1. **State the problem:** We have three point charges in the x-y plane: - Charge 1: $q_1 = 11.2$ mC at $(-9.4 \text{ mm}, 0)$ - Charge 2: $q_2 = -19.8$ $\mu$C at the origin $(0,0)$ - Charge 3: $q_3 = -15.4$ mC at $(6.8 \text{ mm}, 3.5 \text{ mm})$ We want to find: (a) The individual forces on charge 2 due to charges 1 and 3. (b) The net force on charge 2 and illustrate the free-body diagram. 2. **Formula and rules:** The electrostatic force between two point charges is given by Coulomb's law: $$F = k_e \frac{|q_i q_j|}{r^2}$$ where $k_e = 8.99 \times 10^9$ N·m$^2$/C$^2$, $q_i$ and $q_j$ are charges, and $r$ is the distance between them. The force direction is along the line connecting the charges, repulsive if charges have the same sign, attractive if opposite. 3. **Convert units:** - $q_1 = 11.2$ mC = $11.2 \times 10^{-3}$ C - $q_2 = -19.8$ $\mu$C = $-19.8 \times 10^{-6}$ C - $q_3 = -15.4$ mC = $-15.4 \times 10^{-3}$ C - Distances in meters: - $r_{12} = 9.4$ mm = $9.4 \times 10^{-3}$ m - $r_{23} = \sqrt{(6.8 \times 10^{-3})^2 + (3.5 \times 10^{-3})^2} = \sqrt{(6.8)^2 + (3.5)^2} \times 10^{-3} = \sqrt{46.24 + 12.25} \times 10^{-3} = \sqrt{58.49} \times 10^{-3} = 7.65 \times 10^{-3}$ m 4. **Calculate force magnitude from charge 1 on charge 2:** $$F_{12} = k_e \frac{|q_1 q_2|}{r_{12}^2} = 8.99 \times 10^9 \times \frac{11.2 \times 10^{-3} \times 19.8 \times 10^{-6}}{(9.4 \times 10^{-3})^2}$$ Calculate denominator: $$(9.4 \times 10^{-3})^2 = 8.836 \times 10^{-5}$$ Calculate numerator: $$11.2 \times 10^{-3} \times 19.8 \times 10^{-6} = 2.2176 \times 10^{-7}$$ So, $$F_{12} = 8.99 \times 10^9 \times \frac{2.2176 \times 10^{-7}}{8.836 \times 10^{-5}} = 8.99 \times 10^9 \times 0.00251 = 22570.5 \text{ N}$$ 5. **Direction of $F_{12}$:** Charge 1 is positive, charge 2 is negative, so force on 2 due to 1 is attractive, pointing from 2 to 1 (left along negative x-axis). Vector form: $$\vec{F}_{12} = (-F_{12}, 0) = (-22570.5, 0) \text{ N}$$ 6. **Calculate force magnitude from charge 3 on charge 2:** $$F_{32} = k_e \frac{|q_3 q_2|}{r_{23}^2} = 8.99 \times 10^9 \times \frac{15.4 \times 10^{-3} \times 19.8 \times 10^{-6}}{(7.65 \times 10^{-3})^2}$$ Calculate denominator: $$(7.65 \times 10^{-3})^2 = 5.85 \times 10^{-5}$$ Calculate numerator: $$15.4 \times 10^{-3} \times 19.8 \times 10^{-6} = 3.0492 \times 10^{-7}$$ So, $$F_{32} = 8.99 \times 10^9 \times \frac{3.0492 \times 10^{-7}}{5.85 \times 10^{-5}} = 8.99 \times 10^9 \times 0.00521 = 46839.3 \text{ N}$$ 7. **Direction of $F_{32}$:** Charge 3 is negative, charge 2 is negative, so force is repulsive, pointing from 3 to 2. Vector from 2 to 3 is $(6.8 \times 10^{-3}, 3.5 \times 10^{-3})$, so vector from 3 to 2 is $(-6.8 \times 10^{-3}, -3.5 \times 10^{-3})$. Unit vector: $$\hat{r}_{32} = \frac{(-6.8, -3.5)}{7.65} = (-0.8895, -0.4575)$$ Force vector: $$\vec{F}_{32} = F_{32} \times \hat{r}_{32} = 46839.3 \times (-0.8895, -0.4575) = (-41656.5, -21427.1) \text{ N}$$ 8. **Net force on charge 2:** $$\vec{F}_{net} = \vec{F}_{12} + \vec{F}_{32} = (-22570.5, 0) + (-41656.5, -21427.1) = (-64227, -21427.1) \text{ N}$$ Magnitude: $$|\vec{F}_{net}| = \sqrt{(-64227)^2 + (-21427.1)^2} = \sqrt{4.126 \times 10^9 + 4.59 \times 10^8} = \sqrt{4.585 \times 10^9} = 67699.3 \text{ N}$$ Direction angle (from x-axis): $$\theta = \tan^{-1} \left( \frac{-21427.1}{-64227} \right) = \tan^{-1}(0.3337) = 18.43^\circ$$ Since both components are negative, force points in the third quadrant, so angle is $180^\circ + 18.43^\circ = 198.43^\circ$ from positive x-axis. --- **Final answers:** - Force on charge 2 by charge 1: $$\vec{F}_{12} = (-22570.5, 0) \text{ N}$$ - Force on charge 2 by charge 3: $$\vec{F}_{32} = (-41656.5, -21427.1) \text{ N}$$ - Net force on charge 2: $$\vec{F}_{net} = (-64227, -21427.1) \text{ N}, \quad |\vec{F}_{net}| = 67699.3 \text{ N}, \quad \theta = 198.43^\circ$$