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

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Knight Calc 5th Edition

Ch 23: The Electric Field

Problem 68c

Chapter 23, Problem 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.

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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.

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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 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.

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.

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.

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Textbook Question

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