Problem 74 - Fundamentals of Physics

Question

A uniform charge density of \(500 \mathrm{nC} / \mathrm{m}^{3}\) is distributed throughout a spherical volume of radius \(6.00 \mathrm{~cm}\). Consider a cubical Gaussian surface with its center at the center of the sphere. What is the electric flux through this cubical surface if its edge length is (a) \(4.00 \mathrm{~cm}\) and \((\mathrm{b}) 14.0 \mathrm{~cm} ?\)

Step-by-Step Solution

Answer

The electric flux is approximately \(3.06 \times 10^{3} \mathrm{~N} \cdot \mathrm{m}^{2}/\mathrm{C}\) for both cube sizes.

Step 1: Calculate the Total Charge in the Sphere

The charge density is given as \( \rho = 500 \mathrm{~nC} / \mathrm{m}^{3} \). The radius of the sphere is \( R = 6.00 \mathrm{~cm} = 0.06 \mathrm{~m} \). The volume of the sphere, \( V \), is given by \( V = \frac{4}{3} \pi R^3 \). Substituting the values, we find the total charge \( Q \):\[ Q = \rho \times V = 500 \times \frac{4}{3} \pi (0.06)^3 \approx 2.71 \times 10^{-8} \mathrm{~C}. \]

Step 2: Apply Gauss's Law for the Gaussian Surfaces

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

Step 3: Calculate Electric Flux for Edge Length 4 cm

The Gaussian cube with edge length \( a = 4.00 \mathrm{~cm} = 0.04 \mathrm{~m} \) is smaller than the sphere. Hence, the cube is completely inside the sphere. This means the entire charge of the sphere is enclosed. The electric flux is given by:\[ \Phi = \frac{Q}{\varepsilon_0} = \frac{2.71 \times 10^{-8}}{8.85 \times 10^{-12}} \approx 3.06 \times 10^{3} \mathrm{~N} \cdot \mathrm{m}^{2}/\mathrm{C}. \]

Step 4: Calculate Electric Flux for Edge Length 14 cm

The Gaussian cube with edge length \( a = 14.0 \mathrm{~cm} = 0.14 \mathrm{~m} \) encapsulates the entire sphere, as the edge length extends beyond the sphere’s radius. Therefore, the entire charge is also enclosed here, leading to the same flux calculation:\[ \Phi = \frac{Q}{\varepsilon_0} = \frac{2.71 \times 10^{-8}}{8.85 \times 10^{-12}} \approx 3.06 \times 10^{3} \mathrm{~N} \cdot \mathrm{m}^{2}/\mathrm{C}. \]

Key Concepts

Understanding Electric Flux

Electric flux is a fundamental concept in electromagnetism that describes how an electric field passes through a given surface. It is represented by \( \Phi \), and according to Gauss's Law, is given by:

• \( \Phi = \frac{Q_{\text{enc}}}{\varepsilon_0} \)

This means the electric flux depends on two key factors:- The amount of charge \( Q_{\text{enc}} \) enclosed by the surface.- The permittivity of free space \( \varepsilon_0 \), a constant that relates the units of charge to the electric field.
If a surface encloses charge, the electric flux is proportional to the amount of that charge. Imagine it as how many electric field "lines" pass through the surface area.
Electric flux is instrumental in determining how electric fields interact with materials and how charges within a system affect those fields.

Exploring Charge Density

Charge density is a measure of electric charge per unit volume, area, or length. It's a critical value specifying how charge is distributed in space.

• Volume charge density \( \rho \) is given in units of charge per volume, such as \( \mathrm{nC/m}^3 \).
• The formula for volume charge density is \( \rho = \frac{Q}{V} \).

In our example, we have a uniform charge density of \( 500 \mathrm{nC/m}^3 \) over a spherical volume.
Understanding charge density is essential for calculating the total charge in any volume, as it allows for determination based on the volume's shape and size.

The Concept of Spherical Volume

A spherical volume refers to the three-dimensional space enclosed within a sphere, defined by its radius. The volume \( V \) of a sphere with radius \( R \) is given by:

• \( V = \frac{4}{3} \pi R^3 \)

This formula helps determine the total volume in which charge is distributed in our problem.
When we know the radius of the sphere and the charge density, computing the total charge becomes straightforward using:

• \( Q = \rho \times V \)

The sphere's symmetry also simplifies the application of Gauss's Law, making it easier to calculate electric flux.

Permittivity of Free Space: A Gateway Constant

The permittivity of free space, \( \varepsilon_0 \), is a fundamental physical constant crucial in electromagnetics.

• \( \varepsilon_0 = 8.85 \times 10^{-12} \mathrm{C}^2/\mathrm{N} \cdot \mathrm{m}^2 \).

It sets the scale of electric forces and fields in a vacuum and is a bridge that connects electric charge and electric field strength.
In Gauss's Law, the permittivity of free space allows us to calculate the electric flux through a surface by relating the enclosed charge to how the electric field spreads out from it. This constant remains unchanged in empty space, ensuring consistent and predictable calculations for electric fields.