1. **State the problem:** We need to find the area of the plates of a capacitor given the plate separation $d = 2.0 \times 10^{-3}$ m and the capacitance $C = 1$ F. 2. **Formula used:** The capacitance of a parallel plate capacitor is given by $$ C = \epsilon_0 \frac{A}{d} $$ where $\epsilon_0$ is the permittivity of free space, $A$ is the area of the plates, and $d$ is the separation between the plates. 3. **Known values:** - $C = 1$ F - $d = 2.0 \times 10^{-3}$ m - $\epsilon_0 = 8.854 \times 10^{-12}$ F/m (permittivity of free space) 4. **Rearrange the formula to solve for $A$:** $$ A = \frac{C \times d}{\epsilon_0} $$ 5. **Substitute the values:** $$ A = \frac{1 \times 2.0 \times 10^{-3}}{8.854 \times 10^{-12}} $$ 6. **Calculate the area:** $$ A = \frac{2.0 \times 10^{-3}}{8.854 \times 10^{-12}} = 2.26 \times 10^{8} \text{ m}^2 $$ 7. **Interpretation:** The area of the plates must be approximately $2.26 \times 10^{8}$ square meters to achieve a capacitance of 1 F with the given plate separation. **Final answer:** $$ \boxed{A = 2.26 \times 10^{8} \text{ m}^2} $$