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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 50a

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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Understand the problem: A cylindrical capacitor consists of two concentric cylindrical conductors. The inner conductor has radius R₆ = 0.50 mm, and the outer conductor has radius Rₐ = 3.5 mm. The potential difference between the conductors is 625 V. We are tasked with calculating the energy stored in a 1.0-m length of the capacitor.
Step 1: Write the formula for the capacitance per unit length of a cylindrical capacitor. The capacitance per unit length (C') is given by: C'=2πεln(Ra), where ε₀ is the permittivity of free space (8.85 × 10⁻¹² F/m), Rₐ is the radius of the outer conductor, and R₆ is the radius of the inner conductor.
Step 2: Substitute the given values into the formula for C'. Convert the radii to meters: Rₐ = 3.5 mm = 3.5 × 10⁻³ m and R₆ = 0.50 mm = 0.50 × 10⁻³ m. Then calculate the natural logarithm term: ln(3.5×10-30.50×10-3).
Step 3: Use the formula for the energy stored in a capacitor. The energy stored (U) is given by: U=12C'V2×l, where V is the potential difference (625 V), l is the length of the capacitor (1.0 m), and C' is the capacitance per unit length calculated in Step 2.
Step 4: Substitute the values of C', V, and l into the energy formula. Perform the necessary calculations to find the energy stored in the capacitor. Ensure all units are consistent (e.g., meters, volts, farads).

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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 electric charge per unit voltage. For cylindrical capacitors, the capacitance can be calculated using the formula C = (2πε₀L) / ln(Rₐ/R₆), where ε₀ is the permittivity of free space, L is the length of the capacitor, and Rₐ and R₆ are the radii of the outer and inner conductors, respectively.
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Energy Stored in a Capacitor

The energy (U) stored in a capacitor can be calculated using the formula U = 1/2 C V², where C is the capacitance and V is the potential difference across the capacitor. This relationship shows how energy is directly proportional to the square of the voltage and the capacitance, highlighting the importance of both factors in energy storage.
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Potential Difference

Potential difference, or voltage, is the difference in electric potential between two points in an electric field. In this context, it is the voltage applied across the cylindrical capacitor, which influences both the charge stored and the energy contained within the capacitor. A higher potential difference results in greater energy storage.
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Related Practice
Textbook Question

What is the capacitance of a pair of circular plates with a radius of 5.0 cm separated by 2.3 mm of mica?

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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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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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The capacitor shown in Fig. 24–34 is connected to an 80.0-V battery. Calculate (and sketch) the electric field everywhere between the capacitor plates. Find both the free charge on each capacitor plate and the induced charge on the faces of the glass dielectric plate.

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Textbook Question

A 3500-pF air-gap capacitor is connected to an 18-V battery. If a piece of mica fills the space between the plates, how much charge will flow from the battery?

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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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