Ch 24: Gauss' Law

Knight Calc - Physics for Scientists and Engineers 5th Edition

Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook

Knight Calc 5th Edition

Ch 24: Gauss' Law

Problem 31

Chapter 24, Problem 31

A 10 nC charge is at the center of a 2.0 m x 2.0 m x 2.0 m cube. What is the electric flux through the top surface of the cube?

Verified step by step guidance

Step 1: Recall Gauss's Law, which states that the electric flux (Φ) through a closed surface is given by Φ = Q / ε₀, where Q is the total charge enclosed and ε₀ is the permittivity of free space (ε₀ ≈ 8.85 × 10⁻¹² C²/(N·m²)).

Step 2: Understand that the charge is symmetrically placed at the center of the cube. Due to symmetry, the electric flux is evenly distributed across all six faces of the cube.

Step 3: Calculate the total electric flux through the entire cube using Gauss's Law: Φ_total = Q / ε₀. Substitute Q = 10 nC (10 × 10⁻⁹ C) into the formula.

Step 4: Divide the total electric flux by 6 to find the flux through one face of the cube, since the flux is evenly distributed across all six faces. Use the formula Φ_face = Φ_total / 6.

Step 5: Express the result in terms of the given values and constants, ensuring the units are consistent throughout the calculation. The flux through the top surface of the cube is Φ_face.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.

Video duration:

Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Gauss's Law

Gauss's Law states that the electric flux through a closed surface is proportional to the charge enclosed within that surface. Mathematically, it is expressed as Φ = Q_enc/ε₀, where Φ is the electric flux, Q_enc is the enclosed charge, and ε₀ is the permittivity of free space. This principle is fundamental in electrostatics and allows for the calculation of electric fields in symmetric charge distributions.

Electric Flux

Electric flux is a measure of the quantity of electric field lines passing through a given surface area. It is defined as the dot product of the electric field vector and the area vector, integrated over the surface. The unit of electric flux is the volt-meter (V·m), and it helps in understanding how electric fields interact with surfaces, particularly in the context of Gauss's Law.

Symmetry in Charge Distribution

In problems involving electric fields and flux, symmetry plays a crucial role in simplifying calculations. For a uniformly distributed charge, such as a point charge at the center of a cube, the electric field is symmetric and uniform across the surfaces of the cube. This symmetry allows us to deduce that the electric flux through each face of the cube is equal, facilitating the calculation of the total flux through any specific surface.

Recommended video:

Guided course

12:39

Solving Symmetric Launch Problems

Related Practice

Textbook Question

A spherically symmetric charge distribution produces the electric field $\vec{E} = (5000 r^{2}) \hat{r}$ N/C, where r is in m. How much charge is inside this 40-cm-diameter spherical surface?

views

Textbook Question

Charges $q_{1} = - 4 Q$ and $q sub 2 equals positive 2 Q$ are located at $𝓍 = - a$ and $𝓍 = + a$ , respectively. What is the net electric flux through a sphere of radius $2 a$ centered (a) at the origin and (b) at $script x equals 2 a$ ?

views

Textbook Question

Find the electric fluxes Φ_A to Φ_E through surfaces A to E in FIGURE P24.29.

views

Textbook Question

A 20-cm-radius ball is uniformly charged to 80 nC. How much charge is enclosed by spheres of radii 5, 10, and 20 cm?

views

Textbook Question

A thin, horizontal, 10-cm-diameter copper plate is charged to 3.5 nC. If the charge is uniformly distributed on the surface, what are the strength and direction of the electric field 0.1 mm above the center of the top surface of the plate?

Textbook Question

A spark occurs at the tip of a metal needle if the electric field strength exceeds $3.0 \times 10^{6}$ N/C, the field strength at which air breaks down. What is the minimum surface charge density for producing a spark?

views