Problem 13 - Fundamentals of Physics

Question

SSM 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 is \(-3.54 \times 10^{-7} \ \text{C}\).

Step 1: Identify Initial Information

We have a cube with sides of length 100 m. The electric field at 200 m altitude is 100 N/C and at 300 m altitude is 60 N/C. The cube is positioned such that it is aligned vertically from 200 m to 300 m.

Step 2: Understand Gauss’s Law

Gauss's Law relates the electric flux through a closed surface to the charge enclosed by that surface. It is given by \( \Phi_E = \int \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\varepsilon_0} \), where \( Q_{\text{enc}} \) is the charge enclosed and \( \varepsilon_0 \) is the vacuum permittivity.

Step 3: Calculate Electric Flux

The electric flux \( \Phi_E \) through the top and bottom surfaces of the cube is given by \( \Phi_E = E_{top} \cdot A_{top} - E_{bottom} \cdot A_{bottom} \). The area of each face \( A \) is \( (100 \, \text{m})^2 = 10000 \, \text{m}^2 \). Therefore, the electric flux is \( \Phi_E = 60 \, \text{N/C} \times 10000 \, \text{m}^2 - 100 \, \text{N/C} \times 10000 \, \text{m}^2 \).

Step 4: Simplify Flux Expression

Substitute values into the flux expression: \( \Phi_E = 60,000 \, \text{N} \cdot \text{m}^2 \text{/C} - 100,000 \, \text{N} \cdot \text{m}^2 \text{/C} = -40,000 \, \text{N} \cdot \text{m}^2 \text{/C} \).

Step 5: Apply Gauss's Law to Find Enclosed Charge

Using Gauss's Law, \( Q_{\text{enc}} = \varepsilon_0 \cdot \Phi_E \). The vacuum permittivity \( \varepsilon_0 \) is approximately \( 8.85 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2 \). So,\( Q_{\text{enc}} = 8.85 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2 \times (-40,000 \, \text{N} \cdot \text{m}^2 \text{/C}) \).

Step 6: Compute the Enclosed Charge

Calculate the charge: \( Q_{\text{enc}} = -3.54 \times 10^{-7} \, \text{C} \). This represents the net amount of charge enclosed within the cube.

Key Concepts

Electric Field

An electric field is a field around charged particles that exerts a force on other charges within it. In this exercise, the electric field is significant, as it indicates how charges are distributed across different altitudes. The electric field at 200 meters is stronger, at 100 N/C, than at 300 meters where it is 60 N/C. This variation suggests a difference in charge concentration that Gauss's Law can help analyze. To take advantage of the electric field:

• Consider the direction: In this example, the electric field points vertically downward, indicating a negative charge component as altitude decreases.
• Magnitude matters: The numeric value of an electric field indicates how strong the force will be if another charge enters the area.
• Electric field uniformity: If the fields differed due to charge structure, one would see different flux values through sections within the field.

Electric Flux

Electric flux, denoted as \( \Phi_E \), measures the flow of the electric field through a surface. It's a critical concept when utilizing Gauss's Law. In our exercise, the electric flux through the cube's top and bottom surfaces indicates the amount of electric field passing through those planes:- Calculate area: The cube's sides are each 100 meters, so the area \( A \) of a face is \( 10,000 \, \text{m}^2 \).- Determine flux: Establish the differences in field strength between top (60 N/C) and bottom (100 N/C).The intervening step:

• Compute flux difference: \( \Phi_E = 60 \, \text{N/C} \times 10,000 \, \text{m}^2 - 100 \, \text{N/C} \times 10,000 \, \text{m}^2 = -40,000 \, \text{N} \cdot \text{m}^2/\text{C} \).

This calculation helps understand the net charge within the cube's interior based on how the electric field changes from one altitude to another.

Charge Density

Charge density provides insight into how charge is spread over an area or volume. It can be visualized as either surface charge density on a flat surface or volume charge density within a three-dimensional space. For this exercise:

• The differential in electric fields at 200 m and 300 m suggests a charge distribution change vertically through the cube.
• Gauss's Law helps to aggregate this difference into an effective net charge across the region.

If the charge density is too high, it may indicate a strong electric field, affecting nearby charges more profoundly. Addressing charge density is crucial for accurately predicting forces and potential impacts in varying conditions across a designated area.

Vacuum Permittivity

Vacuum permittivity, symbolized as \( \varepsilon_0 \), is a physical constant that quantifies the ability of the vacuum to permit electric field lines. It is a measure of the electric resistance of a vacuum. \( \varepsilon_0 \) roughly equals \( 8.85 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2 \).In our exercise, vacuum permittivity is essential because:

• It appears in Gauss's Law, which calculates the net enclosed charge using \( Q_{\text{enc}} = \varepsilon_0 \times \Phi_E \).
• It helps standardize electric flux calculations across different scenarios by linking the fundamental properties of the physical universe.

The resulting enclosed charge calculation, influenced by vacuum permittivity, provides clarity on the overall charge distribution within the cube when surface electric fields are known.