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Chapter 22: Problem 56

A Uniformly Charged Slab. 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
At \(x=0\), \(E = 0\). Inside the slab \((-d < x < d)\), \(E = \frac{\rho x}{\varepsilon_0}\). Outside the slab \((x > d\) or \(x < -d)\), \(E = \frac{\rho (2d)}{\varepsilon_0}\).

Step by step solution

01

Understanding the Problem

The problem involves a uniformly charged insulating slab with a uniform charge density \( \rho \). The thickness of the slab is \(2d\) with faces at \(x = d\) and \(x = -d\), and extends infinitely in the \(y\) and \(z\) directions. We are tasked with finding the electric field both inside and outside the slab.
02

Symmetry and Electric Field at Center

Due to symmetry, the electric field contributions from each half of the slab (from \(x = 0\) to \(x = d\) and from \(x = -d\) to \(x = 0\)) cancel each other out at the center of the slab \((x = 0)\). This is because the slab is uniformly charged and the contributions from equally spaced points on either side of the center are equal and opposite. Thus, the electric field at the center \((x=0)\) is zero.
03

Setting Up Gauss's Law

We use Gauss's Law, which states \( \Phi = \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{enc}}{\varepsilon_0} \), where \( \Phi \) is the electric flux, \( \mathbf{E} \) is the electric field, \( d\mathbf{A} \) is the infinitesimal area vector, \( Q_{enc} \) is the charge enclosed, and \( \varepsilon_0 \) is the permittivity of free space. We choose a Gaussian surface in the form of a rectangular box aligned with the slab, extending symmetrically above and below the \(x=0\) plane to evaluate the electric field in different regions.
04

Electric Field Inside the Slab (\(-d < x < d\))

Inside the slab, the symmetry ensures that the electric field points perpendicular to the slab. Consider a Gaussian surface, a rectangular parallelepiped. In this region, the charge enclosed \(Q_{enc} = \rho A x\), where \(A\) is the area of one face of the parallelepiped and \(x\) is the distance from the center. Gauss's law gives \( EA = \frac{\rho Ax}{\varepsilon_0} \). Therefore, \( E = \frac{\rho x}{\varepsilon_0} \), directed along the \(x\)-axis.
05

Electric Field Outside the Slab (\(x > d\) and \(x < -d\))

Outside the slab, consider a Gaussian surface that entirely encloses the slab, thus \(Q_{enc} = \rho A(2d)\). Using Gauss's law, \( EA = \frac{\rho A(2d)}{\varepsilon_0} \), the electric field simplifies to \( E = \frac{\rho (2d)}{\varepsilon_0} \). This field is constant outside the slab and directed outward.
06

Combining Results

The electric field due to the slab is zero at the center \((x = 0)\). For \(-d < x < d\), the electric field's magnitude is \( E = \frac{\rho x}{\varepsilon_0} \), directed along the \(x\)-axis. For points outside the slab \((x > d\) and \(x < -d)\), the field is \( E = \frac{\rho (2d)}{\varepsilon_0} \), pointing outward.

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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 represents the influence that a charge exerts on other charges around it. It's essentially a force field. When an object with an electric charge is placed at a point in space, it experiences a force due to the presence of another electric charge. In a scenario like a uniformly charged slab, the electric field can tell us how the electrical forces are acting throughout space.
  • Electric field vectors show the direction of the force a positive test charge would feel.
  • The strength of the electric field at a point in space is represented by vector length or density.
The electric field inside the slab is zero at its center because the contributions from opposite sides cancel each other out. Outside the slab, the field remains constant and decreases with distance.
Uniformly Charged Slab
A uniformly charged slab is a thick piece of insulating material balanced with a uniform positive charge throughout its volume. For this slab, the thickness is considered as the distance between the two planes at positions \(x=-d\) and \(x=d\). What's notable here:
  • A uniform charge density, \( \rho \), means the charge is evenly distributed throughout the slab.
  • Given the slab's infinite size in the \(y\) and \(z\) directions, it's modeled as having indefinite width and height. This simplifies calculations because it's easier to consider its symmetry.
This symmetry is pivotal in determining the electric field across the slab.
Electric Flux
Electric flux is a measure of the amount of electric field passing through a surface. Imagine electric field lines passing through a space: electric flux quantifies how many of these lines go through a given surface area. The formula for electric flux is \( \Phi = \oint \mathbf{E} \cdot d\mathbf{A} \), where \( \mathbf{E} \) is the electric field, and \( d\mathbf{A} \) is the area vector.
  • For a uniformly charged slab, we analyze how the electric field penetrates through a selected Gaussian surface.
  • This approach helps in understanding how the field distributes outside and inside the slab.
  • Gauss's Law aids in calculating the resultant field through the integral of electric flux over the Gaussian surface.
Symmetry in Physics
Symmetry plays a vital role in simplifying complex problems in physics, and this exercise excellently demonstrates it. The symmetry in the uniformly charged slab allows one to determine parts of the field without tedious calculations. Here's why:
  • Symmetry ensures that the contributions to the electric field from opposite sides can cancel or add up predictably.
  • In the center of the slab, the symmetrically distributed charges mean the net field cancels out to zero.
  • Outside, symmetry ensures the net field is outward and constant, as computations show.
Recognizing symmetry helps to boil down issues into simpler forms, revealing patterns and solutions inherently present in the system.

