$ Simplify numerator: $$(x^2 + d^2)^{3/2} - 3 x^2 (x^2 + d^2)^{1/2} = (x^2 + d^2)^{1/2} (x^2 + d^2 - 3x^2) = (x^2 + d^2)^{1/2} (d^2 - 2x^2)$$ Setting derivative zero: $$d^2 - 2x^2 = 0 \implies x = \frac{d}{\sqrt{2}} = \frac{0.17}{\sqrt{2}} \approx 0.12 \text{ m}$$ 10. Check limits: - At $x=0$, $F_{total}=0$ (force zero) - At $x \to \infty$, $F_{total} \to 0$ 11. Testing a value greater than $0.12$ m shows force is positive but decreases past this maximum point. 12. **Answers:** (i) Minimum force magnitude at $x=0$ m, (ii) Maximum force magnitude at $x\approx 0.12$ m, (iii) Minimum magnitude $F_{min}=0$ N, (iv) Maximum magnitude: $$F_{max} = 2 k_e \frac{q_1 q_3 x}{(x^2 + d^2)^{3/2}} = 2 \times 8.99 \times 10^9 \times \frac{(3.20 \times 10^{-19})(6.40 \times 10^{-19}) \times 0.12}{(0.12^2 + 0.17^2)^{3/2}}$$ Calculate denominator: $$0.12^2 + 0.17^2 = 0.0144 + 0.0289 = 0.0433$$ $$0.0433^{3/2} = 0.0433 \times \sqrt{0.0433} = 0.0433 \times 0.208 = 0.00901$$ Calculate numerator: $$(3.20 \times 10^{-19})(6.40 \times 10^{-19}) \times 0.12 = 2.4576 \times 10^{-37}$$ Calculate $F_{max}$: $$2 \times 8.99 \times 10^9 \times \frac{2.4576 \times 10^{-37}}{0.00901} = 1.798 \times 10^{10} \times 2.727 \times 10^{-35} = 4.90 \times 10^{-25} \text{ N}$$ 13. **Problem 5:** Number of charges in 1 kg of MgCl$_2$. 14. Molar mass of MgCl$_2$: $$M = 24.305 + 2 \times 35.453 = 95.211 \text{ g/mol}$$ 15. Moles in 1 kg = 1000 g: $$n = \frac{1000}{95.211} \approx 10.5 \text{ mol}$$ 16. Number of formula units: $$N = n \times N_A = 10.5 \times 6.022 \times 10^{23} = 6.32 \times 10^{24}$$ 17. Each formula unit has 3 charges (1 Mg$^{2+}$ and 2 Cl$^-$). 18. Total charges: $$Q = 3 \times N = 3 \times 6.32 \times 10^{24} = 1.896 \times 10^{25} \text{ charges}$$ --- **Final answers:** (i) $x=0$ m (minimum force), (ii) $x \approx 0.12$ m (maximum force), (iii) $F_{min}=0$ N, (iv) $F_{max} = 4.90 \times 10^{-25}$ N, (v) Number of charges in 1 kg MgCl$_2$ is $1.896 \times 10^{25}$.