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 (b) \(14.0 \mathrm{~cm} ?\)

Step-by-Step Solution

Answer

(a) Flux for 4.00 cm: 36.2 N·m²/C; (b) Flux for 14.0 cm: 1.53 × 10³ N·m²/C

Step 1: Understand the problem

We need to find the electric flux through a cubical Gaussian surface inside and outside a uniformly charged sphere. The charge density is given, and the problem asks for the flux with two different cube sizes.

Step 2: Recall Gauss's Law

Gauss's Law states that the electric flux \( \Phi_E \) through a closed surface is equal to the charge enclosed \( Q_{enc} \) divided by the electric constant \( \varepsilon_0 \): \[ \Phi_E = \frac{Q_{enc}}{\varepsilon_0} \].

Step 3: Calculate enclosed charge for edge 4.00 cm

The side of the cube is \(4.00\text{ cm} = 0.04\text{ m}\). This cube is within the sphere, so we calculate the volume \( V_{cube} = (0.04)^3 \). The enclosed charge is \( Q_{enc} = \rho V_{cube} \), where \( \rho = 500\text{ nC/m}^3 \). The enclosed charge is thus \( 500 \times 10^{-9} \times (0.04)^3 \) C.

Step 4: Calculate flux for edge 4.00 cm

Using Gauss's Law and \(Q_{enc} = 500 \times 10^{-9} \times 0.04^3\), the flux is \( \Phi_E = \frac{Q_{enc}}{\varepsilon_0} = \frac{500 \times 10^{-9} \times 0.04^3}{8.85 \times 10^{-12}}\) N·m²/C.

Step 5: Calculate enclosed charge for edge 14.0 cm

The side of the cube is \(14.0\text{ cm} = 0.14\text{ m}\), which is larger than the sphere's radius, so it encloses the entire sphere. The sphere's volume \( V_{sphere} = \frac{4}{3} \pi (0.06)^3 \). Therefore, \( Q_{enc} = \rho V_{sphere} = 500 \times 10^{-9} \times \frac{4}{3} \pi (0.06)^3 \) C.

Step 6: Calculate flux for edge 14.0 cm

Using Gauss's Law for the entire sphere, the flux is \( \Phi_E = \frac{Q_{enc}}{\varepsilon_0} = \frac{500 \times 10^{-9} \times \frac{4}{3} \pi (0.06)^3}{8.85 \times 10^{-12}}\) N·m²/C.

Key Concepts

Electric Flux

Electric flux is a measure of the electric field passing through a given surface.
Imagine it as a flow of electric field lines through an area. The net flux depends on the field's strength, the area size, and their orientation.
The more field lines pass through, perpendicularly to the surface, the greater the electric flux.Gauss's Law, a key principle in electromagnetism, calculates this flux.
It's expressed as:

• \( \Phi_E = \frac{Q_{enc}}{\varepsilon_0} \)

Here, \( \Phi_E \) is the electric flux, \( Q_{enc} \) is the enclosed charge, and \( \varepsilon_0 \) is the permittivity of free space, a constant.For a closed surface, the direction matters.
If field lines exit, the flux is positive; if they enter, it's negative.
This conceptual tool allows straightforward computation of the electric field, even in complex scenarios.

Charge Density

Charge density describes how electric charge is distributed in space.
It is measured in terms of charge per unit volume, typically expressed in \( ext{nC/m}^3 \) or another similar unit.In this problem, we have a uniformly charged sphere with a charge density \( \rho = 500 \text{ nC/m}^3 \).
This means every cubic meter of the sphere contains 500 nanocoulombs of charge.
Uniform distribution implies that this charge is spread evenly throughout the sphere.Charge density is crucial for calculating the enclosed charge in a Gaussian surface.
By multiplying the charge density \( \rho \) by the volume of the object, we find the total charge inside.
For different volumes, just adjust the volume part of the calculation to match what you are evaluating.

Gaussian Surface

A Gaussian surface is an imaginary closed surface used to apply Gauss's Law.
This surface helps us simplify the calculation of electric fields in complex situations. In this exercise, the Gaussian surfaces are cubes, situated within and outside of a spherical charge distribution.
Choosing the right Gaussian surface shape helps exploit symmetries in the problem, like spheres or cylinders. The cube of different sizes serves as the Gaussian surface here.
When calculating the electric flux through these surfaces, we first check if the Gaussian surface is entirely within, entirely outside, or partially inside the charge distribution.
This determines how much charge is enclosed, affecting our calculations. A key insight is realizing when the surface encloses all the charge, simplifying the problem.
The symmetry and the choice of Gaussian surface shape play vital roles in applying Gauss's Law efficiently.

Enclosed Charge

Enclosed charge is the total charge contained within a Gaussian surface.
It is pivotal for determining the electric flux through this surface by Gauss’s Law.When dealing with different sized Gaussian surfaces and spatial charge distributions, calculating the enclosed charge varies:

• For a cube entirely inside the sphere, compute the charge by the cube's volume.
• For a cube surrounding the entire sphere, the charge is the total charge of the sphere.

In the exercise, for a cube with an edge of \( 4.00 \text{ cm} \), only the part inside the sphere contributes.
Compute the cube's volume and multiply by the charge density to find \( Q_{enc} \).For a cube with an edge of \( 14.0 \text{ cm} \), it encloses the entire sphere.
Here, \( Q_{enc} \) is simply the total charge of the whole sphere, derived from its known volume and charge density.
Understanding this concept simplifies applying Gauss’s Law to compute the electric flux effectively.