Problem 34 - University Physics with Modern Physics

Question

A cube has sides of length \(L =\) 0.300 m. One corner is at the origin (Fig. E22.6). The nonuniform electric field is given by \(\overrightarrow{E} =\) (-5.00 N/C \(\cdot\) m)\(x\hat{\imath}\) + (3.00 N/C \(\cdot\) m)\(z \hat{k}\). (a) Find the electric flux through each of the six cube faces \(S_1, S_2, S_3, S_4, S_5\), and \(S_6\) . (b) Find the total electric charge inside the cube.

Step-by-Step Solution

Answer

Zero electric flux for S1, S3, S5. Non-zero flux for S2 and S6; total charge is found using Gauss's Law.

Step 1: Understand the Problem

We need to find the electric flux through each face of a cube located at the origin in a non-uniform electric field, and then find the total electric charge inside the cube.

Step 2: Identify Electric Field Components

The electric field is given by \( \overrightarrow{E} = (-5.00 \text{ N/C} \cdot \text{ m})x\hat{\imath} + (3.00 \text{ N/C} \cdot \text{ m})z \hat{k} \). This means the field has components in the \(x\) and \(z\) directions.

Step 3: Flux Through Faces Perpendicular to X-axis

For faces perpendicular to \(x\), i.e., \(S_1\) and \(S_2\), the normal vectors are \( \hat{\imath} \) and \( -\hat{\imath} \). The electric field component affecting these faces is \( -5.00 x \). Calculate flux using \( \Phi = \int \overrightarrow{E} \cdot d\overrightarrow{A} \).

Step 4: Flux Through Faces Perpendicular to Y-axis

For faces perpendicular to \(y\), i.e., \(S_3\) and \(S_4\), normal vectors are \( \hat{\jmath} \) and \( -\hat{\jmath} \). Since there is no \(y\) component in \( \overrightarrow{E} \), the electric flux through these faces is zero: \( \Phi = 0 \).

Step 5: Flux Through Faces Perpendicular to Z-axis

For faces perpendicular to \(z\), i.e., \(S_5\) and \(S_6\), normal vectors are \( \hat{k} \) and \( -\hat{k} \). The electric field component is \(3.00z\). Calculate the flux: \( \Phi = \int \overrightarrow{E} \cdot d\overrightarrow{A} \).

Step 6: Find Electric Flux for Each Face

Calculate the electric flux across each relevant face using the given field equations and the geometry of the cube. Essentially, the flux through \(S_1\) is zero, \(S_2\) is \(-2.25 \text{ Nm}^2/\text{C}\), \(S_5\) is \(0\), and \(S_6\) is \(0.81 \text{ Nm}^2/\text{C}\).

Step 7: Calculate Total Electric Flux

Sum the electric fluxes from all six faces. From previous calculations, the total flux coming in from \(S_2\) and \(S_6\) affects total flux non-trivially.

Step 8: Use Gauss's Law to Find Charge

Using Gauss's Law, \( \Phi_{total} = \frac{Q_{inside}}{\varepsilon_0} \), where \( \varepsilon_0 = 8.854 \times 10^{-12} \text{ C}^2/\text{Nm}^2 \). Calculate \(Q_{inside}\) using the total flux found from sum of all faces.

Key Concepts

Non-Uniform Electric Field

An electric field is said to be non-uniform when its magnitude and direction change from one point to another. In the given problem, our electric field is non-uniform, meaning it varies depending on the position along the cube. This particular field can be expressed as \( \overrightarrow{E} = (-5.00 \text{ N/C} \cdot \text{ m})x\hat{\imath} + (3.00 \text{ N/C} \cdot \text{ m})z \hat{k} \). This equation means that the field strength depends on the coordinates \(x\) and \(z\), with no component along the \(y\) axis.

Understanding how non-uniform fields work is crucial because it causes variations in electric flux through the cube's different faces. Unlike a uniform field, where the strength and direction stay constant, non-uniform fields demand that we carefully calculate the flux for every face, noting how changes occur due to different field components.

• In the \(x\)-direction, the field strength is \(-5.00x\), using \(\hat{\imath}\).
• In the \(z\)-direction, field strength is \(3.00z\), using \(\hat{k}\).

Understanding the underlying non-uniformity helps us comprehend why electric flux values differ so vastly across the faces of the cube.

Gauss's Law

Gauss's Law is a fundamental principle in electromagnetism. It connects the electric flux through a closed surface to the charge contained within it. Formally, Gauss’s Law is written as \( \Phi_{total} = \frac{Q_{inside}}{\varepsilon_0} \), where \( \Phi_{total} \) is the total electric flux through the surface, \( Q_{inside} \) is the net charge enclosed, and \( \varepsilon_0 \) is the permittivity of free space, valued at approximately \( 8.854 \times 10^{-12} \text{ C}^2/\text{Nm}^2 \).

Using Gauss's Law to connect electric flux and charge allows us to calculate how much electric charge resides within a certain boundary. Here, after summing the fluxes through all the cube’s faces, we find it necessary to assess the internal charge via Gauss's Law. This is the principle that allows us to deduce \( Q_{inside} \) given we have found \( \Phi_{total} \).

• Gauss’s Law holds for any closed surface, making it highly versatile and applicable in different geometric scenarios.
• This law is particularly valuable when the symmetry of the problem (like our cube) simplifies the calculations of flux.

Understanding and applying Gauss’s Law is similar to looking at how the net field arises from sources within if you imagine the entire structure as a combination of multiple field lines passing through.

Electric Charge

Electric charge is a property of matter that causes it to experience a force when placed in an electric field. Charges can be positive or negative and their interaction forms the basis of many physical phenomena. In the context of this problem, we calculate the total electric charge \( Q_{inside} \) found using Gauss’s Law. Once we have the total electric flux determined from the surface of the cube, relating it back to the amount of enclosed charge becomes straightforward.

Charges cause electric fields and their distribution within a space determines how these fields behave. Here, the presence of a charge within the cube influences the shifting field across the faces of the cube. A clear understanding of how this works is crucial for electric flux calculations which essentially trace how the field interacts with surfaces it's exposed to.

• Electric charges can attract or repel each other: like charges repel, opposite charges attract.
• Quantifying charge in enclosed spaces reveals how charged particles influence fields on boundaries.

Appreciating the nature of electric charge and its interaction across surfaces offers valuable insights into electromagnetic fields and flux.