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Chapter 22: Problem 54

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.

Short Answer

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

Step by step solution

01

- 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.
02

- 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.
03

- Calculate Charge Enclosed

The charge density is \( \rho \), thus the charge enclosed by the Gaussian surface is \( Q_{\text{enc}} = \rho \times 2xA \).
04

- 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 \[ E = \frac{\rho \cdot x}{\varepsilon_0} \].
05

- 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 卤x sides. For points \( x > 0 \), \( \vec{E} \) points in the positive x-direction; for \( x < 0 \), in the negative x-direction.

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Key Concepts

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

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鈥攅nabling predictions about how the field behaves on either side of the symmetry plane, aligning field vectors perpendicularly to the slab surface.

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Most popular questions from this chapter

22.42. A Sphere in a Sphere. A solid conducting sphere carry- ing charge \(q\) has radius a. It is inside a concentric hollow conducting sphere with inner radius \(b\) and outer radius \(c\) . The hollow sphere has no net charge. (a) Derive expressions for the electric-field magnitude in terms of the distance \(r\) from the center for the regions \(rc .\) (b) Graph the magnitude of the electric field as a function of \(r\) from \(r=0\) to \(r=2 c\) (c) What is the charge on the inner surface of the hollow sphere? (d) On the outer surface? (e) Represent the charge of the small sphere by four plus signs. Sketch the field lines of the system within a spherical volume of radius 2 c.

22.52. Thomson's Model of the Atom. In the early years of the 20 th century, a leading model of the structure of the atom was that of the English physicist. I. Thomson (the discoverer of the electron). In Thomson's model, an atom consisted of a sphere of positively charged material in which were embedded negatively charged electrons, like chocolate chips in a ball of cookie dough. Consider such an atom consisting of one electron with mass \(m\) and charge \(-e\) , which may be regarded as a point charge, and a uniformly charged sphere of charge \(+e\) and radius \(R\) (a) Explain why- - the equilibrium position of the electron is at the center of the nucleus. (b) In Thomson's model, it was assumed that the positive material provided little or no resistance to the motion of the electron. If the electron is displaced from equilibrium by a distance less than \(R,\) show that the resulting motion of the electron will be simple harmonic, and calculate the frequency of oscillation. (Hint: Review the definition of simple harmonic motion in Section 13.2 . If it can be shown that the net force on the electron is of this form, then it follows that the motion is simple harmonic. Conversely, if the net force on the electron does not follow this form, the motion is not simple harmonic.) (c) By Thomson's time, it was known that excited atoms emit light waves of only certain frequencies. In his model, the frequency of emitted light is the same as the oscillation frequency of the electron or electrons in the atom. What would the radius of a Thomson-model atom have to be for it to produce red light of frequency \(4.57 \times 10^{14} \mathrm{Hz}\) . Compare your answer to the radii of real atoms, which are of the order of \(10^{-10} \mathrm{m}\) (see Appendix F for data about the electron). (d) If the electron were displaced from equilibrium by a distance greater than \(R\) , would the electron oscillate? Would its motion be simple harmonic? Explain your reasoning. (Historical note: In 1910 , the atomic nucleus was discovered, proving the Thomson model to be incorrect. An atom's positive charge is not spread over its volume as Thomson sup- posed, but is concentrated in the tiny nucleus of radius \(10^{-14}\) to \(10^{-13} \mathrm{m} .\)

22.36. A long line carying a uniform linear charge density \(+50.0 \mu C / m\) runs parallel to and 10.0 \(\mathrm{cm}\) from the surface of a large, flat plastic sheet that has a uniform surface charge density of \(-100 \mu \mathrm{C} / \mathrm{m}^{2}\) on one side. Find the location of all points where an \(\alpha\) particle would feel no force due to this arrangement of charged objects.

22.15. A point charge of \(+5.00 \mu C\) is located on the \(x\) -axis at \(x=4.00 \mathrm{m},\) next to a spherical surface of radius 3.00 \(\mathrm{m}\) centered at the origin. (a) Calculate the magnitude of the electric field at \(x=3.00 \mathrm{m} .\) (b) Calculate the magnitude of the electric field at \(x=-3.00 \mathrm{m} .\) (c) According to Gauss's law, the net flux through the sphere is zero because it contains no charge. Yet the field due to the external charge is much stronger on the near side of the sphere (i.e., at \(x=3.00 \mathrm{m}\) ) than on the far side (at \(x=-3.00 \mathrm{m}\) ). How, then, can the flux into the sphere (on the near side) equal the flux out of it (on the far side)? Explain. A sketch will help.

22.48. A solid conducting sphere with radius \(R\) carries a positive total charge \(Q .\) The sphere is surrounded by an insulating shell with inner radius \(R\) and outer radius 2\(R\) . The insulating shell has a uniform charge density \(\rho\) (a) Find the value of \(\rho\) so that the net charge of the entire system is zero. (b) If \(\rho\) has the value found in part (a), find the electric field (magnitude and direction) in each of the regions \(02 R\) . Show your results in a graph of the radial component of \(\overrightarrow{\boldsymbol{E}}\) as a function of \(\boldsymbol{r}\) . (c) As a general rule, the electric field is discontinuous only at locations where there is a thin sheet of charge. Explain how your results in part (b) agree with this rule.
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