1. **Problem Statement:** Find the area of the parallel plates of a 1.0-F capacitor separated by 1.0 mm, given the permittivity of free space $\varepsilon_0$ in F/m. 2. **General Formula:** The capacitance $C$ of a parallel plate capacitor is given by: $$ C = \varepsilon_0 \frac{A}{d} $$ where: - $C$ is the capacitance in farads (F), - $\varepsilon_0$ is the permittivity of free space (F/m), - $A$ is the area of the plates (m$^2$), - $d$ is the separation between the plates (m). 3. **Derived Formula:** Rearranging to solve for area $A$: $$ A = \frac{C \times d}{\varepsilon_0} $$ 4. **Detailed Solution:** Given: - $C = 1.0$ F - $d = 1.0$ mm = $1.0 \times 10^{-3}$ m - $\varepsilon_0 = 8.85 \times 10^{-12}$ F/m (standard value) Substitute values: $$ A = \frac{1.0 \times 1.0 \times 10^{-3}}{8.85 \times 10^{-12}} $$ Calculate numerator and denominator: $$ A = \frac{1.0 \times 10^{-3}}{8.85 \times 10^{-12}} $$ Simplify: $$ A = 1.0 \times 10^{-3} \times \frac{1}{8.85 \times 10^{-12}} = 1.0 \times 10^{-3} \times 1.12994 \times 10^{11} $$ Multiply: $$ A = 1.12994 \times 10^{8} \text{ m}^2 $$ 5. **Final Answer:** $$ \boxed{A = 1.13 \times 10^{8} \text{ m}^2} $$ 6. **Implication:** This very large area indicates that to achieve a capacitance of 1 farad with a 1 mm separation, the plates must be extremely large, reflecting the practical difficulty of making large capacitance with simple parallel plates.