Problem 32 - University Physics with Modern Physics

Question

12.32. A cube has sides of length \(L .\) It is placed with one corner at the origin as shown in Fig. 22.32 . The electric field is uniform and given by \(\overrightarrow{\boldsymbol{E}}=-B \hat{\boldsymbol{i}}+\boldsymbol{C} \hat{\boldsymbol{j}}-\boldsymbol{D} \hat{\boldsymbol{k}},\) where \(\boldsymbol{B}, \boldsymbol{C},\) and \(\boldsymbol{D}\) are positive constants. (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 electric flux through the entire cube.

Step-by-Step Solution

Answer

The net electric flux through the cube is 0.

Step 1: Understand the Position of Faces

The cube is aligned with one corner at the origin. Each face of the cube is perpendicular to one of the coordinate axes. Face \(S_1\) is along the yz-plane, \(S_2\) along the plane \(x = L\), \(S_3\) along the zx-plane, \(S_4\) along the plane \(y = L\), \(S_5\) along the xy-plane, and \(S_6\) along the plane \(z = L\).

Step 2: Calculate Flux Through Face S1 (yz-plane)

Face \(S_1\) is perpendicular to the x-axis. The area vector for this face is \( \hat{\boldsymbol{i}} L^2 \). The electric field component along \(\hat{\boldsymbol{i}}\) is \(-B\). The flux through \(S_1\) is given by \( \Phi_{1} = \overrightarrow{\boldsymbol{E}} \cdot \overrightarrow{\boldsymbol{A}}_1 = (-B)L^2 \).

Step 3: Calculate Flux Through Face S2 (x = L)

Face \(S_2\) is parallel to the yz-plane, thus its outward normal is \(-\hat{\boldsymbol{i}}\). This flux is \( \Phi_{2} = \overrightarrow{\boldsymbol{E}} \cdot \overrightarrow{\boldsymbol{A}}_2 = -(\overrightarrow{\boldsymbol{E}} \cdot \hat{\boldsymbol{i}} \overrightarrow{L^2}) = B L^2 \).

Step 4: Calculate Flux Through Face S3 (zx-plane)

Face \(S_3\) has an area vector \( \hat{\boldsymbol{j}} L^2 \). The electric field component in the \(\hat{\boldsymbol{j}}\) direction is \(C\). The flux is \( \Phi_{3} = C L^2 \).

Step 5: Calculate Flux Through Face S4 (y = L)

Face \(S_4\) has an outward normal \(-\hat{\boldsymbol{j}}\), so the flux is \( \Phi_{4} = -C L^2 \).

Step 6: Calculate Flux Through Face S5 (xy-plane)

Face \(S_5\) has an area vector \( \hat{\boldsymbol{k}} L^2 \). The electric field component in the \(\hat{\boldsymbol{k}}\) direction is \(-D\). Flux is \( \Phi_{5} = -D L^2 \).

Step 7: Calculate Flux Through Face S6 (z = L)

Face \(S_6\) has normal \(\hat{\boldsymbol{k}}\), making the flux \( \Phi_{6} = D L^2 \).

Step 8: Sum of Flux Through All Faces

The total flux through the entire cube is the algebraic sum of the flux through all its faces: \( \Phi_{\text{total}} = \Phi_1 + \Phi_2 + \Phi_3 + \Phi_4 + \Phi_5 + \Phi_6 = -BL^2 + BL^2 + CL^2 - CL^2 -DL^2 + DL^2 = 0 \).

Key Concepts

Gauss's Law

Gauss's Law is a fundamental concept in electromagnetism. It relates the electric flux passing through a closed surface to the charge enclosed within that surface. Mathematically, it is stated as: \[ \Phi = \frac{Q_{\text{enclosed}}}{\varepsilon_0} \]Where:

• \( \Phi \) is the electric flux through a closed surface.
• \( Q_{\text{enclosed}} \) is the total charge inside the surface.
• \( \varepsilon_0 \) is the permittivity of free space.

The law acts like a bridge connecting the internal charge distribution to the external electric field. For the given exercise, since no charge is enclosed within the cube, the total electric flux through the cube is zero. This exemplifies how Gauss's Law remains consistent in situations of symmetry and uniform fields.

Uniform Electric Field

In the exercise given, the electric field is described as uniform. This means that the field has the same magnitude and direction at every point in the region considered. A uniform electric field can be represented by a constant vector, as in the exercise: \[ \overrightarrow{E} = -B \hat{i} + C \hat{j} - D \hat{k} \]
Here,

• \( \hat{i}, \hat{j}, \hat{k} \) are unit vectors along the x, y, and z axes respectively.
• \( B, C, \) and \( D \) are constants that determine the components of the electric field.

The uniformity of the field simplifies calculations of electric flux, as the field doesn't vary across the surfaces of the cube. This simplification is evident in the solution steps, where the same field value is used across each of the cube's faces.

Vector Calculus

Vector calculus is heavily used in electromagnetism, helping to solve problems involving fields that change over space. In the given exercise, the dot product, a fundamental tool in vector calculus, helps in calculating flux.The dot product of two vectors \( \overrightarrow{A} \) and \( \overrightarrow{B} \) is given by:\[ \overrightarrow{A} \cdot \overrightarrow{B} = |\overrightarrow{A}| |\overrightarrow{B}| \cos{\theta} \]Here,

• \( |\overrightarrow{A}| \) and \( |\overrightarrow{B}| \) are magnitudes of the vectors.
• \( \theta \) is the angle between them.

In the exercise, this operation computes the flux, as \( \overrightarrow{E} \cdot \overrightarrow{A} \). The dot product translates the influence of the field along a surface's normal, capturing the effective field contribution to the flux.

Electric Field Components

Electric field components break down a field vector into parts that lie along coordinate axes. For example, the field \( \overrightarrow{E} = -B \hat{i} + C \hat{j} - D \hat{k} \) has components:

• \( -B \) along the x-axis
• \( C \) along the y-axis
• \( -D \) along the z-axis

These components are crucial for understanding how the electric field interacts with surfaces. Each component affects only the respective parallel face of the cube:- The \( x \)-component (-B) affects faces parallel to the yz-plane.- The \( y \)-component (C) determines influence on faces along zx-plane.- The \( z \)-component (-D) applies to faces along the xy-plane.Breaking fields into components simplifies calculations, allowing straightforward application of vector operations like the dot product.