To make a parallel plate capacitor, you have available two flat plates of aluminum (area \(120 \mathrm{cm}^{2}\) ), a sheet of paper (thickness $=0.10 \mathrm{mm}, \kappa=3.5),\( a sheet of glass (thickness \)=2.0 \mathrm{mm}, \kappa=7.0),\( and a slab of paraffin (thickness \)=10.0 \mathrm{mm}, \kappa=2.0) .$ (a) What is the largest capacitance possible using one of these dielectrics? (b) What is the smallest?

Using the formulas from Steps 1, 2, and 3 for each dielectric: \(C_\text{paper} = 3.5 \cdot (8.85 \times 10^{-12} \mathrm{F/m}) \cdot \frac{1.2 \times 10^{-2} \mathrm{m}^2}{1.0 \times 10^{-4} \mathrm{m}} = 3.53 \times 10^{-9} \mathrm{F}\) \(C_\text{glass} = 7.0 \cdot (8.85 \times 10^{-12} \mathrm{F/m}) \cdot \frac{1.2 \times 10^{-2} \mathrm{m}^2}{2.0 \times 10^{-3} \mathrm{m}} = 1.89 \times 10^{-8} \mathrm{F}\) \(C_\text{paraffin} = 2.0 \cdot (8.85 \times 10^{-12} \mathrm{F/m}) \cdot \frac{1.2 \times 10^{-2} \mathrm{m}^2}{1.0 \times 10^{-2} \mathrm{m}} = 2.13 \times 10^{-10} \mathrm{F}\) Comparing the capacitance values to determine the largest and smallest capacitance: \(C_\mathrm{max} = \mathrm{max}(3.53 \times 10^{-9} \mathrm{F}, 1.89 \times 10^{-8} \mathrm{F}, 2.13 \times 10^{-10} \mathrm{F}) = 1.89 \times 10^{-8} \mathrm{F}\) (using glass as the dielectric) \(C_\mathrm{min} = \mathrm{min}(3.53 \times 10^{-9} \mathrm{F}, 1.89 \times 10^{-8} \mathrm{F}, 2.13 \times 10^{-10} \mathrm{F}) = 2.13 \times 10^{-10} \mathrm{F}\) (using paraffin as the dielectric) The largest capacitance is \(1.89 \times 10^{-8} \mathrm{F}\) using glass as the dielectric, and the smallest capacitance is \(2.13 \times 10^{-10} \mathrm{F}\) using paraffin as the dielectric.