Problem 59 - Fundamentals of Physics

Question

Charge of uniform volume density \(\rho=1.2 \mathrm{nC} / \mathrm{m}^{3}\) fills an infinite slab between \(x=-5.0 \mathrm{~cm}\) and \(x=+5.0 \mathrm{~cm} .\) What is the magnitude of the electric field at any point with the coordinate (a) \(x=4.0 \mathrm{~cm}\) and (b) \(x=6.0 \mathrm{~cm} ?\)

Step-by-Step Solution

Answer

(a) 5.42 N/C, (b) 6.78 N/C.

Step 1: Understanding the Problem

We have a uniformly charged infinite slab with charge density \(\rho = 1.2 \, \text{nC/m}^3\) spread uniformly between \(x = -5.0 \, \text{cm}\) and \(x = +5.0 \, \text{cm}\). We are asked to find the electric field at specific points outside and within this charged slab.

Step 1: Applying Gauss's Law

We will use Gauss's Law, which states that the electric flux through a closed surface is equal to \(\frac{1}{\varepsilon_0}\) times the enclosed charge. The electric field \(E\) due to an infinite slab at a distance \(x\) from its center is given by \(E = \frac{\rho \cdot x}{\varepsilon_0}\) if \(-5 \leq x \leq 5\) cm, and \(E = \frac{\rho \cdot L}{\varepsilon_0}\) where \(L\) is the half-thickness for points beyond this range. Here, \(L = 5 \, \text{cm} = 0.05 \, \text{m}\).

Step 2: Calculating the Electric Field at \(x=4.0 \, \text{cm}\)

Since \(4.0 \, \text{cm}\) is within the slab, use \(E = \frac{\rho \cdot x}{\varepsilon_0}\). Convert \(x\) to meters: \(x = 0.04 \, \text{m}\). Substitute \(\rho = 1.2 \, \text{nC/m}^3 = 1.2 \times 10^{-9} \, \text{C/m}^3\) and \(\varepsilon_0 = 8.85 \times 10^{-12} \, \text{F/m}\). The equation becomes \(E = \frac{1.2 \times 10^{-9} \times 0.04}{8.85 \times 10^{-12}}\). Calculate to get \(E \approx 5.42 \, \text{N/C}\).

Step 3: Calculating the Electric Field at \(x=6.0 \, \text{cm}\)

Since \(6.0 \, \text{cm}\) is outside the slab, use \(E = \frac{\rho \cdot L}{\varepsilon_0}\). Replace \(L = 0.05 \, \text{m}\). Substitute \(\rho = 1.2 \times 10^{-9} \, \text{C/m}^3\) and \(\varepsilon_0 = 8.85 \times 10^{-12} \, \text{F/m}\). The equation becomes \(E = \frac{1.2 \times 10^{-9} \times 0.05}{8.85 \times 10^{-12}}\). Calculate to find \(E \approx 6.78 \, \text{N/C}\).

Key Concepts

Gauss's Law

Gauss's Law is a fundamental principle in electrostatics, which relates the electric field emanating from a charged object to the charge enclosed within a closed surface. The law states that the total electric flux through a closed surface is proportional to the charge enclosed by that surface. Mathematically, Gauss's Law is expressed as: \[ \Phi_E = \oint_S \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{ ext{enc}}}{\varepsilon_0} \]where \( \Phi_E \) is the electric flux, \( \mathbf{E} \) is the electric field, \( d\mathbf{A} \) is a differential area element of the surface, and \( \varepsilon_0 \) is the permittivity of free space.

In applications such as an infinite slab, Gauss's Law simplifies the calculation of the electric field because it leverages the symmetry of the problem. By choosing a Gaussian surface that matches the symmetry, like a box through the slab, calculating the electric field becomes a straightforward process. This is because the charge enclosed within such a surface is easily determined, allowing for direct calculation.

Charge Density

Charge density (\( \rho \)) is a measure of how much electric charge is accumulated in a particular region of space. It is defined as the amount of charge per unit volume and is expressed in units such as Coulombs per cubic meter (C/m³).

Understanding charge density is key when dealing with distributed charges, like that within an infinite slab. In the given exercise, the slab has a uniform charge density, meaning the charge per unit volume remains constant throughout the slab.

This uniformity allows us to use formulas that rely on a consistent distribution of charge, making the calculations for the electric field more manageable. For the infinite slab, this uniform distribution results in a straightforward application of Gauss's Law, as the enclosed charge within any Gaussian surface can be easily calculated from the volume and charge density.

Infinite Slab

An infinite slab is an idealized concept in physics where a flat layer of material extends infinitely in two dimensions. The practical relevance of this concept comes from its simplicity in calculating electric fields.

When dealing with an infinite slab, it is assumed to have a consistent thickness but extends infinitely in the other dimensions. This idealization results in a uniform electric field in regions of space close to the slab and allows for the use of symmetry to simplify calculations.

• Within the slab: The electric field is proportional to the distance from the center of the slab, derived from the charge density and depth within the slab.
• Outside the slab: The electric field becomes constant because the effective area contributing charge remains the same as the distance increases beyond the edge of the slab.

These properties make infinite slabs an excellent model for understanding basic principles in electromagnetism and are especially useful for approximating solutions in finite systems where edge effects are negligible.