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Most popular questions from this chapter

The gravitational force between two point masses separated by a distance \(r\) is proportional to \(1 / r^{2},\) just like the electric force between two point charges. Because of this similarity between gravitational and electric interactions, there is also a Gauss's law for gravitation. (a) Let \(\vec{g}\) be the acceleration due to gravity caused by a point mass \(m\) at the origin, so that \(\vec{g}=-\left(G m / r^{2}\right) \hat{r}\) . Consider a spherical Gaussian surface with radius \(r\) centered on this point mass, and show that the flux of \(\vec{g}\) through this surface is given by $$\oint \vec{g} \cdot d \vec{A}=-4 \pi G m$$ (b) By following the same logical steps used in Section 22.3 to obtain Gauss's law for the electric field, show that the flux of \(\vec{g}\) through any closed surface is given by $$\oint \vec{g} \cdot d \vec{A}=-4 \pi G M_{\mathrm{encl}}$$ where \(M_{\text { encl }}\) is the total mass enclosed within the closed surface.

The electric field at a distance of 0.145 m from the surface of a solid insulating sphere with radius 0.355 m is 1750 N/C.(a) Assuming the sphere's charge is uniformly distributed, what is the charge density inside it? (b) Calculate the electric field inside the sphere at a distance of 0.200 m from the center.

The nuclei of large atoms, such as uranium, with 92 protons, can be modeled as spherically symmetric spheres of charge. The radius of the uranium nucleus is approximately \(7.4 \times 10^{-15} \mathrm{m}\) (a) What is the electric field this nucleus produces just outside its surface? (b) What magnitude of electric field does it produce at the distance of the electrons, which is about \(1.0 \times 10^{-10} \mathrm{m} ?\) (c) The electrons can be modeled as forming a uniform shell of negative charge. What net electric field do they produce at the location of the nucleus?

A small conducting spherical shell with inner radius \(a\) and outer radius \(b\) is concentric with a larger conducting spherical shell with inner radius \(c\) and outer radius \(d\) (Fig. P22.47). The inner shell has total charge \(+2 q,\) and the outer shell has charge \(+4 q .\) (a) Calculate the electric field (magnitude and direction) in terms of \(q\) and the distance \(r\) from the common center of the two shells for (i) \(r < a ;\) (ii) \(a < r < b ;\) (iii) \(b < r < c ;\) (iv) \(c < r < d\) ; (v) \(r > d .\) Show your results in a graph of the radial component of \(\vec{\boldsymbol{E}}\) as a function of \(r .\) (b) What is the total charge on the (i) inner surface of the small shell; (ii) outer surface of the small shell; (ii) inner surface of the large shell; (iv) outer surface of the large shell?

A square insulating sheet 80.0 cm on a side is held horizontally. The sheet has 7.50 nC of charge spread uniformly over its area. (a) Calculate the electric field at a point 0.100 mm above the center of the sheet. (b) Estimate the electric field at a point 100 m above the center of the sheet. (c) Would the answers to parts (a) and (b) be different if the sheet were made of a conducting material? Why or why not?
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