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Chapter 23: Problem 23

Figure 23-34 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 \(\vec{E}=+23.0 \hat{\mathrm{i}}-2.00 y^{2} \hat{\mathrm{j}}-16.0 \hat{\mathrm{k}} \mathrm{N} / \mathrm{C}\), with \(y\) in meters. What is the net charge contained by the cube?

Short Answer

Expert verified
The net charge enclosed by the cube is zero.

Step by step solution

01

Understand the Problem

We are given a cubic Gaussian surface and an electric field in terms of its vector components. The task is to calculate the net charge enclosed by the cube using Gauss's Law.
02

Recall Gauss's Law

Gauss's Law states that the electric flux through a closed surface is equal to the net charge enclosed divided by the permittivity of free space: \(\Phi = \frac{Q_{\text{enc}}}{\varepsilon_0}\). Thus, \(Q_{\text{enc}} = \varepsilon_0 \Phi\).
03

Calculate Electric Flux

Electric flux \( \Phi \) is the integral of \( \vec{E} \cdot d\vec{A} \) over the closed surface. For each face of the cube, calculate the integral: \( \Phi = \sum \int \vec{E} \cdot d\vec{A}\). Calculate separately for each face taking into account the orientation and the electric field at that face.
04

Electric Flux - Top and Bottom Faces Contribution

For the top and bottom faces parallel to the xy-plane, \( \vec{A} \) vectors are \(\pm \, k\) direction. Since the k-component of the electric field is constant \(-16\), the flux contribution from these faces cancels each other as they are equal and opposite.
05

Electric Flux - Front and Back Faces Contribution

For the faces lying in the xz-plane, the \( j \)-component of the field is zero at y = constant. Hence, no contribution from the \( x=- \hat{k} \) and \( x=+ \hat{k} \) faces since \( E_y = 0 \).
06

Electric Flux - Left and Right Faces Contribution

For the faces lying in the yz-plane, the \( i \)-component of the electric field contributes as it is constant. Calculate the total flux for these faces using \(23.0\) as the constant electric field component in \(23.0 \hat{i}\): \( \Phi = 23.0 \times ( ext{surface area})\).
07

Compute Enclosed Charge

Having calculated the total electric flux, use Gauss's Law: \( Q_{\text{enc}} = \varepsilon_0 \Phi \). Compute this value using \(\varepsilon_0 = 8.85 \times 10^{-12} \text{C}^2\text{N}^鈭�1\text{m}^鈭�2 \).

