Problem 13 - Fundamentals of Physics

Question

The electric field in a certain region of Earth's atmosphere is directed vertically down. At an altitude of \(300 \mathrm{~m}\) the field has magnitude \(60.0 \mathrm{~N} / \mathrm{C} ;\) at an altitude of \(200 \mathrm{~m},\) the magnitude is \(100 \mathrm{~N} / \mathrm{C} .\) Find the net amount of charge contained in a cube \(100 \mathrm{~m}\) on edge, with horizontal faces at altitudes of 200 and \(300 \mathrm{~m}\).

Step-by-Step Solution

Answer

The net charge enclosed by the cube is \(1.419 \times 10^{-5} \mathrm{~C}.\)

Step 1: Understand the Problem

We are given electric field magnitudes at two different altitudes and need to find the net charge within a cube between these altitudes. The relevant concept is Gauss's law, which relates the electric field surrounding a closed surface to the charge enclosed by that surface.

Step 2: Apply Gauss's Law

Gauss's Law states \( \Phi_E = \frac{Q_{enc}}{\varepsilon_0} \), where \( \Phi_E \) is the electric flux through a surface, \( Q_{enc} \) is the charge enclosed, and \( \varepsilon_0 \) is the permittivity of free space, \( 8.85 \times 10^{-12} \mathrm{~C}^2/\mathrm{N} \cdot \mathrm{m}^2 \). The cube's top face at 300 m and bottom face at 200 m will contribute flux due to electric field.

Step 3: Calculate the Electric Flux

The electric flux through the top face is \( \Phi_{300} = E_{300} \times A \) and through the bottom face is \( \Phi_{200} = E_{200} \times A \), where \( A \) is the area of a face of the cube: \( A = (100 \mathrm{~m})^2 = 10^4 \mathrm{~m}^2 \). Hence, \( \Phi_{300} = 60.0 \mathrm{~N} / \mathrm{C} \times 10^4 \mathrm{~m}^2 \) and \( \Phi_{200} = -100.0 \mathrm{~N} / \mathrm{C} \times 10^4 \mathrm{~m}^2 \) (negative because the field is directed downwards at 200 m).

Step 4: Calculate the Net Electric Flux

The net electric flux \( \Phi_{net} \) through the cube is \( \Phi_{300} - \Phi_{200} = (60.0 \mathrm{~N} / \mathrm{C} \times 10^4 \mathrm{~m}^2) - (-100.0 \mathrm{~N} / \mathrm{C} \times 10^4 \mathrm{~m}^2) = 60.0 \times 10^4 + 100.0 \times 10^4 \mathrm{~N} \cdot \mathrm{m}^2/\mathrm{C} \).

Step 5: Solve for the Charge Enclosed

Using Gauss's law, \( Q_{enc} = \Phi_{net} \times \varepsilon_0 \). Substituting the values, \( Q_{enc} = (60.0 \times 10^4 + 100.0 \times 10^4) \mathrm{~N} \cdot \mathrm{m}^2/\mathrm{C} \times 8.85 \times 10^{-12} \mathrm{~C}^2/\mathrm{N} \cdot \mathrm{m}^2 \). Calculate \( Q_{enc} = 1.419 \times 10^{-5} \mathrm{~C} \).

Key Concepts

Electric Field

The electric field is a region around a charged object, where other charged objects experience a force. This force can attract or repel the objects based on their charges. In simple terms, you can think of the electric field as the "push or pull" region that the charge creates in the space around it.

The electric field is usually represented by the symbol \( E \) and is measured in newtons per coulomb (\( \mathrm{N} / \mathrm{C} \)). The direction of the electric field represents the path a positive charge would take if placed in the field.

• In our given problem, the electric field points vertically downwards in Earth's atmosphere, indicating that a positive charge would move downward.
• The field strength is 60.0 N/C at 300 meters and increases to 100 N/C at 200 meters, signifying a stronger force closer to the Earth's surface.

Understanding how the electric field operates is crucial, as it affects how charges interact within the field, leading to calculations involving electric flux and total enclosed charge.

Electric Flux

Electric flux relates to the amount of electric field that "flows" through a given area. It's an important concept in understanding how the electric field interacts with surfaces. Think of electric flux as how much of the field is "cut" by a surface, similar to how wind might flow through an open window.

Mathematically, electric flux is symbolized as \( \Phi_E \), and it depends on both the electric field strength (\( E \)) and the area (\( A \)) through which it passes. The formula is given by \( \Phi_E = E \times A \), where \( E \) is the electric field, and \( A \) is the area.

• In the provided exercise, the electric field interacts with the cube's top and bottom surfaces, each 100 m by 100 m, resulting in significant flux values for each.
• The net flux is crucial as it determines the charge enclosed within the cube, aiding in applying Gauss's Law effectively.

Analyzing electric flux helps us understand how much of the field's effects penetrate through different surfaces, giving insight into enclosed charge distributions.

Charge Distribution

Charge distribution is a depiction of how electric charges are arranged in a given space or on an object. The way charges are spread out affects how strong the electric field will be in specific areas and, subsequently, the electric flux through surfaces.

Understanding charge distribution helps in determining the electric field's characteristics such as its direction and intensity. In our exercise, the distribution of charge between altitudes results in varying electric field strengths.

When tackling problems involving Gauss's Law, as in this exercise, recognizing how charges are distributed in a space is essential. This understanding helps deduce:

• the relationship between electric field variations at different points
• the calculation of net electric flux
• the total charge enclosed within a region

Exploring charge distribution provides a clearer picture of how charges create electric fields and their resulting impact on surrounding environments.

Permittivity of Free Space

The permittivity of free space, often represented as \( \varepsilon_0 \), is a fundamental constant used in the study of electromagnetism. It describes how electric fields interact in a vacuum, serving as a measure of how well the vacuum "permits" electric field lines to pass through.

The value of the permittivity of free space is approximately \( 8.85 \times 10^{-12} \mathrm{~C}^2/ ext{N} \cdot \mathrm{m}^2 \). This constant plays a crucial role in Gauss's Law, especially when calculating the electric flux and charge relationships.

• In the context of the given problem, \( \varepsilon_0 \) enables the conversion of net electric flux to the enclosed charge using the formula \( Q_{enc} = \Phi_{net} \times \varepsilon_0 \).
• Essentially, it provides the groundwork for calculating how much charge exists within certain boundaries, based on observed electric field variations.

Without the permittivity of free space, predicting interactions between charges in a vacuum or even approximating those in gases like the Earth's atmosphere would be far more complex.