Question

Two circular metal plates of radius \(0.610 \mathrm{~m}\) and thickness \(7.10 \mathrm{~mm}\) are used in a parallel plate capacitor. A gap of \(2.10 \mathrm{~mm}\) is left between the plates, and half of the space (a semicircle) between them 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: To find the total capacitance, follow these steps: 1. Calculate the area of each semi-circle: $$A_{semi-circle} = \frac{1}{2} \pi r^2$$ 2. Find the capacitance of the dielectric-filled half: $$C_{dielectric} = \kappa \varepsilon_{0} \frac{A_{semi-circle}}{d}$$ 3. Find the capacitance of the air-filled half: $$C_{air} = \varepsilon_{0} \frac{A_{semi-circle}}{d}$$ 4. Add the capacitance values from Step 2 and Step 3 to find the total capacitance: $$C_{total} = C_{air} + C_{dielectric}$$

Step-by-step solution

Calculate the area of each semi-circle

To find the capacitance of each half of the capacitor, we need to determine the area of each semi-circle. For a circle with a radius of \(0.610~\mathrm{m}\), the area of a semi-circle will be: $$ A_{semi-circle} = \frac{1}{2} \pi r^2 = \frac{1}{2}\pi(0.610)^2 $$

Find the capacitance of the dielectric-filled half

Now, we can find the capacitance of the dielectric-filled half using the formula $$ C_{dielectric} = \kappa \varepsilon_{0} \frac{A_{semi-circle}}{d} $$ where \(\kappa = 11.1\), \(\varepsilon_{0} = 8.854\times 10^{-12}~\mathrm{F/m}\), \(A_{semi-circle}\) is the area found in Step 1, and \(d = 2.10\times 10^{-3}~\mathrm{m}\) is the gap between the plates.

Find the capacitance of the air-filled half

For the air-filled half, the capacitance can be found using the formula $$ C_{air} = \varepsilon_{0} \frac{A_{semi-circle}}{d} $$ where \(\varepsilon_{0} = 8.854\times 10^{-12}~\mathrm{F/m}\), \(A_{semi-circle}\) is the area found in Step 1, and \(d = 2.10\times 10^{-3}~\mathrm{m}\) is the gap between the plates.

Find the overall capacitance

Since the capacitance of the air-filled half and the dielectric-filled half are in parallel, the overall capacitance can be found using the formula $$ C_{total} = C_{air} + C_{dielectric} $$ Calculating the values from Step 2 and Step 3 and adding them together will give the final capacitance value.

Key concepts

Dielectric Materials

Understanding dielectric materials is crucial when studying capacitors. Dielectric materials are insulating substances that do not conduct electricity but can support an electrostatic field. In the context of a parallel plate capacitor, placing a dielectric material between the plates increases the capacitor's capacity to store electric charge. This property is due to the dielectric material's ability to be polarized by the applied electric field, effectively reducing the field inside the material and allowing the capacitor to store more charge for a given voltage. The dielectric constant or relative permittivity, denoted as \(\kappa\), represents how much better a dielectric material can store charge compared to a vacuum. For example, a \(\kappa\) of 11.1 means the material can store 11.1 times more charge than a vacuum under the same conditions. Dielectrics are used in various applications, including capacitors, cables, and insulators, to enhance performance and prevent electrical breakdown.

Electric Permittivity

The concept of electric permittivity is central to the functioning of capacitors. It is a measure of how much resistance is encountered when forming an electric field in a medium. The permittivity of free space, or vacuum permittivity, is denoted by \(\varepsilon_0\) and has a value of approximately \(8.854 \times 10^{-12}~\mathrm{F/m}\). This constant is essential for capacitance calculations as it sets the baseline for measuring the effectiveness of dielectric materials. When a dielectric is introduced, its ability to polarize and reduce the electric field is factored in by multiplying \(\varepsilon_0\) by the material's relative permittivity \(\kappa\) to get its absolute permittivity. The higher the permittivity of a material, the greater the charge a capacitor can store. This implies that both the nature of the dielectric material and the design of the capacitor influence its storage capacity.

Parallel Plate Capacitor

A parallel plate capacitor consists of two conductive plates separated by a specific distance, often filled with a dielectric material. The capacity of this type of capacitor to store charge is directly proportional to the area of the plates and the permittivity of the dielectric and is inversely proportional to the distance between the plates. When a voltage is applied across the plates, an electric field is established, and an equal but opposite charge accumulates on each plate. These capacitors are widely used due to their simple structure and dependability. Their applications range from electronic circuits to energy storage systems. The design's simplicity makes it ideal for explaining fundamental concepts of electrostatics and capacitance.

Capacitance Calculations

Capacitance calculations involve determining the amount of charge a capacitor can hold at a given voltage. The formula for the capacitance \(C\) of a parallel plate capacitor is given by \(C = \kappa \varepsilon_{0} \frac{A}{d}\), where \(\kappa\) is the dielectric constant, \(\varepsilon_{0}\) is the permittivity of free space, \(A\) is the area of one of the plates, and \(d\) is the distance between the plates. In the exercise provided, the capacitance is calculated for a capacitor with a semi-circular dielectric material and air in the other half. Given that these two halves produce capacitances in parallel, the overall capacitance is the sum of the capacitances of each half. This type of calculation is fundamental in designing capacitors for specific applications and for understanding the energy storage limits of a system. Accurate capacitance calculations ensure optimal performance in electronic circuits and other technological applications where capacitors play a crucial role.