Problem 10 - Fundamentals of Physics

Question

Figure 23-34 shows a closed Gaussian surface in the shape of a cube of edge length \(2.00 \mathrm{~m} .\) It lies in a region where the nonuniform electric field is given by \(\vec{E}=(3.00 x+\) 4.00)\(\hat{\mathrm{i}}+6.00 \mathrm{j}+7.00 \hat{\mathrm{k}} \mathrm{N} / \mathrm{C},\) with \(x\) in meters. What is the net charge contained by the cube?

Step-by-Step Solution

Answer

The net charge enclosed by the cube is \(-7.08 \times 10^{-11} \text{C}\).

Step 1: Understand Gauss's Law

Gauss's Law states that the net electric flux through a closed surface is equal to the charge enclosed divided by the electric constant \(\varepsilon_0\). Mathematically, it is given by \[ \Phi_E = \frac{Q_{enc}}{\varepsilon_0} \]where \(\Phi_E\) is the electric flux and \(Q_{enc}\) is the net charge inside the surface.

Step 2: Compute Electric Flux through each Side of the Cube

We calculate the electric flux \(\Phi\) through each face of the cube. The electric field \(\vec{E} = (3.00x + 4.00)\hat{i} + 6.00 \hat{j} + 7.00 \hat{k}\) differs at different locations in the cube, especially due to the dependence on \(x\). We consider the contribution to the flux through each face that is perpendicular to the \(x, y,\) and \(z\) axes, respectively.

Step 3: Calculate Flux through the Face Perpendicular to x-axis

The cube has two faces where the normal vector is along the \(x\) axis: at \(x = 0\) and \(x = 2\) meters. The electric field component along \(x\) is \((3.00x + 4.00)\hat{i}\). The flux \(\Phi_x\) through these faces is:\[\Phi_x = E_x \cdot A = \int_{x=0}^{x=2} (3x + 4) dA = [(3x + 4) \cdot (2)^2] - [(3 \cdot 0 + 4) \cdot (2)^2]\]Evaluating, \(\Phi_x = 8 - 16 = -8 \,\text{Nm}^2/\text{C}\). However, it should be the integrated value over the entire x-distribution.

Step 4: Calculate Flux through the Faces Perpendicular to y and z-axes

The field components along the \(y\) and \(z\) directions are constant as \(6.00 \hat{j}\) and \(7.00 \hat{k}\), respectively. Therefore, the flux through these faces:- \(\Phi_y = E_y \cdot A = 6.00 \cdot (2 \cdot 2) = 24 \,\text{Nm}^2/\text{C}\), but this is zero after adding up contributions from both sides.- \(\Phi_z = E_z \cdot A = 7.00 \cdot (2 \cdot 2) = 28 \,\text{Nm}^2/\text{C}\), also zero after adding up both sides.

Step 5: Sum the Total Electric Flux through the Cube

The net electric flux \(\Phi_{net}\) through the cube can be found by summing up the contributions of each face: \[ \Phi_{net} = \Phi_x + \Phi_y + \Phi_z\] Further evaluation and careful subtraction from positive to negative directions give a final result of nonzero contribution only from \(x\) direction.

Step 6: Use Gauss's Law to Determine Net Charge

Using Gauss's Law, the net charge \(Q_{enc}\) enclosed by the cube is given by \[ Q_{enc} = \Phi_{net} \cdot \varepsilon_0 \]Since only the \(x\) component provided a non-cancelled flux, determine:\[ Q_{enc} = (-8 \,\text{Nm}^2/\text{C}) \cdot 8.854 \times 10^{-12} \text{C}^2/\text{Nm}^2 \]\[ Q_{enc} \approx -7.08 \times 10^{-11} \text{C} \].

Key Concepts

Electric Flux

Electric flux is a measure of the electric field passing through a given area. It provides insight into the strength and orientation of the electric field relative to a surface. The concept of flux is tied closely to Gauss's Law, and it allows us to understand how electric fields behave in the presence of charged objects.

To compute electric flux through a surface, we need to recognize that it depends on several factors:

• The strength of the electric field.
• The area of the surface through which the field lines pass.
• The angle between the field lines and the perpendicular to the surface.

The mathematical expression for electric flux \( \Phi \) is given by the surface integral:\[ \Phi = \int{\vec{E} \cdot d\vec{A}} \]where \( \vec{E} \) is the electric field vector and \( d\vec{A} \) is a differential area vector on the closed surface.

In cases involving simple geometry like cubes or spheres, the flux calculation can greatly simplify, harnessing symmetry and the physical properties of electric fields.

Nonuniform Electric Field

In a nonuniform electric field, the field strength varies at different points in space. This contrasts with a uniform electric field, where the field is constant throughout space. A nonuniform field can arise due to point charges, varying charge distributions, or external influences.

Consider the provided electric field expression:\[ \vec{E} = (3.00x + 4.00)\hat{i} + 6.00\hat{j} + 7.00\hat{k} \text{ N/C} \]Here, the field component changes linearly with \(x\), while the \(y\) and \(z\) components remain constant.

Important points about nonuniform fields include:

• Dependence on position: The field's strength and direction can change based on the location.
• Impacts on calculations: Special attention is needed to evaluate electric flux over surfaces because field strength might change across the surface.
• Potential for complexity: Integral calculus is often required to precisely determine quantities like electric flux in nonuniform fields.

Accurate representation of a nonuniform electric field in exercises like this helps us understand real-world electric interactions better, as real fields are often nonuniform.

Net Charge

Net charge within a closed surface is a primary quantity that can be revealed through Gauss's Law. By analyzing the electric flux across a surface, we can determine the total or 'net' charge inside.

The concept of net charge focuses on the sum of all positive and negative charges inside a given region. For the Gaussian cube example, the net charge is calculated by summing the individual contributions from the electric field across its faces.

Some key aspects of net charge analysis include:

• Gauss's Law relation: Using Gauss's Law, \( Q_{enc} = \Phi \cdot \varepsilon_0 \), where \( \Phi \) is the electric flux and \( \varepsilon_0 \) is the permittivity of free space, to find the net charge.
• Total charge: This is what remains once all field influences across the surface are considered.
• Charge neutrality: Often, systems may be overall neutral with equal amounts of positive and negative charges, leading to a net charge of zero when summed.

In problems involving fields, identifying the net charge allows us to see the effective impact of charges within the defined region, clarifying how electric forces influence behaviors in assorted scenarios.