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Chapter 24: Problem 8

24.8. A \(5.00-\) pF, parallel-plate, air-filled capacitor with circular plates is to be used in a circuit in which it will be subjected to potentials of up to \(1.00 \times 10^{2} \mathrm{V}\) . The electric field between the plates is to be no greater than \(1.00 \times 10^{4} \mathrm{N} / \mathrm{C}\) . As a budding electrical engineer for Live-Wire Electronics, your tasks are to (a) design the capacitor by finding what its physical dimensions and separation must be; (b) find the maximum charge these plates can hold.

Short Answer

Expert verified
The capacitor has a plate separation of 0.01 m, a radius of 0.0134 m, and can hold a maximum charge of 5.00 x 10^-10 C.

Step by step solution

01

Understand the Problem

We need to design a parallel-plate capacitor with a capacitance of \(5.00\,\mathrm{pF}\) that can handle voltages up to \(100\,\mathrm{V}\) without exceeding an electric field of \(10,000\,\mathrm{N/C}\). We must find the plate radius and separation, and the maximum charge it can hold.
02

Use Capacitance Formula

The formula for capacitance \(C\) of a parallel-plate capacitor is \(C = \frac{\varepsilon_0 A}{d}\), where \(\varepsilon_0\) is the permittivity of free space (\(8.85 \times 10^{-12}\,\mathrm{F/m}\)), \(A\) is the plate area, and \(d\) is the plate separation.
03

Relate Electric Field to Voltage and Separation

The electric field \(E\) between the plates is related to the voltage \(V\) and the separation \(d\) by \(E = \frac{V}{d}\). Given \(E = 10,000\,\mathrm{N/C}\) and \(V = 100\,\mathrm{V}\), solve for \(d\).
04

Solve for Plate Separation

Rearrange the electric field formula: \(d = \frac{V}{E} = \frac{100\,\mathrm{V}}{10,000\,\mathrm{N/C}} = 0.01\,\mathrm{m}\).
05

Calculate Plate Area

Substitute \(C = 5.00 \times 10^{-12}\,\mathrm{F}\) and \(d = 0.01\,\mathrm{m}\) into the capacitance formula: \(5.00 \times 10^{-12} = \frac{8.85 \times 10^{-12} \cdot A}{0.01}\). Solve for \(A\).
06

Area Calculation

\(A = \frac{5.00 \times 10^{-12} \times 0.01}{8.85 \times 10^{-12}} = 5.65 \times 10^{-4}\,\mathrm{m^2}\).
07

Calculate Plate Radius

For circular plates, use \(A = \pi r^2\) to solve for \(r\): \(r = \sqrt{\frac{A}{\pi}} = \sqrt{\frac{5.65 \times 10^{-4}}{\pi}} \approx 0.0134\,\mathrm{m}\).
08

Calculate Maximum Charge

The charge \(Q\) is given by \(Q = C \cdot V = 5.00 \times 10^{-12}\,\mathrm{F} \times 100\,\mathrm{V} = 5.00 \times 10^{-10}\,\mathrm{C}\).

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Key Concepts

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

Electric Field
The electric field is a concept that helps us understand forces within electrical systems. It is defined as the force per unit charge that a charged object exerts in the space around it. For a parallel-plate capacitor, the electric field is uniform between the plates.

In simple terms, this field acts as the 'push' that drives electrical charge from one plate to another. This is crucial because it affects how much voltage a capacitor can handle. The electric field is directly related to the voltage and the distance between the plates by the formula:
  • \( E = \frac{V}{d} \)
where \(E\) is the electric field, \(V\) is the voltage, and \(d\) is the separation between the plates.

For the given problem, limiting the electric field to no more than \(10,000\, \mathrm{N/C}\) ensures that the capacitor operates safely without breaking down.
Parallel-Plate Capacitor
A parallel-plate capacitor consists of two conductive plates separated by an insulating material. This configuration is widely used in electrical circuits due to its simplicity and effectiveness in storing electrical energy.

