A rod of length L lies along the y-axis with its center at the origin. The rod has a nonuniform linear charge density λ=a∣y∣ λ=a|y|, where a is a constant with the units C/m2. Find the electric field strength of the rod at distance x on the x-axis.

Ch 23: The Electric Field

Knight Calc - Physics for Scientists and Engineers 5th Edition

Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Non è quello che usi tu?Cambia libro di testo

Tutti i libri di testo

Knight Calc 5th Edition

Ch 23: The Electric Field

Problema 68c

Capitolo 23, Problema 68c

A rod of length L lies along the y-axis with its center at the origin. The rod has a nonuniform linear charge density $λ = a ∣ y ∣$ , where a is a constant with the units C/m². Find the electric field strength of the rod at distance x on the x-axis.

Guida verificata passo dopo passo

Understand the problem: The rod lies along the y-axis, centered at the origin, with a nonuniform linear charge density λ = a|y|. We need to calculate the electric field strength at a point on the x-axis at a distance x from the origin. The electric field contributions from each infinitesimal charge element on the rod will be integrated to find the total field.

Set up the coordinate system: Consider an infinitesimal charge element dq on the rod at a position y along the y-axis. The linear charge density is λ = a|y|, so dq = λ dy = a|y| dy. The distance from this charge element to the point on the x-axis is r = √(x² + y²).

Determine the electric field contribution: The infinitesimal electric field dE due to dq at the point on the x-axis has a magnitude given by dE = (1 / (4πε₀)) * (dq / r²). Substituting dq = a|y| dy and r = √(x² + y²), we get dE = (1 / (4πε₀)) * (a|y| dy) / (x² + y²).

Resolve the components of the electric field: The electric field has both x and y components. Due to symmetry, the y-components of the electric field from charges at +y and -y cancel out. Only the x-component contributes to the total field. The x-component of dE is dEx = dE * (x / r) = (1 / (4πε₀)) * (a|y| dy * x) / (x² + y²)^(3/2).

Integrate to find the total electric field: Integrate dEx over the length of the rod from y = -L/2 to y = L/2. The integral is Ex = ∫[(1 / (4πε₀)) * (a|y| x) / (x² + y²)^(3/2)] dy, where the limits of integration are y = -L/2 to y = L/2. Solve this integral to find the total electric field strength at the point on the x-axis.

Risposta video verificata per un problema simile:

Questa soluzione video è stata consigliata dai nostri tutor come utile per risolvere questo problema.

Durata del video:

10m

Concetti chiave

Ecco i concetti essenziali che devi comprendere per rispondere correttamente alla domanda.

Electric Field

The electric field is a vector field that represents the force exerted by an electric charge on other charges in its vicinity. It is defined as the force per unit charge experienced by a positive test charge placed in the field. The direction of the electric field is away from positive charges and towards negative charges, and its strength diminishes with distance from the charge.

Video consigliato:

Percorso guidato

03:16

Intro to Electric Fields

Linear Charge Density

Linear charge density (λ) is defined as the amount of electric charge per unit length along a line, typically measured in coulombs per meter (C/m). In this problem, the charge density is nonuniform, given by λ = a|y|, indicating that the charge density varies with the position along the y-axis. This variation affects the calculation of the electric field at a point in space.

Video consigliato:

Percorso guidato

8:13

Intro to Density

Superposition Principle

The superposition principle states that the total electric field created by multiple charges is the vector sum of the electric fields produced by each charge individually. This principle is crucial when dealing with continuous charge distributions, such as the rod in this problem, as it allows us to integrate the contributions of infinitesimal charge elements along the length of the rod to find the resultant electric field at a specific point.

Video consigliato:

Percorso guidato

03:32

Superposition of Sinusoidal Wave Functions

Pratica correlata

Domanda del libro di testo

An infinitely long sheet of charge of width L lies in the xy-plane between x = -L /2 and x = L /2. The surface charge density is h. Draw a graph of field strength E versus x for x > L /2.

448

views

Domanda del libro di testo

A proton orbits a long charged wire, making $1.0 \times 10^{6}$ revolutions per second. The radius of the orbit is $1.0$ cm. What is the wire's linear charge density?

1927

views

Domanda del libro di testo

A rod of length $L$ lies along the $y$ -axis with its center at the origin. The rod has a nonuniform linear charge density $λ = a ∣ y ∣$ , where a is a constant with the units C/m². Draw a graph of $λ$ versus $y$ over the length of the rod.

363

views

Domanda del libro di testo

One type of ink-jet printer, called an electrostatic ink-jet printer, forms the letters by using deflecting electrodes to steer charged ink drops up and down vertically as the ink jet sweeps horizontally across the page. The ink jet forms 30-μm-diameter drops of ink, charges them by spraying 800,000 electrons on the surface, and shoots them toward the page at a speed of 20 m/s . Along the way, the drops pass through two horizontal, parallel electrodes that are 6.0 mm long, 4.0 mm wide, and spaced 1.0 mm apart. The distance from the center of the electrodes to the paper is 2.0 cm. To form the tallest letters, which have a height of 6.0 mm, the drops need to be deflected upward (or downward) by 3.0 mm. What electric field strength is needed between the electrodes to achieve this deflection? Ink, which consists of dye particles suspended in alcohol, has a density of 800 kg/m³.

1681

views

rank

Domanda del libro di testo

A rod of length $L$ lies along the $y$ -axis with its center at the origin. The rod has a nonuniform linear charge density $λ=a|y|$ , where a is a constant with the units C/m². Determine the constant a in terms of $L$ and the rod's total charge $Q$ .

1860

views

Domanda del libro di testo

An electric field can induce an electric dipole in a neutral atom or molecule by pushing the positive and negative charges in opposite directions. The dipole moment of an induced dipole is directly proportional to the electric field. That is, $\vec{p} = α \vec{E}$ , where α is called the polarizability of the molecule. A bigger field stretches the molecule farther and causes a larger dipole moment. An ion with charge q is distance r from a molecule with polarizability α. Find an expression for the force _{$\vec{E}$ ion on dipole}.

2209

views