Calculating the Electric Field Due to a Half-Infinite Line Charge: An Advanced Guide

Understanding the electric field produced by charge distributions is a fundamental concept in electrostatics. Consider a half-infinite line charge of uniform density (lambda) extending along the y-axis from the origin to (y infty). This article delves into the mathematical derivation to find the electric field at a point (P(x, 0)) using the principles of superposition and Coulomb's law.

Step-by-Step Guide

Step 1: Set Up the Problem

We consider a point (P(x, 0)) in the x-y plane and a charge distribution along the y-axis from the origin to infinity with a constant linear charge density (lambda). The goal is to calculate the electric field at point (P(x, 0)).

Step 2: Consider a Differential Element

A small charge element (dq) is located at (y), where (0 leq y leq infty). This charge element can be expressed as:

(dq lambda dy)

Step 3: Determine the Distance and Direction

The distance (r) from the charge element to point (P) is given by:

(r sqrt{x^2 y^2})

The electric field (dmathbf{E}) due to the charge element at point (P) is directed away from the charge (assuming (lambda > 0)) and has a magnitude given by Coulomb's law:

(dE frac{1}{4pi varepsilon_0} frac{dq}{r^2} frac{1}{4pi varepsilon_0} frac{lambda dy}{x^2 y^2})

Step 4: Resolve the Electric Field into Components

The electric field (dmathbf{E}) has components in both the x-direction and the y-direction. These components can be determined using trigonometry:

- The angle (theta) between the line connecting (dq) and point (P) and the x-axis is given by:

(tan theta frac{y}{x})

Thus, the components are:

(dmathbf{E}_x dE cos theta dE frac{x}{sqrt{x^2 y^2}} frac{1}{4pi varepsilon_0} frac{lambda dy}{x^2 y^2} frac{x}{sqrt{x^2 y^2}} frac{lambda x dy}{4pi varepsilon_0 (x^2 y^2)^{3/2}})

(dmathbf{E}_y dE sin theta dE frac{y}{sqrt{x^2 y^2}} frac{1}{4pi varepsilon_0} frac{lambda dy}{x^2 y^2} frac{y}{sqrt{x^2 y^2}} frac{lambda y dy}{4pi varepsilon_0 (x^2 y^2)^{3/2}})

Step 5: Integrate to Find Total Electric Field

X-Component

The x-component of the electric field is:

(E_x int_0^{infty} dmathbf{E}_x int_0^{infty} frac{lambda x dy}{4pi varepsilon_0 (x^2 y^2)^{3/2}})

Let (u x^2 y^2), then (du 2y dy) and (dy frac{du}{2sqrt{u - x^2}}). The limits change from (y 0) to (y infty) which corresponds to (u x^2) to (u infty). The integral becomes:

(E_x frac{lambda x}{4pi varepsilon_0} int_{x^2}^{infty} frac{1}{u^{3/2}} du frac{lambda x}{4pi varepsilon_0} left[-frac{2}{sqrt{u}}right]_{x^2}^{infty} frac{lambda x}{4pi varepsilon_0} cdot frac{2}{x} frac{lambda}{2pi varepsilon_0})

Y-Component

The y-component of the electric field (E_y) integrates in a similar manner but due to symmetry, the contributions from positive and negative y will cancel out:

(E_y 0)

Final Result

The total electric field (mathbf{E}) at point (P(x, 0)) is:

(mathbf{E} left(frac{lambda}{2pi varepsilon_0}, 0, 0right))

This indicates that the electric field due to the half-infinite line charge points in the positive x-direction and has a magnitude of (frac{lambda}{2pi varepsilon_0}).