1. **State the problem:** We have a capacitor with capacitance $C = 1.0$ F and plate separation $d = 1.00$ mm $= 1.00 \times 10^{-3}$ m. We need to find the area $A$ of the plates. 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, approximately $8.85 \times 10^{-12}$ F/m. 3. **Rearrange the formula to solve for area $A$:** $$ A = \frac{C d}{\epsilon_0} $$ 4. **Substitute the known values:** $$ A = \frac{1.0 \times 1.00 \times 10^{-3}}{8.85 \times 10^{-12}} $$ 5. **Calculate the area:** $$ A = \frac{1.00 \times 10^{-3}}{8.85 \times 10^{-12}} = 1.13 \times 10^{8} \text{ m}^2 $$ 6. **Interpretation:** The area of the plates is approximately $1.13 \times 10^{8}$ square meters, which is very large, indicating that a 1.0 F capacitor with 1 mm separation requires very large plates. **Final answer:** $$ \boxed{A = 1.13 \times 10^{8} \text{ m}^2} $$