Textbook Question
A 2.0-cm-diameter parallel-plate capacitor with a spacing of 0.50 mm is charged to 200 V. What are (a) the total energy stored in the electric field and (b) the energy density?
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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 39a
The electric field in a region of space is Ex = −1000x^2 V/m, where x is in meters. Graph Ex versus x over the region −1 m ≤ x ≤1 m.
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Understand the given electric field expression: The electric field is given as \( E_x = -1000x^2 \) V/m, where \( x \) is the position in meters. This means the electric field depends on the square of \( x \), and it is negative, indicating direction.
Identify the range for \( x \): The problem specifies the range \( -1 \leq x \leq 1 \) meters. This will be the domain for the graph.
Determine the behavior of \( E_x \): Since \( E_x = -1000x^2 \), the electric field is always negative (due to the \( -1000 \) factor) and symmetric about \( x = 0 \) because \( x^2 \) is an even function. The maximum magnitude of \( E_x \) occurs at \( x = \pm 1 \).
Calculate key points for the graph: Evaluate \( E_x \) at specific values of \( x \) within the range, such as \( x = -1, -0.5, 0, 0.5, 1 \). For example, at \( x = 0 \), \( E_x = -1000(0)^2 = 0 \); at \( x = 1 \), \( E_x = -1000(1)^2 = -1000 \) V/m.
Plot the graph: On the horizontal axis, represent \( x \) from \( -1 \) to \( 1 \). On the vertical axis, represent \( E_x \) values. Plot the calculated points and connect them with a smooth curve, showing a parabolic shape opening downward (negative values) and symmetric about \( x = 0 \).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Electric Field
The electric field is a vector field that represents the force per unit charge experienced by a positive test charge placed in the field. It is defined mathematically as E = F/q, where E is the electric field, F is the force, and q is the charge. In this case, the electric field varies with position, specifically as a function of x, indicating how the force changes as the position changes.
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Intro to Electric Fields
Graphing Functions
Graphing functions involves plotting points on a coordinate system to visualize the relationship between variables. In this context, we are graphing the electric field Ex as a function of position x. Understanding how to interpret the graph helps in analyzing the behavior of the electric field across the specified range, revealing trends such as symmetry and the nature of the field's variation.
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Quadratic Functions
A quadratic function is a polynomial function of degree two, typically expressed in the form f(x) = ax^2 + bx + c. The given electric field Ex = -1000x^2 V/m is a quadratic function where the coefficient of x^2 is negative, indicating that the graph will be a downward-opening parabola. Recognizing the characteristics of quadratic functions is essential for accurately graphing and interpreting the electric field's behavior.
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Related Practice
Textbook Question
50 pJ of energy is stored in a 2.0 cm×2.0 cm×2.0 cm region of uniform electric field. What is the electric field strength?
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Textbook Question
The electric field in a region of space is Ex = −1000x2 V/m, where x is in meters. What is the potential difference between xi = −20 cm and xf = 30 cm?
Textbook Question
An infinitely long cylinder of radius R has linear charge density λ. The potential on the surface of the cylinder is V0, and the electric field outside the cylinder is Er = λ/2πϵ0r . Find the potential relative to the surface at a point that is distance r from the axis, assuming r>R.
Textbook Question
The electric field in a region of space is Ex=5000x V/m , where x is in meters. Find an expression for the potential V at position x. As a reference, let V=0 V at the origin.
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Textbook Question
Use the on-axis potential of a charged disk from Chapter 25 to find the on-axis electric field of a charged disk.
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