Domanda del libro di testo
An electric dipole at the origin consists of two charges ±q spaced distance s apart along the y-axis. What is the field Ē on the bisecting axis? Does your result agree with Equation 23.11?
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Ch 26: Potential and Field
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796
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TOC.tableOfContents
- Ch 01: Concepts of Motion35
- Ch 02: Kinematics in One Dimension75
- Ch 03: Vectors and Coordinate Systems58
- Ch 04: Kinematics in Two Dimensions60
- Ch 05: Force and Motion23
- Ch 06: Dynamics I: Motion Along a Line53
- Ch 07: Newton's Third Law27
- Ch 08: Dynamics II: Motion in a Plane52
- Ch 09: Work and Kinetic Energy56
- Ch 10: Interactions and Potential Energy50
- Ch 11: Impulse and Momentum56
- Ch 12: Rotation of a Rigid Body71
- Ch 13: Newton's Theory of Gravity52
- Ch 14: Fluids and Elasticity44
- Ch 15: Oscillations55
- Ch 16: Traveling Waves37
- Ch 17: Superposition39
- Ch 18: A Macroscopic Description of Matter53
- Ch 19: Work, Heat, and the First Law of Thermodynamics60
- Ch 20: The Micro/Macro Connection53
- Ch 21: Heat Engines and Refrigerators34
- Ch 22: Electric Charges and Forces40
- Ch 23: The Electric Field39
- Ch 24: Gauss' Law32
- Ch 25: The Electric Potential63
- Ch 26: Potential and Field57
- Ch 27: Current and Resistance42
- Ch 28: Fundamentals of Circuits39
- Ch 29: The Magnetic Field47
- Ch 30: Electromagnetic Induction60
- Ch 31: Electromagnetic Fields and Waves32
- Ch 32: AC Circuits63
- Ch 33: Wave Optics32
- Ch 34: Ray Optics57
- Ch 35: Optical Instruments30
- Ch 36: Special Relativity38
- Ch 37: The Foundations of Modern Physics25
- Ch 38: Quantization49
- Ch 39: Wave Functions and Uncertainty31
- Ch 40: One-Dimensional Quantum Mechanics35
- Ch 41: Atomic Physics18
- Ch 42: Nuclear Physics27
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Non è quello che usi tu?Cambia libro di testo
Capitolo 26, Problema 81a
Find an expression for the capacitance of a spherical capacitor, consisting of concentric spherical shells of radii R1 (inner shell) and R2 (outer shell).
Guida verificata passo dopo passo1
Start by recalling the formula for the capacitance of a capacitor: \( C = \frac{Q}{\Delta V} \), where \( Q \) is the charge on the capacitor and \( \Delta V \) is the potential difference between the two shells.
Determine the electric field \( E \) between the two spherical shells using Gauss's law: \( \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q}{\varepsilon_0} \). For a spherical shell, the electric field at a distance \( r \) from the center is \( E = \frac{Q}{4 \pi \varepsilon_0 r^2} \).
Find the potential difference \( \Delta V \) between the two shells by integrating the electric field from \( R_1 \) to \( R_2 \): \( \Delta V = \int_{R_1}^{R_2} E \, dr = \int_{R_1}^{R_2} \frac{Q}{4 \pi \varepsilon_0 r^2} \, dr \). Perform the integration to get \( \Delta V = \frac{Q}{4 \pi \varepsilon_0} \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \).
Substitute \( \Delta V \) into the capacitance formula \( C = \frac{Q}{\Delta V} \): \( C = \frac{Q}{\frac{Q}{4 \pi \varepsilon_0} \left( \frac{1}{R_1} - \frac{1}{R_2} \right)} \). Simplify the expression to get \( C = 4 \pi \varepsilon_0 \left( \frac{1}{\frac{1}{R_1} - \frac{1}{R_2}} \right) \).
Simplify further to express the capacitance in terms of \( R_1 \) and \( R_2 \): \( C = \frac{4 \pi \varepsilon_0 R_1 R_2}{R_2 - R_1} \). This is the final expression for the capacitance of the spherical capacitor.
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Concetti chiave
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Capacitance
Capacitance is a measure of a capacitor's ability to store charge per unit voltage. It is defined as the ratio of the electric charge (Q) stored on one conductor to the potential difference (V) between the conductors. The unit of capacitance is the farad (F), where 1 farad equals 1 coulomb per volt.
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Percorso guidato
08:02
Capacitors & Capacitance (Intro)
Electric Field
The electric field (E) is a vector field that represents the force per unit charge experienced by a positive test charge placed in the field. For a spherical capacitor, the electric field between the two shells can be derived from Gauss's law, which relates the electric field to the charge enclosed by a Gaussian surface. The electric field is crucial for understanding how the capacitor stores energy.
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Percorso guidato
03:16Intro to Electric Fields
Gauss's Law
Gauss's Law states that the electric flux through a closed surface is proportional to the charge enclosed within that surface. It is mathematically expressed as ∮E·dA = Q_enc/ε₀, where E is the electric field, dA is the differential area, Q_enc is the enclosed charge, and ε₀ is the permittivity of free space. This law is essential for deriving the electric field in spherical capacitors and ultimately calculating their capacitance.
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Percorso guidato
10:20Gauss' Law
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Consider a uniformly charged sphere of radius R and total charge Q. The electric field Eout outside the sphere (r≥R) is simply that of a point charge Q. In Chapter 24, we used Gauss’s law to find that the electric field Ein inside the sphere (r≤R) is radially outward with field strength . What is the ratio Vcenter/Vsurface?
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