Question

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?

Short answer

Answer: The capacitance of the parallel plate capacitor is approximately \(29.77 \times 10^{-9} \mathrm{F}\).

Step-by-step solution

Calculate the area of the plates

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: \(A_{air} = A_{dielectric} = \frac{1}{2} A_{total}\)

Calculate the capacitance of air and dielectric sections separately

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 \(C_{dielectric} = \epsilon_r \epsilon_0\frac{A_{dielectric}}{d}\)

Use the formula for capacitors in parallel

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

Calculate the final capacitance

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 \(29.77 \times 10^{-9} \mathrm{F}\).

Key concepts

Parallel Plate Capacitor

A parallel plate capacitor consists of two conductive plates placed parallel to each other, often separated by a small gap. The basic equation for the capacitance of a parallel plate capacitor is:\[ C = \frac{\epsilon A}{d} \]where:

• \(C\) is the capacitance
• \(\epsilon\) is the permittivity of the material between the plates
• \(A\) is the area of one of the plates
• \(d\) is the separation distance between the plates.

For the problem at hand, each plate is circular, and we must calculate the area considering the radius given. The function of parallel plates is to store electric charges and create an electric field in the gap between the plates.

Dielectric Material

A dielectric material is an insulating substance that increases the capacitance of a capacitor when placed between its plates. The dielectric is characterized by a dielectric constant, \(\kappa\), which quantifies how much it can increase the capacitance compared to a vacuum. In this exercise, half of the capacitor is filled with a dielectric material, enhancing its capacitance due to the dielectric constant of 11.1.When calculating the contribution of the dielectric section to the total capacitance, the formula used is:\[ C_{dielectric} = \epsilon_r \epsilon_0\frac{A_{dielectric}}{d} \]where:

• \(\epsilon_r\) is the relative permittivity of the dielectric material, equivalent to \(\kappa\)
• \(\epsilon_0\) is the permittivity of free space
• \(A_{dielectric}\) is the area of the plates covered by the dielectric.

This material helps in increasing the charge storage capacity of the capacitor without altering its physical size.

Electric Permittivity

Electric permittivity is a fundamental property of materials that describes how an electric field affects, and is affected by, a dielectric medium. The permittivity of free space, denoted as \(\epsilon_0\), is a constant value approximately \(8.854 \times 10^{-12} \mathrm{F/m}\).In essence, permittivity determines the ability of a material to store electrical energy in an electric field. For capacitors, specifically those containing dielectrics, the effective permittivity is indicated by the product \(\epsilon_r \epsilon_0\), where \(\epsilon_r\) refers to the relative permittivity or dielectric constant. This relative permittivity is greater than 1 and indicates how much more effective the material is at storing charge compared to a vacuum.In this setup, using air alongside a dielectric enhances the overall capacitance by combining sections with different permittivities. Thus, understanding permittivity is crucial for designing capacitors to optimize energy storage.