Problem 56 - College Physics

Question

\(\bullet\) A point charge 8.00 \(\mathrm{nC}\) is at the center of a cube with sides of length 0.200 \(\mathrm{m}\) What is the electric flux through (a) the surface of the cube, (b) one of the six faces of the cube?

Step-by-Step Solution

Answer

(a) Flux through cube: \(9.04 \times 10^2 \text{ N} \cdot \text{m}^2/\text{C}\). (b) Flux through one face: \(1.51 \times 10^2 \text{ N} \cdot \text{m}^2/\text{C}\).

Step 1: Understand the Problem

We have a point charge of 8.00 nC placed at the center of a cube with a side length of 0.200 m. We are asked to find the electric flux through (a) the entire surface of the cube, and (b) one of the individual faces of the cube.

Step 2: Apply Gauss's Law

According to Gauss's Law, the total electric flux \( \Phi \) through a closed surface surrounding a charge \( q \) is given by \( \Phi = \frac{q}{\varepsilon_0} \), where \( \varepsilon_0 \) is the vacuum permittivity \( (8.85 \times 10^{-12} \text{ C}^2/\text{N} \cdot \text{m}^2) \).

Step 3: Calculate Total Electric Flux (a)

Using Gauss's Law, the electric flux through the surface of the cube is \[ \Phi_{\text{total}} = \frac{8.00 \times 10^{-9} \text{ C}}{8.85 \times 10^{-12} \text{ C}^2/\text{N} \cdot \text{m}^2} \approx 9.04 \times 10^2 \text{ N} \cdot \text{m}^2/\text{C}. \]

Step 4: Determine Electric Flux through One Face (b)

The cube has six identical faces, and since the charge is symmetrically placed at the center, the electric flux is evenly distributed over each face. Thus, the flux through one face is \( \frac{1}{6} \) of the total flux: \[ \Phi_{\text{face}} = \frac{9.04 \times 10^2 \text{ N} \cdot \text{m}^2/\text{C}}{6} \approx 1.51 \times 10^2 \text{ N} \cdot \text{m}^2/\text{C}. \]

Key Concepts

Gauss's Law

Gauss's Law is a core principle in electromagnetism, which relates the distribution of electric charge to the resulting electric field. It's a simple yet powerful tool used to calculate electric flux. The law states that the total electric flux across a closed surface is equal to the charge enclosed divided by the vacuum permittivity. Mathematically, it is expressed as:

• \( \Phi = \frac{q}{\varepsilon_0} \)

Here, \( \Phi \) represents the electric flux, \( q \) is the enclosed charge, and \( \varepsilon_0 \) is the permittivity of the vacuum.
This law is immensely useful for symmetrical situations such as spherical, cylindrical, and planar symmetries. In this specific exercise, Gauss's Law is leveraged to determine the electric flux through a cube surrounding a point charge, simplifying the process by avoiding direct calculation of the electric field.

Point Charge

A point charge refers to an idealized model of a charged particle, treated as if all its charge is concentrated at a single point in space. In practice, point charges are useful for calculations involving electric fields and forces that would otherwise be too complex to handle with distributed charge.
In this exercise, the point charge is 8.00 nC and is placed at the center of the cube. Introducing a point charge centrally simplifies the calculations of electric flux across the surfaces of the cube. Because of the symmetry, it allows the electric flux to be uniformly distributed, as described by Gauss's Law.
Such simplifications are crucial for efficiently solving problems in electrostatics, especially when dealing with symmetrical geometries.

Cube Geometry

Cube geometry involves understanding a three-dimensional object with six square faces, twelve equal edges, and eight vertices. This geometry significantly impacts how we'd calculate electric properties like flux when a charge, like our point charge, is placed at the center.
For this exercise, we have a cube with side lengths of 0.200 m. The geometry makes it an 'ideal' case for applying Gauss's Law due to its symmetry about the center.

• Each of the six faces of the cube receives an equal proportion of the total electric flux.
• This geometry ensures simplification: if we calculate the total flux through the cube, we can equally distribute that across its six faces.

Therefore, by using symmetry inherent in cube geometry, it becomes easier to break down complex electrostatics problems.

Vacuum Permittivity

Vacuum permittivity, symbolized as \( \varepsilon_0 \), is a fundamental physical constant involved in the formulation of the laws of electromagnetism. It describes how electric fields interact with a vacuum and is crucial for calculations involving electric flux and fields.

• The value of vacuum permittivity is \( 8.85 \times 10^{-12} \text{ C}^2/\text{N} \cdot \text{m}^2 \).
• In the context of Gauss's Law, \( \varepsilon_0 \) facilitates calculating the total electric flux across a closed surface given a known charge.

Incorporating \( \varepsilon_0 \) in this exercise, we can determine the electric flux through the cube’s entire surface area by dividing the charge by the permittivity.
Understanding \( \varepsilon_0 \) is crucial as it is used as a bridge between the charge and the resulting electric field, directly affecting how flux calculations are made in theoretical and practical applications.