The capacitance \(C\) of a parallel-plate capacitor is determined by the plate area \(A\), the separation between the plates \(d\), and the material between them. The formula to calculate capacitance is:
  • \( C = \frac{\varepsilon \times A}{d} \)
Here, \(\varepsilon\) represents the permittivity of the insulating material. For air, this is typically referred to as \(\varepsilon_0\), the permittivity of free space.

In this exercise, the primary aim is to design a capacitor to fit certain specifications, using these fundamental relationships to find the necessary dimensions.
Permittivity of Free Space
Permittivity of free space, symbolized as \( \varepsilon_0 \), is a constant that characterizes how electric fields interact with the vacuum of space. Its value is \( 8.85 \times 10^{-12} \, \mathrm{F/m} \) (farads per meter). This parameter is critical in calculating a capacitor's capacity for storing electrical energy.

The permittivity affects the capacitance of a capacitor. High permittivity allows a capacitor to store more energy at a given voltage. In formulas, we see this value as a part of the equation for capacitance \( C = \frac{\varepsilon_0 A}{d} \).

By substituting \(\varepsilon_0\) in calculations, we determine how physically large or small a capacitor will be for a given capacitance and plate separation.
Voltage
Voltage is the difference in electric potential energy between two points in a circuit. Often described as 'electric pressure', voltage pushes the electric current through a circuit.

In a parallel-plate capacitor, voltage correlates with the electric field and the plate separation:
  • \( V = E \times d \)
where \(V\) is voltage, \(E\) is the electric field strength, and \(d\) is the distance between the plates. For the given capacitor design, voltage restrictions ensure the durability and functionality of the capacitor.

It is crucial to ensure that the voltage does not exceed a certain limit, as it can cause the dielectric material between the plates to break down, rendering the capacitor unusable. This exercise uses a maximum voltage of \(100 \mathrm{V}\) to calculate the necessary plate dimensions.

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Most popular questions from this chapter

24.20. Two parallel plate vacuum capacitors have plate spacings \(d_{1}\) and \(d_{2}\) and equal plate areas \(A\) . Show that when the capacitors are connected in series, the eqnivalent capacitance is the same as for a single capacitor with plate area \(A\) and spacing \(d_{1}+d_{2}\) .

24.13. A spherical capacitor contains a charge of 3.30 \(\mathrm{nC}\) when connected to a potential difference of 220 \(\mathrm{V}\) . If its plates are sepa- rated by vacuum and the inner radius of the outer shell is \(4.00 \mathrm{cm},\) calculate: (a) the capacitance; (b) the radius of the inner sphere; (c) the electric field just outside the surface of the inner sphere.

24.65. A parallel-plate capacitor with only air between the plates is charged by connecting it to a battery. The capacitor is then disconnected from the battery, without any of the charge leaving the plates. (a) A voltmeter reads 45.0 \(\mathrm{V}\) when placed across the capacitor. When a dielectric is inserted between the plates, completely filling the space, the voltmeter reads 11.5 \(\mathrm{V}\) . What is the dielectric constant of this material? (b) What will the voltmeter read if the dielectric is now pulled partway out so it fills only one-third of the space between the plates?

24.37. You have two identical capacitors and an external potential source. (a) Compare the total energy stored in the capacitors when they are connected to the applied potential in series and in parallel. (b) Compare the maximum amount of charge stored in each case. (c) Energy storage in a capacitor can be limited by the inaximuni electric field between the plates. What is the ratio of the electric field for the series and parallel combinations?

24.45. When a \(360-n F\) air capacitor \(\left(1 n F=10^{-9} \mathrm{F}\right)\) is connected to a power supply, the energy stored in the capacitor is \(1.85 \times 10^{-5}\) J. While the capacitor is kept connected to the power supply, a slab of dielectric is inserted that completely fills the space between the plates. This increases the stored energy by \(2.32 \times 10^{-5} \mathrm{J}\) (a) What is the potential difference between the capacitor plates?(b) What is the dielectric constant of the slab?
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