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Chapter 22: Problem 34

What Aflat, square surface with sides of length \(L\) is described by the equations $$ x-L \quad(0 \leq y \leq L, 0 \leq z \leq L) $$ (a) Draw this square and show the \(x\) - \(y\) - and \(z\) -axes. (b) Find the electric flux through the square due to a positive point charge \(q\) located at the origin \((x=0, y=0, z=0)\) . (Hint: Think of the square as part of a cube centered on the origin.)

Short Answer

Expert verified
The square is on the plane \(x=L\) with electric flux \(\frac{q}{6\varepsilon_0}\).

Step by step solution

01

Understanding the Square's Location

To describe the square surface using the provided equation \(x = L\), it means that this surface is a plane parallel to the \(y\)-\(z\) plane, located at \(x = L\). Given the constraints \(0 \leq y \leq L\) and \(0 \leq z \leq L\), it forms a square in the \(y\)-\(z\) plane, but shifted to \(x = L\). Thus, the square is parallel to the \(yz\)-plane and intersects the \(x\)-axis at \(x=L\).
02

Drawing the Square and Axes

To draw the square, picture a three-dimensional coordinate system with origin at the point \((0,0,0)\). The square has its sides parallel to the \(y\)-axis and the \(z\)-axis, located at \(x = L\). Draw the \(x\), \(y\), and \(z\) axes as straight lines passing through the origin. The square is a flat surface parallel to these axes in the \(y\)-\(z\) plane but offset by \(L\) units along the \(x\)-axis.
03

Visualizing the Cube

Visualize the square as one face of a cube. The cube is centered at the origin \((0, 0, 0)\) with each side of length \(2L\). The square you're considering is the side of the cube at \(x = L\), while the origin at \((0, 0, 0)\) lies at the center of the cube.
04

Calculate the Electric Flux

Electric flux \(\Phi\) through a closed surface surrounding a point charge \(q\) is given by Gauss's law: \(\Phi = \frac{q}{\varepsilon_0}\), where \(\varepsilon_0\) is the permittivity of free space. In this case, the cube with an edge length \(2L\) forms a closed surface, and the given square is one of the six faces of this cube. Thus, each face of the cube contributes equally to the total flux due to symmetry. This means the flux through the single square is \(\Phi_{face} = \frac{q}{6\varepsilon_0}\).

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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 through a closed surface to the charge enclosed by that surface. According to Gauss's Law, the total electric flux through a closed surface, known as a Gaussian surface, is given by the equation: \[ \Phi = \frac{q}{\varepsilon_0} \] where
  • \( \Phi \) is the electric flux,
  • \( q \) is the total charge enclosed within the surface,
  • \( \varepsilon_0 \) is the permittivity of free space.
This law is particularly useful when dealing with symmetrical charge distributions. In the case of the problem, the Gaussian surface is visualized as a cube centered at the origin, encompassing the charge \( q \). Each face of the cube, including our square, will have an equal amount of flux passing through it because of the cube's symmetry.
Cube Visualization
Visualizing the square as part of a cube is a pivotal step in solving the problem. The square given, defined at \( x = L \), forms one face of an imaginary cube embedded in three-dimensional space, with its center aligned at the origin (0, 0, 0). The cube has edges of length \( 2L \), extending symmetrically around the origin.
The square face located at \( x = L \) evenly shares the total electric flux with the other five cube faces.
Consider this visual:
  • Six identical square faces form the entire cube.
  • The point charge \( q \) sitting at the center ensures symmetry.
  • Each face contributes equally to the electric flux calculation.
Such visualization helps to comprehend how Gauss鈥檚 Law is applied across symmetrical surfaces like cubes.
Electric Field
The electric field is another key concept needed to understand electric flux. It is the force per unit charge exerted on a small positive test charge placed in the vicinity of a charge distribution. For a point charge \( q \), this electric field \( E \) at a distance \( r \) from the charge is given by:\[ E = \frac{kq}{r^2} \]where
  • \( k \) is Coulomb's constant,
  • \( q \) is the charge,
  • \( r \) is the distance from the charge to the point of interest.
Even though we are computing flux using Gauss's Law, it's crucial to understand that the field lines represent the electric field through the surfaces. In this exercise, the symmetry allows the computation of flux without explicitly calculating the electric field at all points on the surface.
Coordinate System
Understanding the coordinate system is essential for solving the problem. In this case, the exercise uses a Cartesian coordinate system where three coordinates \( (x, y, z) \) define any point in space.
The square is located in the \( y-z \)-plane at \( x = L \), indicating a plane parallel to the \( y-z \) axes but displaced along the \( x \)-axis.
  • Think of the origin (0, 0, 0) as the intersection point for all three axes.
  • The positive \( x \)-direction is where the square's face at \( x = L \) lies.
  • Each point on the square has coordinates \( (L, y, z) \) with \( 0 \leq y \leq L \) and \( 0 \leq z \leq L \).
This systematic representation enables visualization of three-dimensional problems, crucial for understanding spatial relations and applying mathematical concepts like Gauss's Law effectively.

