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Chapter 24: Problem 38

Planar Slab A planar slab of thickness \(d\) has a uniform volume charge density \(\rho\). Find the magnitude of the electric field at all points in space both (a) within and (b) outside the slab, in terms of \(x\), the distance measured from the central plane of the slab.

Short Answer

Expert verified
Within the slab: \( E = \frac{\rho x}{2 \epsilon_0} \).Outside the slab: \( E = \frac{\rho d}{2 \epsilon_0} \).

Step by step solution

01

Define the parameters

Identify the given parameters from the problem. Thickness of the slab is denoted as \(d\), the uniform volume charge density is \(\rho\), the position relative to the central plane is \(x\).
02

Concept of planar slab and symmetry

Due to the symmetry of the problem, the electric field will only depend on the distance \(x\) from the central plane and will be directed perpendicularly to the plane of the slab.
03

Apply Gauss's Law for within the slab

For points within the slab (\text{for } -d/2 \leq x \leq d/2 \text{ }), use a Gaussian pill-box that encloses charge density. The enclosed charge \(Q_{enc} = \rho Ax\), where \(A\) is the area of the pill-box and \(x\) is half of the pill-box thickness. Applying Gauss's law, \[ E \cdot 2A = \frac{\rho Ax}{\epsilon_0} \]Thus, \[ E = \frac{\rho x}{2 \epsilon_0}\]
04

Apply Gauss's Law for outside the slab

For points outside the slab (\text{for } x> d/2 \text{ or } x< -d/2 \text{ }), use a Gaussian surface that encloses the entire slab thickness. The enclosed charge now is \(Q_{enc} = \rho Ad\). Applying Gauss's law, \[ E \cdot A = \frac{\rho Ad}{\epsilon_0} \]Thus, \[ E = \frac{\rho d}{2 \epsilon_0} \]
05

Synthesize the general result

Combine the results for within and outside the slab. Therefore, the electric field (1) within the slab is given by \( E = \frac{\rho x}{2 \epsilon_0} \), and (2) outside the slab is given by \( E = \frac{\rho d}{2 \epsilon_0} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gauss's Law
Gauss's Law is a vital principle in electromagnetism. It relates the electric flux passing through a closed surface to the charge enclosed within that surface. Mathematically, Gauss's Law is expressed as \[\oint_{\text{surface}} \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\epsilon_0}.,\] where \mathbf{E}\ is the electric field, \mathbf{A}\ is the vector area of the surface, \Q_{\text{enc}}\ is the enclosed charge, and \epsilon_0\ is the permittivity of free space. This law helps us calculate the electric field when symmetry simplifies the setup. For the planar slab problem, we use Gauss's Law to determine the electric field both inside and outside the slab. We imagine a Gaussian surface (a pill-box shape) around regions of interest. Within the slab, we consider the partial volume, whereas outside we consider the whole volume of the slab.
Symmetry in Electric Fields
Symmetry in electric fields simplifies complex problems significantly. For a planar slab with a uniform volume charge density, the symmetry is planar. This means that the electric field will be perpendicular to the slab and uniform in magnitude at any given distance \(x\) from the central plane. Due to this planar symmetry, the determination of the electric field becomes more straightforward. The electric field on each side of the central plane mirrors each other. This allows us to use Gaussian surfaces effectively, as the electric field on both ends of the surface will be equal and opposite, canceling out any tangential components.
Uniform Volume Charge Density
Uniform volume charge density means that the charge is distributed evenly throughout the volume of the slab. If \rho\ is the charge density, then for any given volume \V\, the charge \[Q = \rho V\]. When dealing with a planar slab, this uniform distribution impacts how the electric field is calculated across different regions.
  • Within the slab: The charge enclosed by a Gaussian surface increases linearly with the distance from the central plane. This results in an electric field that also increases linearly within the slab, given by \[ E = \frac{\rho x}{2 \epsilon_0}\].
  • Outside the slab: The enclosed charge remains constant, equal to \rho Ad\. Here, the electric field is uniform and can be calculated using the formula \[ E = \frac{\rho d}{2 \epsilon_0}\].
    Understanding uniform volume charge density is crucial as it underpins the linear and uniform properties of the electric field in this problem.

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Most popular questions from this chapter

Charge Is Distributed Uniformly Charge is distributed uniformly throughout the volume of an infinitely long cylinder of radius \(R\). (a) Show that, at a distance \(r\) from the cylinder axis (for \(rR\)

If/Can If the electric field in a region of space is zero, can you conclude there are no electric charges in that region? Explain.

Geiger Counter Figure \(24-33\) shows a Geiger counter, a device used to detect ionizing radiation (radiation that causes ionization of atoms). The counter consists of a thin, positively charged central wire surrounded by a concentric, circular, conducting cylinder with an equal negative charge. Thus, a strong radial electric field is set up inside the cylinder. The cylinder contains a low-pressure inert gas. When a particle of radiation enters the device through the cylinder wall, it ionizes a few of the gas atoms The resulting free electrons (labelled e) are drawn to the positive wire. However, the electric field is so intense that, between collisions with other gas atoms, the free electrons gain energy sufficient to ionize these atoms also. More free electrons are thereby created, and the process is repeated until the electrons reach the wire. The resulting "avalanche" of electrons is collected by the wire generating a signal that is used to record the passage of the original particle of radiation. Suppose that the radius of the central wire is \(25 \mu \mathrm{m}\), the radius of the cylinder \(1.4 \mathrm{~cm}\), and the length of the tube \(16 \mathrm{~cm}\). If the electric field component \(E_{r}\) at the cylinder's inner wall is \(+2.9 \times\) \(10^{4} \mathrm{~N} / \mathrm{C}\), what is the total positive charge on the central wire?

Uniform Volume Charge Density Figure 24-29a shows a spherical shell of charge with uniform volume charge density \(\rho .\) Plot \(E\) due to the shell for distances \(r\) from the center of the shell ranging from zero to \(30 \mathrm{~cm}\). Assume that \(\rho=1.0 \times 10^{-6} \mathrm{C} / \mathrm{m}^{3}, a=10 \mathrm{~cm}\), and \(b=20 \mathrm{~cm}\).

Two Concentric Cylinders Two long, charged, concentric cylinders have radii of \(3.0\) and \(6.0 \mathrm{~cm}\). Assume the outer cylinder is hollow. The charge per unit length is \(5.0 \times 10^{-6} \mathrm{C} / \mathrm{m}\) on the inner cylinder and \(-7.0 \times 10^{-6} \mathrm{C} / \mathrm{m}\) on the outer cylinder. Find the electric field at (a) \(r=4.0 \mathrm{~cm}\) and (b) \(r=8.0 \mathrm{~cm}\), where \(r\) is the radial distance from the common central axis.
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