Domanda del libro di testo
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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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 38b
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.
Guida verificata passo dopo passo1
Step 1: Recall the relationship between electric field and electric potential. The electric field is the negative gradient of the electric potential: \( E_x = -\frac{dV}{dx} \).
Step 2: Substitute the given electric field \( E_x = 5000x \) into the equation \( E_x = -\frac{dV}{dx} \). This gives \( -\frac{dV}{dx} = 5000x \).
Step 3: Rearrange the equation to solve for \( dV \): \( dV = -5000x \, dx \).
Step 4: Integrate both sides to find the potential \( V \). The integral of \( dV \) is \( V \), and the integral of \( -5000x \, dx \) is \( -2500x^2 \) (using the power rule for integration).
Step 5: Apply the boundary condition \( V = 0 \) at \( x = 0 \) to determine the constant of integration. Since \( V = -2500x^2 + C \), and \( V = 0 \) when \( x = 0 \), \( C = 0 \). Thus, the expression for the potential is \( V(x) = -2500x^2 \).
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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. It is defined mathematically as the negative gradient of the electric potential (V), indicating how the potential changes with position. In this case, the electric field is given as Ex = 5000x V/m, which shows that the field strength varies linearly with the position x.
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Percorso guidato
03:16
Intro to Electric Fields
Electric Potential
Electric potential (V) is a scalar quantity that represents the potential energy per unit charge at a point in an electric field. It is related to the electric field by the equation V = -∫E·dx, where the integral is taken along a path from a reference point to the point of interest. In this problem, we need to find the expression for V at position x, using the given electric field and the reference point where V = 0 V at the origin.
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Percorso guidato
07:33Electric Potential
Integration in Physics
Integration is a fundamental mathematical tool used in physics to find quantities that accumulate over a continuous range, such as area under a curve or total work done. In the context of electric fields and potentials, integration allows us to calculate the potential difference by summing the contributions of the electric field over a distance. For this problem, we will integrate the electric field expression to derive the potential function V(x).
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11:43Finding Moment Of Inertia By Integrating
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