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Most popular questions from this chapter

22.21. A very long uniform line of charge has charge per unit length 4.80\(\mu \mathrm{C} / \mathrm{m}\) and lies along the \(x\) -axis. A second long uniform line of charge has charge per unit length \(-2.40 \mu \mathrm{C} / \mathrm{m}\) and is parallel to the \(x\) -axis at \(y=0.400 \mathrm{m}\) . What is the net clectric field (magnitude and direction) at the following points on the \(y\) -axis: (a) \(y=0.200 \mathrm{m}\) and \((\mathrm{b}) y=0.600 \mathrm{m} ?\)

22.31. A negative charge \(-Q\) is placed inside the cavity of a hol- low metal solid. The outside of the solid is grounded by connecting a conducting wire between it and the earth. (a) Is there any excess charge induced on the inner surface of the piece of metal? If so, find its sign and magnitude. (b) Is there any excess charge on the outside of the piece of metal? Why or why not?(c) Is there an electric field in the cavity? Explain. (d) Is there an electric field within the metal? Why or why not? Is there an electric field outside the piece of metal? Explain why or why not. (e) Would someone outside the solid measure an electric field due to the charge \(-Q ?\) Is it reasonable to say that the grounded conductor has shielded the region from the ciffects of the charge \(-Q ?\) In principle, could the same thing be done for gravity? Why or why not?

22.25. The electric field at a distance of 0.145 \(\mathrm{m}\) from the surface of a solid insulating sphere with radius 0.355 \(\mathrm{m}\) is 1750 \(\mathrm{N} / \mathrm{C}\) . (a) Assuming the sphere's charge is uniformly distributed, what is the charge density inside it? ( 6 ) Calculate the electric field inside the sphere at a distance of 0.200 \(\mathrm{m}\) from the center.

22.19. How many excess electrons must be added to an isolated spherical conductor 32.0 \(\mathrm{cm}\) in diameter to produce an electric field of 1150 \(\mathrm{N} / \mathrm{C}\) just outside the surface?

22.15. A point charge of \(+5.00 \mu C\) is located on the \(x\) -axis at \(x=4.00 \mathrm{m},\) next to a spherical surface of radius 3.00 \(\mathrm{m}\) centered at the origin. (a) Calculate the magnitude of the electric field at \(x=3.00 \mathrm{m} .\) (b) Calculate the magnitude of the electric field at \(x=-3.00 \mathrm{m} .\) (c) According to Gauss's law, the net flux through the sphere is zero because it contains no charge. Yet the field due to the external charge is much stronger on the near side of the sphere (i.e., at \(x=3.00 \mathrm{m}\) ) than on the far side (at \(x=-3.00 \mathrm{m}\) ). How, then, can the flux into the sphere (on the near side) equal the flux out of it (on the far side)? Explain. A sketch will help.
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