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Key Concepts

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

Electric Flux
Electric flux is an essential concept when dealing with electric fields and closed surfaces. It represents the number of electric field lines passing through a given surface. To picture this, imagine an imaginary surface like a net that catches the field lines. The more lines that pass through, the greater the electric flux.
  • The formula for electric flux through a surface is given by the integral of the electric field \((\vec{E})\) dotted with the differential area \(d\vec{A}\): \( \Phi = \int \vec{E} \cdot d\vec{A} \.\)
  • This is a vector operation, meaning the area element is perpendicular to the surface, pointing outward.
Electric flux varies with the strength of the electric field and the angle at which the field lines intersect the surface. If the fields are perpendicular, all lines are through, maximizing flux.
Understanding electric flux is key to solving problems involving Gauss鈥檚 Law.
Gaussian Surface
A Gaussian surface is an imaginary closed surface used to simplify the application of Gauss's Law. These surfaces are often chosen to match the symmetry of the electric field around a charge distribution, making the calculations simpler. In our scenario, the Gaussian surface is a cube. This is ideal because the electric field components simplify when applied to symmetric shapes like cubes or spheres.
  • The purpose of the Gaussian surface is to easily calculate the net electric flux, leading to finding the net charge inside.
  • For an area with uniform symmetry, like a uniform cube in a uniform electric field, calculating the surface integral of the field becomes more straightforward.
Choosing the right Gaussian surface means aligning the surface with the characteristics of the electric field. Thus, it optimizes the calculation process when using Gauss's Law.
Net Charge
The net charge enclosed by a Gaussian surface determines the flux through the surface. According to Gauss's Law, the flux is directly proportional to the net charge.Using Gauss's Law, the formula for net charge is: \[ Q_{\text{enc}} = \varepsilon_0 \Phi \]where \(\varepsilon_0\) is the permittivity of free space with a value of \(8.85 \times 10^{-12} \text{C}^2/\text{N} \, \text{m}^2\).
  • In our exercise, after deriving the total flux through the cube, we multiply by \(\varepsilon_0\) to find the enclosed net charge.
  • This correlation is key to using Gauss's Law in practical situations, linking abstract concepts of field lines with the tangible quantity, charge.
Understanding the net charge is vital to knowing how charges affect the fields around them.
Electric Field
The electric field describes how strong and what direction an electric force would act on a charge if it were placed in the field.Given by the vector \(\vec{E}\), it has components that determine how the field influences charges in space:\(\vec{E} = +23.0 \hat{\mathrm{i}}-2.00 y^{2} \hat{\mathrm{j}}-16.0 \hat{\mathrm{k}} \ \text{N/C}.\)
  • The vector components describe how the field varies in different directions 鈥� linear in the x-direction, quadratic in the y, and constant in the z.
  • An electric field's properties are crucial in determining how the electric flux varies over a surface.
In essence, understanding an electric field and its components is foundational to calculating associated flux and eventual charge distribution. It connects electric forces with spatial distribution, allowing us to understand how imaginary surfaces "see" the field.

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Most popular questions from this chapter

The electric field in a certain region of Earth's atmosphere directed vertically down. At an altitude of \(300 \mathrm{~m}\) the field has mag nitude \(75.0 \mathrm{~N} / \mathrm{C} ;\) at an altitude of \(200 \mathrm{~m}\), the magnitude is \(210 \mathrm{~N} / \mathrm{C}\) Find the net amount of charge contained in a cube \(100 \mathrm{~m}\) on edg with horizontal faces at altitudes of 200 and \(300 \mathrm{~m}\).

Figure 23-28 shows a section of a long, thin-walled metal tube of radius \(R=2.50 \mathrm{~cm}\), with a charge per unit length of \(\lambda=2.00 \times 10^{-8} \mathrm{C} / \mathrm{m}\). What is the magnitude \(E\) of the electric field at radial distance (a) \(r=R / 2.00\) and (b) \(r=2.00 R ?\) (c) Graph \(E\) versus \(r\) for the range \(r=0\) to \(2.00 R\).

Figure \(23-43\) shows a closed Gaussian surface in the shape of a cube of edge length \(1.50 \mathrm{~m}\). It lies in a region where the nonuniform electric field is given by \(\vec{E}=(3.00 x+4.00) \hat{\mathrm{i}}+6.00 \hat{\mathrm{j}}+7.00 \hat{\mathrm{k}} \mathrm{N} / \mathrm{C}\), with \(x\) in meters. What is the net charge contained by the cube?

Figure 23-42 shows, in cross section, two solid spheres with uniformly distributed charge throughout their volumes. Each has radius \(R\). Point \(P\) lies on a line connecting the centers of the spheres, at radial distance \(R / 4.00\) from the center of sphere 1 . If the net electric field at point \(P\) is zero, what is the ratio \(q_{2} / q_{1}\) of the total charges?

The volume charge density of a solid nonconducting sphere of radius \(R=5.60 \mathrm{~cm}\) varies with radial distance \(r\) as given by \(\rho=\left(35.4 \mathrm{pC} / \mathrm{m}^{3}\right) r / R\). (a) What is the sphere's total charge? What is the field magnitude \(E\) at (b) \(r=0\), (c) \(r=R / 3.00\), and (d) \(r=R\) ?
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