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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.

Short Answer

Expert verified
The net electric flux through the cube is 0.

Step by step solution

01

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 \(z = L\).
02

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 \( \Phi_{1} = \overrightarrow{\boldsymbol{E}} \cdot \overrightarrow{\boldsymbol{A}}_1 = (-B)L^2 \).
03

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 \( \Phi_{2} = \overrightarrow{\boldsymbol{E}} \cdot \overrightarrow{\boldsymbol{A}}_2 = -(\overrightarrow{\boldsymbol{E}} \cdot \hat{\boldsymbol{i}} \overrightarrow{L^2}) = B L^2 \).
04

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 \( \Phi_{3} = C L^2 \).
05

Calculate Flux Through Face S4 (y = L)

Face \(S_4\) has an outward normal \(-\hat{\boldsymbol{j}}\), so the flux is \( \Phi_{4} = -C L^2 \).
06

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 \( \Phi_{5} = -D L^2 \).
07

Calculate Flux Through Face S6 (z = L)

Face \(S_6\) has normal \(\hat{\boldsymbol{k}}\), making the flux \( \Phi_{6} = D L^2 \).
08

Sum of Flux Through All Faces

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

22.23. A hollow, conducting sphere with an outer radius of 0.250 \(\mathrm{m}\) and an inner radius of 0.200 \(\mathrm{m}\) has a uniform surface charge density of \(+6.37 \times 10^{-6} \mathrm{C} / \mathrm{m}^{2} .\) A charge of \(-0.500 \mu \mathrm{C}\) is now introduced into the cavity inside the sphere. (a) What is the new charge density on the outside of the sphere? (b) Calculate the strength of the electric field just outside the sphere. (c) What is the electric fiux through a spherical surface just inside the inner surface of the sphere?

22.29. An infinitely long cylindrical conductor has radius \(R\) and uniform surface charge density \(\sigma\) . (a) In terms of \(\sigma\) and \(R\) , what is the charge per unit length \(\lambda\) for the cylinder? (b) In terms of \(\sigma\) , what is the magnitude of the electric field produced by the charged cylinder at a distance \(r>R\) from its axis? (c) Express the result of part (b) in terms of \(\lambda\) and show that the electric field outside the cylinder is the same as if all the charge were on the axis. Compare your result to the result for a line of charge in Example 22.6 (Section \(22.4 ) .\)

22\. 13. A \(9.60-\mu\) C point charge is at the center of a cube with sides of length 0.500 \(\mathrm{m}\) (a) What is the electric flux through one of the six faces of the cube? (b) How would your answer to part (a) change if the sides were 0.250 \(\mathrm{m}\) long? Explain.

22.20. The electric field 0.400 \(\mathrm{m}\) from a very long uniform line of charge is 840 \(\mathrm{N} / \mathrm{C}\) . How much charge is contained in a \(2.00-\mathrm{cm}\) section of the line?

22.24. A point charge of \(-2.00 \mu \mathrm{C}\) is located in the center of a spherical cavity of radius 6.50 \(\mathrm{cm}\) inside an insulating charged solid. The charge density in the solid is \(\rho=7.35 \times 10^{-4} \mathrm{C} / \mathrm{m}^{3} .\) Calculate the electric field inside the solid at a distance of 9.50 \(\mathrm{cm}\) from the center of the cavity.

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