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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
Chapter 23, Problem 20

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.]
Diagram of a cylindrical capacitor illustrating the corona discharge region with ionized air and labeled dimensions.

Verified step by step guidance
1
Step 1: Understand the problem. The goal is to calculate the potential difference (V) required to create a corona discharge region in a cylindrical capacitor. The electric field in the corona discharge region must reach the dielectric strength of air (Eₛ ≈ 3.0 × 10⁶ N/C). The inner radius of the capacitor is R₆ = 0.10 mm, the outer radius is Rₐ = 10.0 cm, and the radius of the corona discharge region is R = 5.0 R₆.
Step 2: Recall the formula for the electric field in a cylindrical capacitor. The electric field at a distance r from the axis of a cylindrical capacitor is given by: E=Vrln(Ra), where V is the potential difference, r is the radial distance, and ln(Rₐ/R₆) is the natural logarithm of the ratio of the outer to inner radii.
Step 3: Set the electric field at the edge of the corona discharge region (r = R) equal to the dielectric strength of air (Eₛ). Substitute r = R = 5.0 R₆ into the formula for E: E=VRln(Ra). This gives: E=V(5R)ln(Ra).
Step 4: Solve for the potential difference V. Rearrange the equation to isolate V: V=ERln(Ra). Substitute R = 5.0 R₆ and E = Eₛ into the equation: V=E(5R)ln(Ra).
Step 5: Substitute the known values into the equation. Use R₆ = 0.10 mm = 0.10 × 10⁻³ m, Rₐ = 10.0 cm = 0.10 m, and Eₛ = 3.0 × 10⁶ N/C. Calculate the natural logarithm term ln(Rₐ/R₆) and substitute all values into the formula for V. This will give the required potential difference to create the corona discharge region.

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Key Concepts

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

Electric Field and Dielectric Strength

The electric field is a vector field around charged particles that exerts force on other charged objects. Dielectric strength is the maximum electric field that a material can withstand without breaking down, which for air is approximately 3.0 x 10⁶ N/C. When the electric field exceeds this strength, air molecules ionize, creating charged particles that can be manipulated in devices like electrostatic precipitators.
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Dielectric Breakdown

Cylindrical Capacitor

A cylindrical capacitor consists of two concentric cylindrical conductors separated by an insulating material. The capacitance depends on the radii of the cylinders and the distance between them. In the context of the precipitator, the inner and outer cylinders create an electric field that can ionize air when a sufficient potential difference is applied, facilitating the charging of dust particles.
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Potential Difference and Ionization

Potential difference, or voltage, is the work done per unit charge to move a charge between two points in an electric field. In the electrostatic precipitator, the potential difference between the inner and outer cylinders must be high enough to create an electric field that exceeds the dielectric strength of air, leading to ionization. This ionization is crucial for charging and collecting dust particles effectively.
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Related Practice
Textbook Question

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?

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Textbook Question

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

Two capacitors connected in parallel produce an equivalent capacitance of 32.9-μF, but when connected in series the equivalent capacitance is only 5.5 μF. What is the individual capacitance of each capacitor?

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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

A cylindrical capacitor (Example 24–2) has Rₐ = 3.5 mm and R₆.= 0.50 mm. The two conductors have a potential difference of 625 V, with the inner conductor at the higher potential. Calculate the energy stored in a 1.0-m length of the capacitor.

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