Question

A single positive point charge, \(q,\) is at one corner of a cube with sides of length \(L,\) as shown in the figure. The net electric flux through the three adjacent sides is zero. The net electric flux through each of the other three sides is a) \(q / 3 \epsilon_{0}\). b) \(q / 6 \varepsilon_{0}\). c) \(q / 24 \epsilon_{0}\) - d) \(q / 8 \epsilon_{0}\)

Short answer

Answer: The net electric flux through each of the other three sides is q / 3ε₀ (Option a).

Step-by-step solution

Determine total electric flux through the cube

Using Gauss's law, we can find the total electric flux through the cube. The flux through the entire cube is equal to the charge enclosed divided by the vacuum permittivity: Φ_total = Q / ε₀ In this case, the enclosed charge is equal to the single point charge at the corner of the cube, so: Φ_total = q / ε₀

Divide total electric flux by the number of other three sides

As mentioned in the analysis, the total electric flux through the entire cube should be equal to the electric flux through the other three sides since the net electric flux through the three adjacent sides is zero. Therefore, we divide the total electric flux by 3 to find the flux through each of the other three sides: Φ_side = (q / ε₀) / 3 Simplifying, we get: Φ_side = q / (3ε₀) So the net electric flux through each of the other three sides is:

Answer

The net electric flux through each of the other three sides is q / 3ε₀ (Option a).

Key concepts

Gauss's Law

Understanding Gauss's law is essential for grasping the principles of electric flux, as it provides a powerful tool for calculating the flux through a closed surface. Gauss's law states that the net electric flux through any closed surface is proportional to the total electric charge enclosed within that surface. Mathematically, this is expressed as:
\[\Phi_{\text{total}} = \frac{Q}{\varepsilon_{0}}\]
where \(\Phi_{\text{total}}\) is the total electric flux, \(Q\) is the total charge enclosed, and \(\varepsilon_{0}\) is the vacuum permittivity. The beauty of Gauss's law lies in its ability to simplify complex electric field calculations, especially when dealing with symmetrical charge distributions. For example, the electric flux through the surface of a sphere with a charged center is easily found by knowing the total charge inside. In the textbook exercise, Gauss's law enables us to conclude that the flux through a cube with a corner charge is evenly distributed through the sides not directly adjacent to the charge.
To improve the exercise, it's beneficial to provide visual aids or illustrations that show how the charge is placed within the cube and how the flux distributes. This enhancement helps students better visualize and understand the concept of electric flux in relation to Gauss's law.

Electric Charge

The concept of electric charge is one of the cornerstones of electromagnetism. Charges are the source of electric fields and are quantified in Coulombs (C). There are two types of charges, positive and negative, and their interaction governs the electric forces in the universe. Charges exert forces on one another, with like charges repelling and opposite charges attracting, as described by Coulomb's law.
In our exercise scenario, we consider a single positive point charge \(q\), which is responsible for creating an electric field around itself. This field extends in all directions and is what generates the electric flux through surfaces like the sides of our cube. Understanding how the location of a charge affects the electric flux through a geometric shape is key — if the point charge were at the center of the cube rather than its corner, the electric flux distribution would be different.
When improving this exercise, highlighting the fundamental principles of charge behavior and the influence of charge placement helps students comprehend why the flux on the cube's adjacent sides is zero while it is nonzero on the other sides.

Vacuum Permittivity

Vacuum permittivity, denoted as \(\varepsilon_{0}\), is a physical constant that plays a crucial role in electromagnetism. It describes how much resistance the vacuum (free space) presents to electric field lines, effectively controlling the strength of the electric field generated by a given charge. Its value is approximately \(8.85 \times 10^{-12} \frac{C^{2}}{N \cdot m^{2}}\). The lower the vacuum permittivity, the stronger the electric field produced by a charge.
This constant is essential in Gauss's law, which relates the electric flux through a surface to the charge inside and the vacuum permittivity. In our cube exercise, we used \(\varepsilon_{0}\) to calculate the electric flux through the cube sides given a single charge. The permittivity of the vacuum sets the proportionality between the electric flux and the charge.
When presenting such exercises, it’s crucial to explain why vacuum permittivity is critical for determining the field's strength and how it's applicable in various situations, not just a vacuum. Additionally, relating vacuum permittivity to real-world applications or more familiar concepts — such as how it affects charging and discharging in capacitors — can enhance students' understanding.