Two circular metal plates of radius \(0.61 \mathrm{~m}\) and thickness \(7.1 \mathrm{~mm}\) are used in a parallel plate capacitor. A gap of 2.1 mm is left between the plates, and half of the space (a semicircle) between the plates is filled with a dielectric for which \(\kappa=11.1\) and the other half is filled with air. What is the capacitance of this capacitor?

First, we need to calculate the area of the semi-circle filled with air and the semi-circle filled with the dielectric material. The total area of the circular plates is: \(A_{total} = \pi \times r^2\) Where \(r\) is the radius of the plates. The area of each semi-circle will be half of the total area: \(A_{air} = A_{dielectric} = \frac{1}{2} A_{total}\)

Now, we'll find the capacitance of the air and dielectric sections separately using the capacitance formula: \(C_{air} = \epsilon_0\frac{A_{air}}{d}\) and \(C_{dielectric} = \epsilon_r \epsilon_0\frac{A_{dielectric}}{d}\)

To find the total capacitance, we'll use the formula for capacitors in parallel: \(C_{total} = C_{air} + C_{dielectric}\)

Substitute the known values into the formulas and calculate the total capacitance: - Calculate \(A_{total}\): \(A_{total} = \pi \times (0.61)^2 = 1.1731 \mathrm{~m^2}\) - Calculate \(A_{air}\) and \(A_{dielectric}\): \(A_{air} = A_{dielectric} = 0.5865 \mathrm{~m^2}\) - Calculate \(C_{air}\): \(C_{air} = (8.854 \times 10^{-12})\frac{0.5865}{2.1 \times 10^{-3}} = 2.4614 \times 10^{-9} \mathrm{F}\) - Calculate \(C_{dielectric}\): \(C_{dielectric} = 11.1(8.854 \times 10^{-12})\frac{0.5865}{2.1 \times 10^{-3}} = 27.3054 \times 10^{-9} \mathrm{F}\) - Calculate \(C_{total}\): \(C_{total} = 2.4614 \times 10^{-9} + 27.3054 \times 10^{-9} = 29.7668 \times 10^{-9} \mathrm{F}\) The capacitance of the parallel plate capacitor is approximately \(29.77 \times 10^{-9} \mathrm{F}\).