Textbook Question
What is the capacitance of the two metal spheres shown in FIGURE EX26.22?
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Ch 26: Potential and Field
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796
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Table of contents
- 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: 9780137344796Not the one you use?Change textbook
Chapter 26, Problem 21a
Two 3.0-cm-diameter aluminum electrodes are spaced 0.50 mm apart. The electrodes are connected to a 100 V battery. What is the capacitance?
Verified step by step guidance1
Step 1: Understand the formula for capacitance of a parallel plate capacitor. The capacitance is given by the formula: , where is the permittivity of the material between the plates, is the area of the plates, and is the separation between the plates.
Step 2: Identify the values given in the problem. The diameter of the electrodes is 3.0 cm, so the radius is 1.5 cm or 0.015 m. The separation between the plates is 0.50 mm or 0.0005 m. The material between the plates is air, so the permittivity is approximately .
Step 3: Calculate the area of the circular plates. The area of a circle is given by . Substitute the radius into the formula to find the area.
Step 4: Substitute the values for , , and into the capacitance formula. Use , the calculated area, and .
Step 5: Simplify the expression to find the capacitance. Ensure the units are consistent throughout the calculation, and express the final capacitance in Farads (F).
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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. It is defined as the ratio of the electric charge (Q) stored on the electrodes to the potential difference (V) across them, expressed as C = Q/V. The unit of capacitance is the farad (F), and it is influenced by the physical characteristics of the capacitor, including the area of the electrodes and the distance between them.
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Parallel Plate Capacitor
A parallel plate capacitor consists of two conductive plates separated by an insulating material (dielectric). The capacitance of such a capacitor can be calculated using the formula C = ε₀(A/d), where ε₀ is the permittivity of free space, A is the area of one of the plates, and d is the separation between the plates. This model simplifies the analysis of capacitors and is fundamental in understanding how they operate.
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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. In the context of capacitors, the electric field between the plates is uniform and can be calculated using E = V/d, where V is the voltage across the plates and d is the separation distance. Understanding the electric field is crucial for analyzing how capacitors store energy and influence charge distribution.
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Related Practice
Textbook Question
Light from the sun allows a solar cell to move electrons from the positive to the negative terminal, doing 2.4×10−19 J of work per electron. What is the emf of this solar cell?
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Textbook Question
A switch that connects a battery to a 10 μF capacitor is closed. Several seconds later you find that the capacitor plates are charged to ±30 μC. What is the emf of the battery?
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Textbook Question
Two 3.0-cm-diameter aluminum electrodes are spaced 0.50 mm apart. The electrodes are connected to a 100 V battery. What is the magnitude of the charge on each electrode?
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Textbook Question
The electric potential along the x-axis is V = 100x2 V, where x is in meters. What is Ex at (a) x=0 m and (b) x=1 m?
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Textbook Question
How much charge does a 9.0 V battery transfer from the negative to the positive terminal while doing 27 J of work?
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