Problem 54 - University Physics with Modern Physics

Question

22.54. A Uniformiy Charged Stab. A slab of insulating material has thickness 2\(d\) and is oriented so that its faces are parallel to the \(y z\) -plane and given by the planes \(x=d\) and \(x=-d\) . The \(y\) - and \(z\) -dimensions of the slab are very large compared to \(d\) and may be treated as essentially infinite. The slab has a uniform positive charge density \(\rho\) . (a) Explain why the electric field due to the slab is zero at the center of the slab \((x=0)\) . (b) Using Gauss's law, find the electric field due to the slab (magnitude and direction) at all points in space.

Step-by-Step Solution

Answer

(a) Symmetrical contributions cancel at \( x = 0 \). (b) \( E = \frac{\rho \cdot x}{\varepsilon_0} \), directed outward from slab.

Step 1 - Analyze Symmetry

The slab is symmetric about the plane at \( x = 0 \), and the electric field lines in an infinite slab with uniform charge density are perpendicular to the slab's surfaces. At the center \( x = 0 \), contributions to the electric field from charges at \( x > 0 \) and \( x < 0 \) cancel each other out due to symmetry. Thus, the net electric field at \( x = 0 \) is zero.

Step 2 - Apply Gauss's Law

Gauss's law states that: \[ \oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\varepsilon_0} \] where \( Q_{\text{enc}} \) is the charge enclosed by the Gaussian surface. We choose a rectangular Gaussian surface inside the slab, with an area \( A \) that extends perpendicular to the \( x \)-axis. This encloses charge, affected by thickness \( 2x \) for a point at distance \( x \) from the center.

Step 3 - Calculate Charge Enclosed

The charge density is \( \rho \), thus the charge enclosed by the Gaussian surface is \( Q_{\text{enc}} = \rho \times 2xA \).

Step 4 - Determine Electric Field Expression

By substituting \( Q_{\text{enc}} = \rho \times 2xA \) into Gauss's law, we solve for the electric field \( E \): \[ E \cdot A = \frac{\rho \times 2xA}{\varepsilon_0} \] Hence, the electric field \( E \) is \[ E = \frac{\rho \cdot x}{\varepsilon_0} \].

Step 5 - Direction of Electric Field

The direction of the electric field is determined by the direction from the slab's center towards the surface, thus \( \vec{E} \) points away from the plane \( x = 0 \) towards both ±x sides. For points \( x > 0 \), \( \vec{E} \) points in the positive x-direction; for \( x < 0 \), in the negative x-direction.

Key Concepts

Electric Field

The electric field is a fundamental concept in electromagnetism, representing the force experienced by a positive test charge placed in the vicinity of other charges. In essence, it is a vector field around a charge where each point in that space can be described by a vector representing the force acting on a positive test charge. This concept helps us understand how charged objects interact with one another.

In the context of a uniformly charged slab, the electric field at any point is influenced by the slab's charge distribution. At the center of the slab, located at the plane where \( x = 0 \), the electric field is zero. This is because the slab is symmetrically charged, and the contributions to the electric field from one side of the center are canceled out by the opposite side.

The electric field direction is always perpendicular to the surface of the slab. As you move away from the center towards the slab surface, the electric field intensity increases, reflecting the symmetric distribution of charges.

Uniform Charge Distribution

A uniform charge distribution implies that the charge is spread evenly across the entire object, in this case, the insulating slab. This means every area of the slab has the same charge density, \(\rho\), simplifying the calculations of electric fields.

In uniformly charged materials, the field produced is consistent and predictable. The charge density plays a key role as it allows us to determine the total charge within any selected region, such as a Gaussian surface for applying Gauss's Law.

Through this uniform distribution, we achieve ease in calculating the electric field since any section of the slab exhibits identical charge characteristics. The concept of charge density, expressed in units of charge per unit volume, bridges our understanding of microscopic charge distribution to macroscopic electric field effects.

Symmetry in Physics

Symmetry is a powerful tool in physics that aids in simplifying complex problems. In the problem of the charged slab, symmetry about the center plane \(x = 0\) allows us to conclude that the net electric field at this point is zero, as field contributions from either side of this plane effectively neutralize each other.

Symmetry not only makes it easier to deduce the behavior of electric fields but is also crucial in applying Gauss's Law efficiently. By recognizing that symmetrical elements of the slab result in balancing field effects, we can reduce complex calculations regarding field vectors and focus on the essential attributes governing electric field magnitude and direction.

In this context, the symmetry simplifies visualization—enabling predictions about how the field behaves on either side of the symmetry plane, aligning field vectors perpendicularly to the slab surface.