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The electric field in a certain region of Earth's atmosphere directed vertically down. At an altitude of \(300 \mathrm{~m}\) the field has mag nitude \(75.0 \mathrm{~N} / \mathrm{C} ;\) at an altitude of \(200 \mathrm{~m}\), the magnitude is \(210 \mathrm{~N} / \mathrm{C}\) Find the net amount of charge contained in a cube \(100 \mathrm{~m}\) on edg with horizontal faces at altitudes of 200 and \(300 \mathrm{~m}\).

Short Answer

Expert verified
The net charge in the cube is approximately \( 1.2 \times 10^{-6} \, \mathrm{C} \).

Step by step solution

01

Understand the Problem

We need to determine the net amount of charge within a volume in Earth's atmosphere. We'll use Gauss's Law to calculate this, considering the electric field values given at two different altitudes.
02

Identify Given Data

From the problem, we know the electric field at an altitude of 300 m is \( E_1 = 75.0 \, \mathrm{N/C} \), and at 200 m is \( E_2 = 210 \, \mathrm{N/C} \). The cube has sides of length \( \Delta h = 100 \, \mathrm{m} \) with its horizontal faces at these altitudes.
03

Recall Gauss's Law

Gauss's Law states that \( \Phi = \oint \mathbf{E} \cdot \mathbf{dA} = \frac{Q_{ ext{enc}}}{\epsilon_0} \), where \( \Phi \) is the electric flux through a closed surface, \( Q_{\text{enc}} \) is the enclosed charge, and \( \epsilon_0 \) is the vacuum permittivity \( (8.85 \times 10^{-12} \, \mathrm{C^2/N\cdot m^2}) \).
04

Calculate the Electric Flux

The flux through the cube is due only to the top and bottom faces since the electric field is vertical. \( \Phi = \mathbf{E}_2 \cdot A - \mathbf{E}_1 \cdot A \), where \( A = \Delta x \times \Delta z = (100)^2 = 10,000 \, \mathrm{m^2} \). Substitute the known values to get \( \Phi = (210)(10,000) - (75)(10,000).\)
05

Compute for Enclosed Charge

Using \( \Phi = \frac{Q_{\text{enc}}}{\epsilon_0} \), rearrange to find \( Q_{\text{enc}} = \Phi \epsilon_0 \). Substitute the flux from the previous step and solve for \( Q_{\text{enc}} \).
06

Final Calculation and Sign

Substitute \( Q_{\text{enc}} \) and evaluate: \( Q_{\text{enc}} = (210,000 - 75,000) \times 8.85 \times 10^{-12} \approx 1.2 \times 10^{-6} \, \mathrm{C} \). The charge is positive, indicating a net positive charge in the region.

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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. It represents the force per unit charge exerted on a charged particle in a particular location. An electric field exists around charged objects and affects any other charges in its vicinity.

Key characteristics of an electric field include its direction and magnitude. The direction of the field is the direction that a positive test charge would move. In this exercise, the electric field is directed vertically downward, which influences the charges within Earth's atmosphere. Its magnitude, given in units of newtons per coulomb (N/C), tells us how strong the field is at a point. The electric field strength decreases as we move from 200 m to 300 m, suggesting a variation in charge distribution in that region.
Electrical Flux
Electrical flux is crucial in understanding how electric fields interact with surfaces. It quantifies the total electric field passing through a surface and is represented by the symbol \( \Phi \).

In mathematical terms, electrical flux is obtained by integrating the electric field over a closed surface, which can be expressed as \( \Phi = \oint \mathbf{E} \cdot \mathbf{dA} \). For a flat surface with uniform field, it reduces to \( \Phi = \mathbf{E} \cdot A \), where \( A \) is the area and \( \mathbf{E} \) is the field strength.
  • The total flux is influenced by both the magnitude of the electric field and the orientation of the surface relative to the field lines.
  • In this example, since the cube's top and bottom sides are parallel to the electric field, they contribute to the flow of electric field lines through the cube.
Enclosed Charge
Enclosed charge refers to the total amount of charge within a closed surface. Calculating enclosed charge is a primary application of Gauss's Law.

According to Gauss's Law, the relation between electrical flux \( \Phi \) through a closed surface and the enclosed charge \( Q_{\text{enc}} \) is given by \( \Phi = \frac{Q_{\text{enc}}}{\epsilon_0} \).
  • In practical terms, you first calculate the flux using the electric fields and surface area.
  • After obtaining the flux, the enclosed charge is determined by multiplying the flux by the vacuum permittivity.
Using this method, we calculated that the enclosed charge in the given problem is approximately \( 1.2 \times 10^{-6} \) C, indicating a net positive charge in the region.
Vacuum Permittivity
Vacuum permittivity, symbolized as \( \epsilon_0 \), is a constant that describes how electric fields interact in a vacuum. It is a fundamental parameter in electrostatics, influencing the force between electric charges and the flux through a surface.

The value of vacuum permittivity is \( 8.85 \times 10^{-12} \, \mathrm{C^2/N \cdot m^2} \). It appears in various electromagnetic formulas, notably Gauss's Law.
  • In Gauss's Law, \( \epsilon_0 \) relates electrical flux to the charge enclosed by a surface. Its significance lies in mediating interactions between electric phenomena in free space or vacuum.
  • This constant is essential in calculating how much charge is influencing a surface through its electric field contributions.

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Most popular questions from this chapter

In Fig. \(23-37\), a butterfly net is in a uniform electric field o magnitude \(E=4.5 \mathrm{mN} / \mathrm{C}\). The rim, a circle of radius \(a=11 \mathrm{~cm}\), is aligned perpendicular to the field. The net contains no net charge. Find the electric flux through the netting.

The electric field just above the surface of the charged conducting drum of a photocopying machine has a magnitude \(E\) of \(1.9 \times 10^{5} \mathrm{~N} / \mathrm{C}\). What is the surface charge density on the drum?

A uniformly charged conducting sphere of \(0.60 \mathrm{~m}\) diameter has surface charge density \(5.7 \mu \mathrm{C} / \mathrm{m}^{2}\). Find (a) the net charge on the sphere and (b) the total electric flux leaving the surface. (c) What is the net flux through a concentric Gaussian sphere of radius \(2.0 \mathrm{~m}\) ?

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 \(3.9 \mu \mathrm{C}\) 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.

A particle of charge \(6.3 \mu \mathrm{C}\) is at the center of a Gaussian cube \(92 \mathrm{~cm}\) on edge. (a) What is the net electric flux through the surface? (b) What is the net flux if the edge length is doubled?

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