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Figure 23-43 shows cross sections through two large, parallel, nonconducting sheets with identical distributions of positive charge with surface charge density \(\sigma=1.77 \times 10^{-22} \mathrm{C} / \mathrm{m}^{2}\). In unit- vector notation, what is \(\vec{E}\) at points (a) above the sheets, (b) between them, and (c) below them?

Short Answer

Expert verified
(a) \( \vec{E} = \frac{1.77 \times 10^{-22}}{8.85 \times 10^{-12}} \hat{k} \) N/C, (b) \( \vec{E} = 0 \), (c) \( \vec{E} = \frac{1.77 \times 10^{-22}}{8.85 \times 10^{-12}} \hat{k} \) N/C.

Step by step solution

01

Understand Gauss's Law for a Plane

For infinitely large parallel sheets with uniform charge density, the electric field due to one sheet is given by \( E = \frac{\sigma}{2\varepsilon_0} \), where \( \sigma \) is the surface charge density and \( \varepsilon_0 \) is the permittivity of free space (approximately \(8.85 \times 10^{-12} \mathrm{C}^2/\mathrm{N} \, \mathrm{m}^2 \)). This is derived from Gauss's law.
02

Calculate Electric Field Above the Sheets

Above both sheets, the contributions to the electric field from each sheet add up since they point in the same direction. So, \( E_{\text{above}} = \frac{\sigma}{2\varepsilon_0} + \frac{\sigma}{2\varepsilon_0} = \frac{\sigma}{\varepsilon_0} \). Substitute \( \sigma = 1.77 \times 10^{-22} \mathrm{C} / \mathrm{m}^2 \) into the equation to find \( E_{\text{above}} \).
03

Calculate Electric Field Between the Sheets

Between the sheets, the contributions from each sheet's electric field cancel each other out because they point in opposite directions. Thus, the net electric field \( E_{\text{between}} = 0 \).
04

Calculate Electric Field Below the Sheets

Below both sheets, like step 2, the electric fields add since they point in the same direction. Hence, \( E_{\text{below}} = \frac{\sigma}{\varepsilon_0} \), the same as \( E_{\text{above}} \). Substituting \( \sigma \) gives the same numeric result as above.

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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 powerful tool in electrostatics for calculating electric fields. It relates the electric field emanating from a closed surface to the charge enclosed by that surface. The law mathematically is expressed as: \[ \Phi_E = \frac{Q_{\text{enc}}}{\varepsilon_0} \] where \( \Phi_E \) represents the electric flux, \( Q_{\text{enc}} \) is the enclosed charge, and \( \varepsilon_0 \) is the permittivity of free space, approximately \(8.85 \times 10^{-12} \mathrm{C}^2/\mathrm{N} \, \mathrm{m}^2 \).

This law becomes incredibly useful for symmetrical charge distributions, such as that found in infinite planes of charge. For an infinite plane with a uniform surface charge density \( \sigma \), it simplifies the complex problem of field calculation to an elegant expression. By using a "Gaussian surface" that encloses a portion of the charge, you can derive the electric field as: \[ E = \frac{\sigma}{2\varepsilon_0} \] for either side of the sheet.

The symmetry of the problem allows us to conclude that the electric field is perpendicular to the surface and is uniform over the infinite plane. This concept forms the basis for understanding how the electric fields above and below a plane interact with the charges.
Electric Field
The electric field is a vector field that represents the force a positive test charge would experience per unit charge at any point in space. In simpler terms, it shows how charged particles affect one another through invisible forces. Mathematically, it is defined as: \[ \vec{E} = \frac{\vec{F}}{q_0} \] where \( \vec{F} \) is the force experienced by a test charge \( q_0 \).

In the context of parallel sheets of charge, the problem becomes easier due to symmetry. With an infinite non-conducting sheet, using Gauss's Law, we determine the electric field produced is \( \frac{\sigma}{2\varepsilon_0} \) from one sheet. When two such sheets are independently assessed, their individual fields combine or cancel depending on the point of observation, due to vector addition.

