"Electric field due to a line of charge A total charge of Q is distributed uniformly on a line segment of length 2L along the y-axis (see figure). The x-component of the electric field at a point (a, 0) is given by
Eₓ(a) = (kQa/2L) ∫-L L dy/(a² + y²)^(3/2),
where k is a physical constant and a > 0.
a. Confirm that Eₓ(a)=kQ / a √(a²+L²)
b. Letting ρ=Q / 2 L be the charge density on the line segment, show that if L → ∞, then Eₓ(a) = 2kρ / a.

Verified step by step guidance

1. Start with the given expression for the x-component of the electric field: \(E_x(a) = \frac{kQa}{2L} \int_{-L}^{L} \frac{dy}{(a^2 + y^2)^{3/2}}\).
2. Recognize that the integral is an even function because the integrand depends on \(y^2\). Therefore, rewrite the integral as: \(\int_{-L}^{L} \frac{dy}{(a^2 + y^2)^{3/2}} = 2 \int_0^{L} \frac{dy}{(a^2 + y^2)^{3/2}}\).
3. Use the substitution \(y = a \tan \theta\), which implies \(dy = a \sec^2 \theta d\theta\). Also, note that \(a^2 + y^2 = a^2 \sec^2 \theta\). Substitute these into the integral to transform it into an integral in terms of \(\theta\).
4. Simplify the integral after substitution: \(\int_0^{L} \frac{dy}{(a^2 + y^2)^{3/2}} = \int_0^{\arctan(L/a)} \frac{a \sec^2 \theta d\theta}{(a^2 \sec^2 \theta)^{3/2}} = \int_0^{\arctan(L/a)} \frac{a \sec^2 \theta d\theta}{a^3 \sec^3 \theta} = \int_0^{\arctan(L/a)} \frac{\sec^2 \theta}{a^2 \sec^3 \theta} d\theta = \frac{1}{a^2} \int_0^{\arctan(L/a)} \cos \theta d\theta\).
5. Evaluate the integral \(\int_0^{\arctan(L/a)} \cos \theta d\theta = \sin \theta \big|_0^{\arctan(L/a)} = \sin(\arctan(L/a))\). Use the right triangle relationship to express \(\sin(\arctan(L/a)) = \frac{L}{\sqrt{a^2 + L^2}}\). Substitute back to get the integral value and then multiply by the constants outside the integral to confirm that \(E_x(a) = \frac{kQ}{a \sqrt{a^2 + L^2}}\).

Key Concepts

Electric Field Due to a Continuous Charge Distribution

The electric field at a point due to a continuous charge distribution is found by integrating the contributions from each infinitesimal charge element. For a line charge, the total field is the integral of the field contributions from each segment of the line, considering their positions and distances relative to the point of interest.

Definite Integral and Substitution in Calculus

Evaluating the electric field involves solving a definite integral with limits from -L to L. Techniques such as substitution simplify the integral, especially when dealing with expressions like (a² + y²)^(3/2). Understanding how to manipulate and evaluate these integrals is essential to derive the closed-form expression for the electric field.

Limit Process and Charge Density Concept

The charge density ρ = Q / 2L represents the uniform charge per unit length. Taking the limit as L approaches infinity models an infinitely long line charge, simplifying the electric field expression. Understanding limits helps analyze how the field behaves for very large charge distributions and connects physical intuition with mathematical results.