Question

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?

Short answer

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.

Step-by-step solution

Calculate the capacitance with paper

The area of the plates \(A = 120 \mathrm{cm}^2 = 1.2 \times 10^{-2} \mathrm{m}^2\). The thickness of the paper \(d = 0.10 \mathrm{mm} = 1.0 \times 10^{-4} \mathrm{m}\). The dielectric constant of the paper is \(\kappa = 3.5\). Using the capacitance formula, we have: \(C_\text{paper} = \kappa_\text{paper} \varepsilon_0 \frac{A}{d} = 3.5 \cdot (8.85 \times 10^{-12} \mathrm{F/m}) \cdot \frac{1.2 \times 10^{-2} \mathrm{m}^2}{1.0 \times 10^{-4} \mathrm{m}}\)

Calculate the capacitance with glass

The thickness of the glass \(d = 2.0 \mathrm{mm} = 2.0 \times 10^{-3} \mathrm{m}\). The dielectric constant of the glass is \(\kappa = 7.0\). Using the capacitance formula, we have: \(C_\text{glass} = \kappa_\text{glass} \varepsilon_0 \frac{A}{d} = 7.0 \cdot (8.85 \times 10^{-12} \mathrm{F/m}) \cdot \frac{1.2 \times 10^{-2} \mathrm{m}^2}{2.0 \times 10^{-3} \mathrm{m}}\)

Calculate the capacitance with paraffin

The thickness of the paraffin \(d = 10.0 \mathrm{mm} = 1.0 \times 10^{-2} \mathrm{m}\). The dielectric constant of the paraffin is \(\kappa = 2.0\). Using the capacitance formula, we have: \(C_\text{paraffin} = \kappa_\text{paraffin} \varepsilon_0 \frac{A}{d} = 2.0 \cdot (8.85 \times 10^{-12} \mathrm{F/m}) \cdot \frac{1.2 \times 10^{-2} \mathrm{m}^2}{1.0 \times 10^{-2} \mathrm{m}}\)

Compare the capacitance values and determine the largest and smallest capacitance

Now that we have the capacitance values for each dielectric, we can compare them to determine the largest and smallest possible capacitance. (a) The largest capacitance will be the maximum of the calculated values for paper, glass, and paraffin: \(C_\mathrm{max} = \mathrm{max}(C_\mathrm{paper}, C_\mathrm{glass}, C_\mathrm{paraffin})\) (b) The smallest capacitance will be the minimum of the same calculated values: \(C_\mathrm{min} = \mathrm{min}(C_\mathrm{paper}, C_\mathrm{glass}, C_\mathrm{paraffin})\) Calculate the capacitance for each dielectric using the formulas from Steps 1, 2, and 3, and compare the results to determine the largest (from Step 4a) and smallest (from Step 4b) capacitance.