Textbook Question
The electric potential is 40 V at point A near a uniformly charged sphere. At point B, 2.0 μm farther away from the sphere, the potential has decreased by 0.16 mV. How far is point A from the center of the sphere?
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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 52b
Two 2.0 cm×2.0 cm metal electrodes are spaced 1.0 mm apart and connected by wires to the terminals of a 9.0 V battery. The wires are disconnected, and insulated handles are used to pull the plates apart to a new spacing of 2.0 mm. What are the charge on each electrode and the potential difference between them?
Verified step by step guidance1
Calculate the initial capacitance of the parallel plate capacitor using the formula: , where is the permittivity of free space, is the area of the plates, and is the separation between the plates.
Determine the charge on each electrode using the relationship , where is the potential difference (9.0 V in this case).
When the plates are pulled apart to a new spacing of 2.0 mm, note that the charge remains constant because the wires are disconnected, meaning no charge can flow.
Recalculate the new capacitance using the updated plate separation in the formula , where is now 2.0 mm.
Determine the new potential difference between the plates using the relationship , where is the charge calculated earlier and is the new capacitance.
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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 C = Q/V, where C is capacitance, Q is the charge stored, and V is the potential difference. In this scenario, the capacitance of the parallel plate capacitor changes as the distance between the plates is altered, affecting the charge on each electrode.
Recommended video:
Guided course
08:02
Capacitors & Capacitance (Intro)
Electric Field
The electric field (E) between two charged plates is defined as the force per unit charge experienced by a positive test charge placed in the field. It is related to the potential difference (V) and the distance (d) between the plates by the equation E = V/d. Understanding the electric field is crucial for determining how the potential difference changes when the distance between the electrodes is modified.
Recommended video:
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03:16Intro to Electric Fields
Charge Conservation
Charge conservation is a fundamental principle stating that the total electric charge in an isolated system remains constant. When the wires are disconnected from the battery, the charge on the electrodes becomes fixed. This principle is essential for calculating the charge on each electrode after the plates are pulled apart, as the total charge will not change despite the alteration in distance.
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05:43Conservation of Charge
Related Practice
Textbook Question
Find expressions for the equivalent capacitance of (a) N identical capacitors C in parallel and (b) N identical capacitors C in series.
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Textbook Question
Two 2.0 cm×2.0 cm metal electrodes are spaced 1.0 mm apart and connected by wires to the terminals of a 9.0 V battery. What are the charge on each electrode and the potential difference between them?
3
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Textbook Question
Two positive point charges q are located on the y-axis at y = ±a. Your answer to part d shows that an electron experiences a linear restoring force, so it will undergo simple harmonic motion. What is the oscillation frequency in GHz for an electron moving between two 1.0 nC charges separated by 2.0 mm?
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Textbook Question
Two positive point charges q are located on the y-axis at y = ±a. Symmetry dictates that the electric field along the x-axis has only an x-component: Ey=Ez=0. Find an expression for Ex if x≪a.
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Textbook Question
Metal sphere 1 has a positive charge of 6.0 nC. Metal sphere 2, which is twice the diameter of sphere 1, is initially uncharged. The spheres are then connected together by a long, thin metal wire. What are the final charges on each sphere?
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