1. **Problem Statement:** We have two parallel, non-conducting plates in air with surface charge densities $\sigma_1 = -5.0 \times 10^{-6}$ C/m$^2$ (plate 1) and $\sigma_2 = -6.0 \times 10^{-6}$ C/m$^2$ (plate 2). We want to find the total electric field at point b, which is 11 cm to the left of plate 2. 2. **Formula and Concepts:** The electric field due to an infinite charged plane with surface charge density $\sigma$ is given by: $$E = \frac{|\sigma|}{2\epsilon_0}$$ where $\epsilon_0 = 8.85 \times 10^{-12}$ C$^2$/N·m$^2$ is the permittivity of free space. The direction of the field from a negatively charged plate points toward the plate. 3. **Calculate the magnitude of the electric fields:** $$E_1 = \frac{|\sigma_1|}{2\epsilon_0} = \frac{5.0 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}} = 2.82 \times 10^{5} \text{ N/C}$$ $$E_2 = \frac{|\sigma_2|}{2\epsilon_0} = \frac{6.0 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}} = 3.39 \times 10^{5} \text{ N/C}$$ 4. **Determine directions at point b:** - Plate 2 is to the right of point b, and since it is negatively charged, its field points toward plate 2, i.e., to the right (positive x-direction). - Plate 1 is further right of plate 2, so from point b, plate 1 is even further right. Its field also points toward plate 1 (negative charge), so also to the right. 5. **Calculate total electric field at point b:** Since both fields point in the same direction (to the right), total field magnitude is: $$E_{total} = E_1 + E_2 = 2.82 \times 10^{5} + 3.39 \times 10^{5} = 6.21 \times 10^{5} \text{ N/C}$$ 6. **Final answer:** The total electric field at point b is: $$\boxed{6.21 \times 10^{5} \text{ N/C, directed to the right}}$$