Ch. 24 - Capacitance, Dielectrics, Electric Energy, Storage

Giancoli Douglas - Physics for Scientists and Engineers 5th edition

Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Not the one you use?Change textbook

Giancoli Douglas 5th edition

Ch. 24 - Capacitance, Dielectrics, Electric Energy, Storage

Problem 13

Chapter 23, Problem 13

To get an idea how big a farad is, suppose you want to make a 1-F air-filled parallel-plate capacitor for a circuit you are building. To make it a reasonable size, suppose you limit the plate area to 1.0 cm². What would the gap have to be between the plates? Is this practically achievable?

Verified step by step guidance

Start by recalling the formula for the capacitance of a parallel-plate capacitor:

C = \frac{ε}{d}

, where

C

is the capacitance,

ε

is the permittivity of the medium (for air,

ε = ε

₀, the permittivity of free space),

A

is the plate area, and

d

is the separation between the plates.

Rearrange the formula to solve for the plate separation

d

d = \frac{ε}{C}

. Substitute

ε = ε

₀ and

ε

₀=8.85×10⁻12 F·m⁻1.

Convert the plate area from cm² to m² for consistency in SI units:

1.0 cm ² = 1.0 \times 10 ⁻ 4 m ²

Substitute the known values into the rearranged formula:

d = ε

₀×AC. Using

ε

₀=8.85×10⁻12,

A = 1.0 \times 10 ⁻ 4

, and

C = 1

Evaluate the result to determine the plate separation

d

. Compare the calculated value to practical dimensions to assess whether such a capacitor is feasible to construct. Note that the separation will likely be extremely small, making it impractical to achieve with current technology.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Capacitance

Capacitance is the ability of a system to store an electric charge per unit voltage. It is measured in farads (F), where one farad is defined as the capacitance of a capacitor that stores one coulomb of charge at one volt. The formula for capacitance in a parallel-plate capacitor is C = ε₀(A/d), where C is capacitance, ε₀ is the permittivity of free space, A is the area of the plates, and d is the distance between them.

Parallel-Plate Capacitor

A parallel-plate capacitor consists of two conductive plates separated by a dielectric material, which can be air or another insulating substance. The electric field between the plates is uniform, and the capacitance depends on the area of the plates and the distance between them. This configuration is commonly used in circuits due to its simplicity and effectiveness in storing charge.

Dielectric Constant

The dielectric constant (or relative permittivity) of a material measures its ability to store electrical energy in an electric field. For air, the dielectric constant is approximately 1. This property affects the capacitance of a capacitor; a higher dielectric constant allows for greater capacitance with the same plate area and separation distance, making it crucial in determining the feasibility of achieving a specific capacitance value.

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Intro To Dielectrics

Related Practice

Textbook Question

Suppose in Fig. 24–27 that C₁ = C₃ = 8.0μF, C₂ = C₄ = 16μF, and Q₃ = 21μC. Determine the voltage V_ba across the combination.

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Textbook Question

(II) A 7.7-μF capacitor is charged by a 185-V battery (Fig. 24–21a) and then is disconnected from the battery. When this capacitor (C₁) is then connected (Fig. 24–21b) to a second (initially uncharged) capacitor, C₂, the final voltage on each capacitor is 15 V. What is the value of C₂? [Hint: Charge is conserved.]

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Textbook Question

Small distances can be measured using a capacitor whose plate separation 𝓍 is variable. Consider an air-filled parallel-plate capacitor with fixed plate area A = 25 mm² separated by a variable distance 𝓍. Assume this capacitor is attached to a capacitance-measuring instrument which can measure capacitance C in the range 1.0 pF to 1000.0 pF with an accuracy of ∆C = 0.1 pF. Define ∆𝓍 to be the accuracy (magnitude) to which 𝓍 can be determined, and determine a formula for ∆𝓍.

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Textbook Question

In an electrostatic air cleaner (“precipitator”), the strong nonuniform electric field in the central region of a cylindrical capacitor (with outer and inner cylindrical radii Rₐ and R₆ ) is used to create ionized air molecules for use in charging dust and soot particles (Fig. 24–22). Under standard atmospheric conditions, if air is subjected to an electric field magnitude that exceeds its dielectric strength Eₛ ≈ 3.0 x 10⁶ N/C, air molecules will dissociate into positively charged ions and free electrons. In a precipitator, the region within which air is ionized (the corona discharge region) occupies a cylindrical volume of radius R that is typically five times that of the inner cylinder. Assume a particular precipitator is constructed with R₆ = 0.10 mm and Rₐ = 10.0 cm. In order to create a corona discharge region with radius R = 5.0 R₆, what potential difference V should be applied between the precipitator’s inner and outer conducting cylinders? [Besides dissociating air, the charged inner cylinder repels the resulting positive ions from the corona discharge region, where they are put to use in charging dust particles, which are then “collected” on the negatively charged outer cylinder.]

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