Problem 11 - University Physics with Modern Physics

Question

A 6.20-\(\mu\)C point charge is at the center of a cube with sides of length 0.500 m. (a) What is the electric flux through one of the six faces of the cube? (b) How would your answer to part (a) change if the sides were 0.250 m long? Explain.

Step-by-Step Solution

Answer

(a) Electric flux through one face is approximately \(1.17 \times 10^5\, \text{N}\cdot\text{m}^2/\text{C}\). (b) Flux is unchanged by cube size.

Step 1: Understanding Gauss's Law

Gauss's Law states that the electric flux \( \Phi \) through a closed surface is equal to the charge \( Q \) enclosed by the surface divided by \( \varepsilon_0 \) (the permittivity of free space): \( \Phi = \frac{Q}{\varepsilon_0} \). Where \( \varepsilon_0 \approx 8.854 \times 10^{-12} \text{ C}^2/\text{N}\cdot\text{m}^2 \).

Step 2: Calculate Total Electric Flux Through the Cube

Since the charge is at the center of the cube, it is enclosed by the cube. Using Gauss's Law, we find the total electric flux through the cube: \( \Phi_{\text{total}} = \frac{6.20 \times 10^{-6} \text{ C}}{8.854 \times 10^{-12} \text{ C}^2/\text{N}\cdot\text{m}^2} \approx 7.00 \times 10^5 \text{ N}\cdot\text{m}^2/\text{C} \).

Step 3: Calculate Electric Flux Through One Face of the Cube

A cube has six faces and the electric flux is uniformly distributed through these faces. Therefore, the electric flux through one face is \( \Phi_{\text{face}} = \frac{\Phi_{\text{total}}}{6} = \frac{7.00 \times 10^5 \text{ N}\cdot\text{m}^2/\text{C}}{6} \approx 1.17 \times 10^5 \text{ N}\cdot\text{m}^2/\text{C} \).

Step 4: Analyze the Effect of Changing Cube Size on Flux

The electric flux through a closed surface only depends on the enclosed charge, not on the size of the surface or shape of the cube. Therefore, the flux through one face of the cube remains the same, regardless of whether the cube side length is 0.500 m or 0.250 m.

Key Concepts

Electric Flux

Electric flux is an important concept in electromagnetism that gives us an idea of the flow of electric field lines through a surface. Think of it like the amount of 'electric energy' moving through a particular area. It is denoted as \( \Phi \) and is measured in newton-meters squared per coulomb (N·m²/C).

Gauss's Law helps us calculate electric flux. According to this law, the total electric flux through any closed surface is directly proportional to the charge enclosed by the surface. This makes it simpler to compute situations where the charge is symmetrically enclosed. In the case of our cube, since the point charge is located at the center, the same amount of electric field lines pass uniformly through each face.

To break it down, the formula we'll use is:

• To find the electric flux, \( \Phi = \frac{Q}{\varepsilon_0} \), where \( Q \) is the enclosed charge.
• For a perfect symmetric shape like a cube and a charge at its center, the electric flux is equally distributed across all faces.

Permittivity of Free Space

The permittivity of free space, symbolized by \( \varepsilon_0 \), is a key physical constant in electromagnetism. It is the ability of a vacuum to allow electric field lines to flow through it and has a defined value of approximately \( 8.854 \times 10^{-12} \text{ C}^2/\text{N}\cdot\text{m}^2 \).

In Gauss's Law, \( \varepsilon_0 \) acts as a proportionality factor. It mediates the relationship between the electric flux and the charge enclosed. This constant tells us how much electric field is produced per unit charge in a vacuum.

• It defines the field strength for a given charge, as the lower the permittivity, the stronger the electric field produced.
• In our exercise, \( \varepsilon_0 \) is used directly when calculating electric flux enclosed due to a point charge inside the cube.

Enclosed Charge

In Gauss's Law, the concept of enclosed charge \( Q \) plays a crucial role. It represents the total charge contained within a closed surface. In many cases, this makes it much easier to calculate complex electromagnetic effects.

For instance, if you know the charge inside a symmetrical shape like a sphere or cube, you can determine the total electric flux emanating through the surface without needing detailed information about the electric field at every single point.

To clarify:

• The "enclosed charge" is the source of the electric field and influences the total amount of electric flux through the surrounding area.
• Regardless of the shape or size of the surface enclosing the charge, according to Gauss's Law, the total electric flux only depends on the amount of enclosed charge.

This is evident in our example where changing the size of the cube does not alter the flux, as long as the charge remains the same inside.