Question

A very long, straight wire has charge per unit length \(3.20 \times 10^{-10}\) C/m. At what distance from the wire is the electric-field magnitude equal to 2.50 N\(/\)C?

Short answer

The distance from the wire is approximately 0.229 meters.

Step-by-step solution

Understanding the Problem

We need to find the distance from a long, straight wire at which the electric field magnitude is equal to 2.50 N/C. The wire has a linear charge density of \( \lambda = 3.20 \times 10^{-10} \) C/m.

Applying Gauss's Law for Electric Fields

According to Gauss's Law, the electric field \( E \) around a long straight wire with linear charge density \( \lambda \) can be given by the formula: \[ E = \frac{\lambda}{2 \pi \varepsilon_0 r} \] where \( \varepsilon_0 \) is the permittivity of free space \( (8.85 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2) \), and \( r \) is the distance from the wire.

Rearranging the Equation

We need to solve for \( r \), so rearrange the equation: \[ r = \frac{\lambda}{2 \pi \varepsilon_0 E} \] Substitute \( E = 2.50 \, \text{N/C} \) and \( \lambda = 3.20 \times 10^{-10} \, \text{C/m} \).

Calculating the Distance

Plug in the values:\[ r = \frac{3.20 \times 10^{-10}}{2 \pi \times 8.85 \times 10^{-12} \times 2.50} \] Simplify to find the distance \( r \).

Performing the Calculation

Calculate the expression:\[ r = \frac{3.20 \times 10^{-10}}{2 \times 3.1416 \times 8.85 \times 10^{-12} \times 2.50} \approx 0.229 \, \text{m} \] Therefore, the distance from the wire is approximately 0.229 meters.

Key concepts

Gauss's Law

Gauss's Law is a fundamental principle in electromagnetism that relates the distribution of electric charge to the resulting electric field. It provides a way to calculate electric field magnitudes by considering a hypothetical closed surface, known as a "Gaussian surface," which encloses the charge.

The law states that the total electric flux through a closed surface is proportional to the enclosed electric charge. Mathematically, it is expressed as:

• \( \Phi_E = \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{enclosed}}{\varepsilon_0} \)

Here, \( \Phi_E \) is the electric flux, \( \mathbf{E} \) is the electric field, \( d\mathbf{A} \) is the differential area vector of the surface, \( Q_{enclosed} \) is the enclosed charge, and \( \varepsilon_0 \) is the permittivity of free space.

For a long, straight, and uniformly charged wire, we often choose a cylindrical Gaussian surface with the wire as the axis. This choice leverages the symmetry of the problem, simplifying calculations due to the uniform radial electric field at every point on the cylindrical surface. By applying Gauss's Law, we derive a specific formula for the electric field around such a wire:

• \( E = \frac{\lambda}{2 \pi \varepsilon_0 r} \)

Where \( \lambda \) is the linear charge density, and \( r \) is the radial distance from the wire.

Linear Charge Density

Linear charge density, often denoted by \( \lambda \), describes how charge is distributed along a line, such as a wire. It is defined as the amount of electric charge per unit length of the line.

This concept is crucial when dealing with problems involving long conductors, as it allows for a simplified description of how charges create electric fields in space. The unit for linear charge density is coulombs per meter (C/m).

For a uniformly charged wire, the linear charge density remains constant along its length. In practical calculations, knowing the value of \( \lambda \) is necessary to apply formulas from electromagnetic theory, such as those derived from Gauss's Law, to find electric field strength or potential.

In the given problem, the wire's linear charge density is \( 3.20 \times 10^{-10} \) C/m. This represents a small amount of charge, which is typical when dealing with highly detailed microscale physical configurations.

Permittivity of Free Space

The permittivity of free space, also known as the electric constant, is a fundamental physical constant important in the study of electromagnetism. It sets the scale for electric forces in a vacuum and is denoted by \( \varepsilon_0 \).

Numerically, \( \varepsilon_0 \) is approximately \( 8.85 \times 10^{-12} \) C²/(N·m²). This constant appears in several fundamental equations, including Gauss's Law, and in the calculation of capacitance and electric field in both free space and materials.

By understanding the role of \( \varepsilon_0 \), one can appreciate how it influences electromagnetic interactions. Specifically, it quantifies how much electric field can permeate a vacuum -- essentially setting the baseline for the strength of electric fields in any given medium.

In the context of solving the problem, \( \varepsilon_0 \) ensures units are consistent and calculations are accurate when computing the electric field at a distance \( r \) from a charged wire, as shown using the formula \( E = \frac{\lambda}{2 \pi \varepsilon_0 r} \). Its presence ensures that the effects of electric forces are correctly scaled to real-world observations.