Question

A slab of insulating material has thickness 2d and is oriented so that its faces are parallel to the yz-plane and given by the planes \(x = d\) and \(x = -d\). The \(y\)- and \(z\)-dimensions of the slab are very large compared to \(d\); treat them 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.

Short answer

(a) The electric field at the slab's center is zero due to symmetry. (b) Inside, \(E = \frac{\rho \, x}{\varepsilon_0}\); outside, \(E = \frac{\rho \, d}{\varepsilon_0}\).

Step-by-step solution

Understand the Electric Field at the Center

The slab has symmetry around its center at x = 0, meaning the distribution of charge is identical on either side of this plane. Therefore, electric fields generated by charges on both sides counterbalance each other due to symmetry. Consequently, at x = 0, the electric field \(E = 0\).

Apply Gauss's Law

Gauss's Law relates the electric flux through a closed surface to the charge enclosed by that surface. The law is given by the equation: \(\Phi_E = \int \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\varepsilon_0}\), where \(\varepsilon_0\) is the vacuum permittivity.

Design a Gaussian Surface

Choose a Gaussian surface as a rectangular box that extends from \(x = -x_0\) to \(x = x_0\), and encloses the charge within these limits. The ends of the box are parallel to the yz-plane such that the electric field is perpendicular to these planes.

Calculate the Enclosed Charge

The charge density is uniform with value \(\rho\), so the enclosed charge \(Q_{enc}\) is given by \(Q_{enc} = \rho \cdot A \cdot 2x_0\), where \(A\) is the area in the yz-plane, and \(2x_0\) is the width of the Gaussian surface.

Calculate the Electric Field Using Gauss's Law

The total electric flux through the Gaussian surface is \(\Phi_E = E \, A + E \, A = 2E \, A\). Applying Gauss's Law gives: \(2E \, A = \frac{\rho \cdot A \cdot 2x_0}{\varepsilon_0}\). Simplifying, \(E = \frac{\rho \, x_0}{\varepsilon_0}\). This shows that the magnitude of the electric field inside the slab is proportional to the distance \(x_0\) from the center.

Determine the Electric Field Outside the Slab

Outside the slab, the entire charge density is enclosed. Using Gauss's Law, \(E \, A = \frac{Q_{enc}}{\varepsilon_0}\), where \(Q_{enc} = \rho \, A \, 2d\). The electric field outside the slab is \(E = \frac{\rho \, d}{\varepsilon_0}\), with direction pointing away from the slab on both sides due to positive charge.

Key concepts

Electric Field

The electric field is a crucial concept in understanding electric forces. It represents how a charged object influences the space around it. In the case of the infinite insulating slab, the electric field is determined by the charge distribution within the slab.
At the center of the slab, specifically at the plane where \( x = 0 \), the electric field is zero. This is due to the symmetric distribution of charge on either side of the center. The fields from each half of the slab cancel each other out exactly at this point.
This symmetry ensures that no net electric field exists at the center, making it a point of equilibrium where the electric force on a charge would be zero.

Charge Density

Charge density \( \rho \) is a measure of how much electric charge is present per unit volume of the material. For an insulating slab, this is uniform, meaning every part of the slab has the same amount of charge.
The concept of charge density is essential when applying Gauss's Law as it helps to calculate the total charge enclosed by a chosen surface.

• Uniform charge density implies that the slab's electric field can be calculated consistently across different regions of the slab.
• In our example, the charge density is a constant \( \rho \), making it straightforward to compute the enclosed charge, which is necessary for determining the electric field using Gauss's law.

Symmetry

Symmetry plays a significant role in determining electric fields, especially in systems with regular shapes and uniform charge distributions.
In our slab, the symmetry around the x-axis allows us to conclude that the electric field at the center of the slab (\( x = 0 \)) is zero.
Here, symmetry implies that the physical situation looks the same on both sides of the center. Thus, any electric field generated by the charges on one side is canceled by the field from charges on the opposite side.

• Such symmetric setups simplify calculation efforts as they allow us to ignore complex field contributions in specific regions.
• When applying Gauss's Law, symmetry helps in the selection of a Gaussian surface that captures the essential characteristics of the field distribution.

Gaussian Surface

A Gaussian Surface is an imaginary closed surface used to apply Gauss's Law conveniently. It is chosen based on symmetry considerations to simplify calculations of electric fields.
For the insulating slab, a Gaussian surface in the form of a rectangular box is adequate. It stretches across the slab from \( x = -x_0 \) to \( x = x_0 \), extending infinitely in the y and z directions.
This choice ensures that the electric field is always perpendicular to the surface it intercepts, making calculation of the electric flux attainable.

• By enclosing charge within the slab, the Gaussian Surface becomes instrumental in using Gauss's Law to find the electric field.
• The effectiveness of this method lies in the ability to relate the total charge enclosed by the surface to the electric field across the surface.
• Ultimately, the Gaussian Surface acts as a bridge between the physical charge distribution and the mathematical application of Gauss's Law.