1. **Problem Statement:** Determine the electric field at the origin due to two point charges: +26 μC at point A (0, 8.0 cm) and -26 μC at point B (8.0 cm, 8.0 cm). 2. **Formula:** The electric field due to a point charge is given by: $$\vec{E} = k \frac{q}{r^2} \hat{r}$$ where $k = 8.99 \times 10^9$ N·m²/C², $q$ is the charge, $r$ is the distance from the charge to the point of interest, and $\hat{r}$ is the unit vector from the charge to the point. 3. **Calculate distances:** - Distance from A to origin: $r_A = 8.0$ cm = 0.08 m - Distance from B to origin: $r_B = \sqrt{(0.08)^2 + (0.08)^2} = \sqrt{0.0128} = 0.113$ m 4. **Calculate electric field magnitudes:** - $E_A = k \frac{|q|}{r_A^2} = 8.99 \times 10^9 \times \frac{26 \times 10^{-6}}{(0.08)^2} = 3.65 \times 10^7$ N/C - $E_B = k \frac{|q|}{r_B^2} = 8.99 \times 10^9 \times \frac{26 \times 10^{-6}}{(0.113)^2} = 1.83 \times 10^7$ N/C 5. **Determine directions:** - $\vec{E}_A$ points away from positive charge A, so downward along negative y-axis: $\vec{E}_A = 0 \hat{i} - 3.65 \times 10^7 \hat{j}$ N/C - $\vec{E}_B$ points toward negative charge B, so from origin to B: vector from origin to B is $\vec{r}_B = 0.08 \hat{i} + 0.08 \hat{j}$, unit vector: $$\hat{r}_B = \frac{0.08 \hat{i} + 0.08 \hat{j}}{0.113} = 0.707 \hat{i} + 0.707 \hat{j}$$ Electric field at origin due to B is toward B (since charge is negative), so: $$\vec{E}_B = E_B \hat{r}_B = 1.83 \times 10^7 (0.707 \hat{i} + 0.707 \hat{j}) = 1.29 \times 10^7 \hat{i} + 1.29 \times 10^7 \hat{j}$$ N/C 6. **Sum the electric fields:** $$\vec{E} = \vec{E}_A + \vec{E}_B = (0 + 1.29 \times 10^7) \hat{i} + (-3.65 \times 10^7 + 1.29 \times 10^7) \hat{j} = 1.29 \times 10^7 \hat{i} - 2.36 \times 10^7 \hat{j}$$ N/C 7. **Calculate magnitude and direction:** $$E = \sqrt{(1.29 \times 10^7)^2 + (-2.36 \times 10^7)^2} = \sqrt{1.66 \times 10^{14} + 5.57 \times 10^{14}} = \sqrt{7.23 \times 10^{14}} = 2.69 \times 10^7 \text{ N/C}$$ Angle $\theta$ with positive x-axis: $$\theta = \tan^{-1} \left( \frac{-2.36 \times 10^7}{1.29 \times 10^7} \right) = \tan^{-1}(-1.83) = -61.5^\circ$$ This means $61.5^\circ$ below the positive x-axis. **Final answer:** The electric field at the origin is $$\boxed{\vec{E} = (1.29 \times 10^7) \hat{i} - (2.36 \times 10^7) \hat{j} \text{ N/C}, \quad E = 2.69 \times 10^7 \text{ N/C}, \quad \theta = 61.5^\circ \text{ below } +x}$$