A parallel plate capacitor has two plates, each of area $3.0 \times 10^{-4} \mathrm{m}^{2},\( separated by \)3.5 \times 10^{-4} \mathrm{m} .$ The space between the plates is filled with a dielectric. When the capacitor is connected to a source of \(120 \mathrm{V}\) rms at \(8.0 \mathrm{kHz}\) an mms current of \(1.5 \times 10^{-4} \mathrm{A}\) is measured. (a) What is the capacitive reactance? (b) What is the dielectric constant of the material between the plates of the capacitor?

First, we need to find the capacitance of the capacitor using the formula \(C = \frac{Q}{V}\), where \(C\) is the capacitance, \(Q\) is the charge on the capacitor and \(V\) is the voltage across the capacitor. We can rewrite the formula as \(C = \frac{I}{\omega V}\), where \(I\) is the rms current, \(\omega\) is the angular frequency, and \(V\) is the rms voltage. Given the rms current \(I = 1.5 \times 10^{-4}\mathrm{A}\), the rms voltage \(V = 120\mathrm{V}\), and the frequency \(f = 8.0\mathrm{kHz}\), we can find the capacitance. First, we need to calculate the angular frequency \(\omega\) using the formula \(\omega = 2\pi f\). \(\omega = 2\pi(8.0 \times 10^{3}\,\text{Hz}) = 16\pi \times 10^{3}\,\text{rad/s}\) Now we can calculate the capacitance: \(C = \frac{I}{\omega V} = \frac{1.5 \times 10^{-4}\,\text{A}}{16\pi \times 10^{3}\,\text{rad/s} \cdot 120\,\text{V}} \approx 3.14 \times 10^{-12}\,\text{F}\)

Now that we have the capacitance, we can find the capacitive reactance using the formula \(X_{C} = \frac{1}{\omega C}\). \(X_{C} = \frac{1}{16\pi \times 10^{3}\,\text{rad/s} \cdot 3.14 \times 10^{-12}\,\text{F}} \approx 2000\,\Omega\) The capacitive reactance of the capacitor is approximately \(2000\,\Omega\).

Now, we'll find the dielectric constant of the material between the plates of the capacitor. We'll use the formula \(C = \epsilon A / d\), where \(C\) is the capacitance, \(\epsilon\) is the permittivity of the dielectric material, \(A\) is the plate area, and \(d\) is the distance between the plates. First, we need to rewrite the formula to find the permittivity of the dielectric material: \(\epsilon = \frac{C \cdot d}{A}\) We're given the plate area \(A = 3.0 \times 10^{-4}\,\text{m}^2\), the distance between the plates \(d = 3.5 \times 10^{-4}\,\text{m}\), and we've found the capacitance \(C \approx 3.14 \times 10^{-12}\,\text{F}\). \(\epsilon = \frac{3.14 \times 10^{-12}\,\text{F} \cdot 3.5 \times 10^{-4}\,\text{m}}{3.0 \times 10^{-4}\,\text{m}^2} \approx 3.64 \times 10^{-8}\,\mathrm{F/m}\) To find the dielectric constant, we need to divide the permittivity of the dielectric material by the permittivity of free space \(\epsilon_0 \approx 8.85 \times 10^{-12}\,\mathrm{F/m}\): \(\text{Dielectric constant} = \frac{\epsilon}{\epsilon_0} = \frac{3.64 \times 10^{-8}\,\mathrm{F/m}}{8.85 \times 10^{-12}\,\mathrm{F/m}} \approx 4116\) The dielectric constant of the material between the plates of the capacitor is approximately \(4116\).