- **Above the Sheets:** Both sheets contribute electric fields that add up because they are in the same direction. Thus, the total field \( \vec{E}_{\text{above}} = \frac{\sigma}{\varepsilon_0} \).- **Between the Sheets:** The fields point in opposite directions and cancel each other completely, resulting in no net field, \( \vec{E}_{\text{between}} = 0 \).- **Below the Sheets:** Like above, the field contributions add up, making \( \vec{E}_{\text{below}} = \frac{\sigma}{\varepsilon_0} \).
Surface Charge Density
Surface charge density \( \sigma \) is a measure of how much electric charge is distributed over a specific area on a surface. It is defined as the charge per unit area and is given in units of Coulombs per square meter (\( \mathrm{C}/\mathrm{m}^2 \)). For our problem, \( \sigma = 1.77 \times 10^{-22} \mathrm{C}/\mathrm{m}^2 \).

It plays a crucial role in determining the electric field produced by charged surfaces, especially infinite planar sheets due to its simplicity and relevance in real-world applications. Knowing \( \sigma \) allows you to directly calculate the electric field. This is primarily because it serves as the primary input in Gauss's law for estimating the field across the sheet:
  • Above and below the plane, the field depends linearly on \( \sigma \).
  • For uniformly distributed charges on a plane, the field only depends on the surface charge density, not its size.
Understanding surface charge density helps in visualizing how strongly charged the surface is and how it influences the space around it. This makes it an essential concept when analyzing any scenario involving electric fields and charged surfaces.

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Most popular questions from this chapter

Space vehicles traveling through Earth's radiation belts can intercept a significant number of electrons. The resulting charge buildup can damage electronic components and disrupt operations. Suppose a spherical metal satellite \(1.3 \mathrm{~m}\) in diameter accumulates \(2.4\) \(\mu \mathrm{C}\) of charge in one orbital revolution. (a) Find the resulting surface charge density. (b) Calculate the magnitude of the electric field just outside the surface of the satellite, due to the surface charge.

Figure \(23-57\) shows a Geiger counter, a device used to detect ionizing radiation, which causes ionization of atoms. A thin, positively charged central wire is surrounded by a concentric, circular, conducting cylindrical shell with an equal negative charge, creating a strong radial electric field. The shell contains a low-pressure inert gas. A particle of radiation entering the device through the shell wall ionizes a few of the gas atoms. The resulting free electrons (e) are drawn to the positive wire. However, the electric field is so intense that, between collisions with 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 inner radius of the shell \(1.4\) \(\mathrm{cm}\), and the length of the shell \(16 \mathrm{~cm}\). If the electric field at the shell's inner wall is \(2.9 \times 10^{4} \mathrm{~N} / \mathrm{C}\), what is the total positive charge on the central wire?

Charge of uniform volume density \(\rho=3.2 \mu \mathrm{C} / \mathrm{m}^{3}\) fills a nonconducting solid sphere of radius \(5.0 \mathrm{~cm}\). What is the magnitude of the electric field (a) \(3.5 \mathrm{~cm}\) and (b) \(8.0 \mathrm{~cm}\) from the sphere's center?

The electric field at point \(P\) just outside the outer surface of a hollow spherical conductor of inner radius \(10 \mathrm{~cm}\) and outer radius \(20 \mathrm{~cm}\) has magnitude \(450 \mathrm{~N} / \mathrm{C}\) and is directed outward. When an unknown point charge \(Q\) is introduced into the center of the sphere, the electric field at \(P\) is still directed outward but is now \(180 \mathrm{~N} / \mathrm{C}\). (a) What was the net charge enclosed by the outer surface before \(Q\) was introduced? (b) What is charge \(Q\) ? After \(Q\) is introduced, what is the charge on the (c) inner and (d) outer surface of the conductor?

A free electron is placed between two large, parallel, nonconducting plates that are horizontal and \(2.3 \mathrm{~cm}\) apart. One plate has a uniform positive charge; the other has an equal amount of uniform negative charge. The force on the electron due to the electric field \(\vec{E}\) between the plates balances the gravitational force on the electron. What are (a) the magnitude of the surface charge density on the plates and (b) the direction (up or down) of \(\vec{E}\) ?

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