A capacitor is constructed from two parallel metal plates, each with an area of \( 2.2 \mathrm{~cm}^{2} \), separated by 1.2 \( \mu \mathrm{~m} \). The space between the plates is filled with an insulating material with a dielectric constant of 26 . What is the capacitance of the capacitor?

To calculate the capacitance of a parallel plate capacitor, you can use the formula: \[ C = \frac{{\varepsilon \cdot A}}{{d}} \] where \(C\) is the capacitance, \(\varepsilon\) is the permittivity of the material (for a dielectric, \(\varepsilon = \varepsilon_0 \cdot \kappa\)), \(A\) is the area of one plate, and \(d\) is the separation between the plates. 1. First, let's find \(\varepsilon_0\), the permittivity of free space, which is approximately \(8.854 \times 10^{-12} \, \mathrm{F/m}\). 2. The dielectric constant \(\kappa\) given is 26. 3. The area \(A\) needs to be converted from cm² to m², so \(2.2 \, \mathrm{cm}^2 = 2.2 \times 10^{-4} \, \mathrm{m}^2\). 4. The distance \(d = 1.2 \, \mu\mathrm{m} = 1.2 \times 10^{-6} \, \mathrm{m}\). Now, substituting in the values: \[ C = \frac{{(8.854 \times 10^{-12} \, \mathrm{F/m}) \cdot (26) \cdot (2.2 \times 10^{-4} \, \mathrm{m}^2)}}{{1.2 \times 10^{-6} \, \mathrm{m}}} \] Calculating this gives: \[ C \approx \frac{{(8.854 \times 10^{-12} \times 26 \times 2.2 \times 10^{-4})}}{{1.2 \times 10^{-6}}} \] \[ C \approx \frac{{5.08 \times 10^{-15}}}{{1.2 \times 10^{-6}}} \approx 4.23 \times 10^{-9} \, \mathrm{F} = 4.23 \, \mathrm{nF} \] Thus, the capacitance of the capacitor is approximately \(4.23 \, \mathrm{nF}\).