Problem 11 - Fundamentals of Physics

Question

co Figure 23-35 shows a closed Gaussian surface in the shape of a cube of edge length \(2.00 \mathrm{~m}\), with one corner at \(x_{1}=5.00 \mathrm{~m}, y_{1}=4.00\) \(\mathrm{m}\). The cube lies in a region where the electric field vector is given by \(\vec{E}=-3.00 \hat{\mathrm{i}}-4.00 \mathrm{y}^{2} \mathrm{j}+3.00 \hat{\mathrm{k}} \mathrm{N} / \mathrm{C}\), with \(y\) in meters. What is the net charge contained by the cube?

Step-by-Step Solution

Answer

The net charge contained in the cube is \(-2.832 \times 10^{-9} \, \mathrm{C}\).

Step 1: Review Gaussian Surface and Electric Field Components

The Gaussian surface is a cube with edge length \(2.00 \, \mathrm{m}\), and the electric field is given by \(\vec{E} = -3.00 \hat{\mathrm{i}} - 4.00 y^2 \hat{\mathrm{j}} + 3.00 \hat{\mathrm{k}}\). This field has three components: \(-3.00\) along the \(x\)-direction, \(-4.00y^2\) along the \(y\)-direction, and \(3.00\) along the \(z\)-direction.

Step 2: Apply Gauss's Law

Gauss's Law states \[\Phi_E = \frac{Q_{\text{enc}}}{\varepsilon_0}\]where \(\Phi_E\) is the electric flux through the closed surface, \(Q_{\text{enc}}\) is the charge enclosed, and \(\varepsilon_0 = 8.85 \times 10^{-12} \, \mathrm{C^2/N \, m^2}\) is the permittivity of free space.

Step 3: Calculate Individual Flux Contributions

For each face of the cube, calculate the flux contribution. The flux due to the \(x\)-component is zero because the field \(-3.00 \hat{\mathrm{i}}\) is constant and the areas have paired opposites. The flux due to the \(z\)-component is also zero for similar reasons. Evaluate the \(y\)-component, noting it varies with \(y\).

Step 4: Calculate the Flux for the \(y\)-Component

The cube's faces at constant \(y\) are at \(y = 4.00\, \mathrm{m}\) and \(y = 6.00\, \mathrm{m}\). Flux through these faces is: \[\Phi_{y,1} = E_j(y_1) \cdot A = (-4.00 \cdot 4.00^2) \cdot (2.00)^2\Phi_{y,2} = E_j(y_2) \cdot A = (-4.00 \cdot 6.00^2) \cdot (2.00)^2\]Calculate each:\[\Phi_{y,1} = -4 \cdot 16 \cdot 4 = -256 \, \mathrm{N \, m^2/C}\Phi_{y,2} = -4 \cdot 36 \cdot 4 = -576 \, \mathrm{N \, m^2/C}\]Then, \(\Phi_{\text{total}} = \Phi_{y,2} - \Phi_{y,1} = -576 + 256 = -320 \, \mathrm{N \, m^2/C}\).

Step 5: Use Total Flux to Find Enclosed Charge

With the total flux calculated, plug into Gauss's Law:\[Q_{\text{enc}} = \Phi_E \times \varepsilon_0 = -320 \, \mathrm{N \, m^2/C} \times 8.85 \times 10^{-12} \, \mathrm{C^2/N \, m^2}\]Calculate \(Q_{\text{enc}}\):\[Q_{\text{enc}} = -320 \, \times 8.85 \times 10^{-12} = -2.832 \times 10^{-9} \, \mathrm{C}\]

Step 6: Final Answer

The net charge contained in the cube is \(-2.832 \times 10^{-9} \, \mathrm{C}\).

Key Concepts

Electric Flux

Electric flux is a crucial concept to understand when studying Gauss's Law. It represents the number of electric field lines passing through a given surface. In a sense, it measures how much of the electric field "flows" through the surface. To calculate electric flux, we use the formula:

• \( \Phi_E = \vec{E} \cdot \vec{A} \)

Here, \( \vec{E} \) is the electric field vector, and \( \vec{A} \) is the area vector perpendicular to the surface. The dot product indicates that only the component of the field that runs perpendicular to the surface contributes to the flux. Thus, the angle between \( \vec{E} \) and \( \vec{A} \) is vital.

When there are multiple components of the electric field, as in this problem, you calculate the flux for each component separately. Then, you add them to find the total electric flux through the entire surface.

Gaussian Surface

A Gaussian surface is an imaginary closed surface used in Gauss's Law to simplify calculations of electric fields. Choosing the right Gaussian surface is essential for solving a problem efficiently. It should align with the symmetry of the charge distribution you're examining. Different scenarios might demand different shapes, such as spheres, cylinders, or cubes.

In our exercise, the Gaussian surface is a cube with edges of length \(2.00 \, \text{m}\), placed conveniently to simplify field calculations. The symmetry of the cube and the uniformity of the electric field over its surfaces make it an excellent choice. The cube's placement and alignment ensure that for certain components of the field, their net contribution can be determined or even deemed zero due to opposing or equal flux cancelation.

Enclosed Charge

The enclosed charge in a Gaussian surface is what we aim to find using Gauss's Law. Gauss's Law states that the total electric flux \( \Phi_E \) passing through a closed surface is proportional to the enclosed charge \( Q_{\text{enc}} \):

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

Here, \(\varepsilon_0\) is the permittivity of free space, given as \(8.85 \times 10^{-12} \, \text{C}^2/\text{N} \, \text{m}^2\).

By calculating the electric flux through the Gaussian surface, we can derive the enclosed charge. In our problem, once the total flux was calculated by summing the contributions from different parts of the Gaussian surface, we used this relation to determine the net charge contained within the cube.

Electric Field Components

Electric field components describe the direction and magnitude of the electric field in different directions. They help in simplifying the computation of electric flux through surfaces. In the given problem, the electric field is denoted by three components: \(-3.00\) in the \(x\)-direction, \(-4.00y^2\) in the \(y\)-direction, and \(3.00\) in the \(z\)-direction.

It is essential to analyze these components individually, as each impacts the flux through the Gaussian surfaces differently. In this case, the constant components along the \(x\) and \(z\) directions mean their contributions to the net flux are zero, due to similar opposite faces on the cube.

The \(y\)-component, however, varies with the square of \(y\), meaning it has a different contribution to the electric flux through the two cube faces perpendicular to the \(y\)-axis. It's crucial to understand how changes in field components affect the overall calculation to correctly solve the problem using Gauss's Law.