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Chapter 24: Problem 60

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

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

Step by step solution

01

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: \(A_{air} = A_{dielectric} = \frac{1}{2} A_{total}\)
02

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 \(C_{dielectric} = \epsilon_r \epsilon_0\frac{A_{dielectric}}{d}\)
03

Use the formula for capacitors in parallel

To find the total capacitance, we'll use the formula for capacitors in parallel: \(C_{total} = C_{air} + C_{dielectric}\)
04

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

A typical AAA battery has stored energy of about 3400 J. (Battery capacity is typically listed as \(625 \mathrm{~mA} \mathrm{~h}\), meaning that much charge can be delivered at approximately \(1.5 \mathrm{~V}\).) Suppose you want to build a parallel plate capacitor to store this amount of energy, using a plate separation of \(1.0 \mathrm{~mm}\) and with air filling the space between the plates. a) Assuming that the potential difference across the capacitor is \(1.5 \mathrm{~V},\) what must the area of each plate be? b) Assuming that the potential difference across the capacitor is the maximum that can be applied without dielectric breakdown occurring, what must the area of each plate be? c) Is either capacitor a practical replacement for the AAA batterv?

Consider a cylindrical capacitor, with outer radius \(R\) and cylinder separation \(d\). Determine what the capacitance approaches in the limit where \(d \ll R\). (Hint: Express the capacitance in terms of the ratio \(d / R\) and then examine what happens as the ratio \(d / R\) becomes very small compared to \(1 .)\) Explain why the limit on the capacitance makes sense.

A parallel plate capacitor is charged with a battery and then disconnected from the battery, leaving a certain amount of energy stored in the capacitor. The separation between the plates is then increased. What happens to the energy stored in the capacitor? Discuss your answer in terms of energy conservation.

A parallel plate capacitor of capacitance \(C\) has plates of area \(A\) with distance \(d\) between them. When the capacitor is connected to a battery of potential difference \(V\), it has a charge of magnitude \(Q\) on its plates. While the capacitor is connected to the battery, the distance between the plates is decreased by a factor of \(3 .\) The magnitude of the charge on the plates and the capacitance are then a) \(\frac{1}{3} Q\) and \(\frac{1}{3} C\). c) \(3 Q\) and \(3 C\). b) \(\frac{1}{3} Q\) and \(3 C\). d) \(3 Q\) and \(\frac{1}{3} C\).

A parallel plate capacitor of capacitance \(C\) is connected to a power supply that maintains a constant potential difference, \(V\). A close-fitting slab of dielectric, with dielectric constant \(\kappa\), is then inserted and fills the previously empty space between the plates. a) What was the energy stored on the capacitor before the insertion of the dielectric? b) What was the energy stored after the insertion of the dielectric? c) Was the dielectric pulled into the space between the plates, or did it have to be pushed in? Explain.
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