Problem 7 - Fundamentals of Physics

Question

A particle of charge \(1.8 \mu \mathrm{C}\) is at the center of a Gaussian cube \(55 \mathrm{~cm}\) on edge. What is the net electric flux through the surface?

Step-by-Step Solution

Answer

The net electric flux is \(2.034 \times 10^5 \text{ N} \cdot \text{m}^2/\text{C}.\)

Step 1: Understanding the Problem

We need to find the net electric flux through a Gaussian surface. The electric flux is defined as the electric field multiplied by the area through which the field lines pass. Here, a charged particle is at the center of a cube.

Step 2: Apply Gauss's Law

Gauss's Law states that the electric flux \( \Phi \) through a closed surface is given by \( \Phi = \frac{Q_{enclosed}}{\varepsilon_0} \), where \( Q_{enclosed} \) is the total charge enclosed and \( \varepsilon_0 \) is the permittivity of free space \( \varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2 \).

Step 3: Calculate Enclosed Charge

The problem states that a charge of \( 1.8 \, \mu \text{C} \) is placed at the center of the cube. Convert \( \mu \text{C} \) to coulombs: \( 1.8 \times 10^{-6} \text{C} \).

Step 4: Use Gauss's Law to Find Electric Flux

Substitute the enclosed charge into Gauss's Law formula: \[ \Phi = \frac{1.8 \times 10^{-6} \text{C}}{8.85 \times 10^{-12} \text{C}^2/\text{N} \cdot \text{m}^2} \].

Step 5: Perform the Calculation

Calculate: \( \Phi = \frac{1.8 \times 10^{-6}}{8.85 \times 10^{-12}} = 2.034 \times 10^{5} \text{ N} \cdot \text{m}^2/\text{C} \).

Step 6: Conclusion

The net electric flux through the surface of the cube is \( 2.034 \times 10^{5} \text{ N} \cdot \text{m}^2/\text{C}. \)

Key Concepts

Gauss's Law

Gauss's Law is both a fundamental law of electromagnetism and a mathematical tool.
It relates the electric flux through a closed surface to the charge enclosed by that surface. Here's how the law works in simple terms:

• Electric Flux: This is a measure of the electric field passing through a surface. It takes the electric field, and multiplies it by the area of the surface.
• Closed Surface: Imagine any 3D shape completely enclosing a space, like a bubble or a box.
• Charge Enclosed: Gauss's Law states that the total flux through the surface is directly proportional to the charge enclosed within it.

The formula for Gauss's Law is: \( \Phi = \frac{Q_{enclosed}}{\varepsilon_0} \).
It simplifies calculating electric fields for symmetrical situations such as spheres or cubes.

Charge Enclosed

The concept of charge enclosed is essential in understanding how electric fields interact with closed surfaces.
In our example, the cube encloses a specific point charge that we need to consider.Here's how to think about it:

• Point Charge: In the exercise, a point charge of \(1.8 \, \mu \text{C}\) is located at the center of the cube. The cube completely surrounds this charge.
• Enclosure: The charge enclosed is exactly the amount inside this closed surface, which in our case is represented by the cube.
• Symmetry: Because the charge is at the center, the problem is simplified. The electric field lines are uniformly diverging from this point.

Determining the charge enclosed is the key first step when applying Gauss's Law for calculating electric flux.

Permittivity of Free Space

The permittivity of free space, denoted \(\varepsilon_0\), is a constant that appears in Gauss's Law and helps relate the electric flux to the enclosed charge.
It essentially quantifies how much electric field can exist in a vacuum.Here are key points to note:

• Value: \(\varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2\). This very small number indicates that a vacuum has very little ability to "permit" electric field lines.
• Role in Gauss's Law: It scales the relationship between the electric field (flux) and the charge enclosed. Without this permittivity, we'd have an unbounded value for flux, leading to incorrect predictions of electric field intensity.
• Universal Constant: It is used universally for vacuum or air, making it a foundational aspect of calculations involving electric fields.

Understanding \(\varepsilon_0\) is crucial for solving problems like the one given, where the flux through a closed surface needs to be calculated accurately.