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

Giancoli Douglas - Physics for Scientists and Engineers 5th edition

Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179당신이 사용하는 게 아니라요?교과서 변경

Giancoli Douglas 5th edition

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

문제 50a

23장, 문제 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.

검증된 단계별 안내

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' = \frac{2 π ε ₀}{ln (\frac{R}{a})}

, 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 (\frac{3.5 \times 10^{- 3}}{0.50 \times 10^{- 3}})

Step 3: Use the formula for the energy stored in a capacitor. The energy stored (U) is given by:

U = \frac{1}{2} C' V 2 \times 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).

비슷한 문제에 대한 검증된 영상 답변:

이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.

영상 길이:

도움이 되었나요?

주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

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.

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.

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.

관련 실천

교과서 질문

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

1695

views

교과서 질문

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

1438

views

교과서 질문

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?

1499

views

교과서 질문

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.

views

교과서 질문

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?

1287

views

교과서 질문

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.]

1382

views