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A rectangular box is uniformly charged with a charge density \(\rho\). The box is extremely long and skinny, and its crosssection is a square with sides of length \(b\). The length is so great in comparison to \(b\) that we can consider it as being infinite. Find the electric field at a point lying on the box's surface, at the midpoint between the two edges. Your answer will involve an integral that is most easily done using computer software.

Short Answer

Expert verified
The electric field at the midpoint is calculated as \(E = \frac{\rho b}{2 \epsilon_0}\).

Step by step solution

01

Identify Known Quantities

We have a rectangular box with an infinite length and a square cross-section with side length \(b\). The box is uniformly charged with a charge density \(\rho\). We need to find the electric field at a point on the box's surface, specifically at the midpoint between the two edges.
02

Symmetry Considerations

Due to the symmetry of the infinite 'line' of charge, the electric field at any point on the surface will only have a component perpendicular to the face of the box. This is because the horizontal components from opposite sides of the infinite length cancel out.
03

Use Gauss’s Law

We can simplify the problem using Gauss's law. Assume a Gaussian surface in the form of a rectangular box that straddles the charged box, with end-caps on opposite faces of the cross-section. The Gaussian surface's sides parallel to the box's length contribute zero flux as electric field lines are parallel to these sides.
04

Calculate Charge Enclosed

The charge enclosed \(Q_{enc}\) by the Gaussian surface is the product of charge density \(\rho\) and the volume of the section of the box inside our Gaussian surface. This volume is given by \(b \cdot b \cdot l = b^2 \cdot l\), where \(l\) is an arbitrary length of the Gaussian surface. Thus, \(Q_{enc} = \rho \cdot b^2 \cdot l\).
05

Apply Gauss's Law

Gauss's Law states \( \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{enc}}{\epsilon_0} \), where \( \mathbf{E} \) is the electric field and \( d\mathbf{A} \) is the area element. Since the box is infinite, \( \mathbf{E} \) is uniform over the opposite faces. Thus, the total flux is \( 2E \cdot (b \cdot l) \).
06

Solve for the Electric Field

Setting the flux equal to \(\frac{Q_{enc}}{\epsilon_0}\), we have \(2E \cdot (b \cdot l) = \frac{\rho \cdot b^2 \cdot l}{\epsilon_0}\). Solving for \(E\), we get \(E = \frac{\rho b}{2 \epsilon_0}\).
07

Use Computer to Evaluate Integral

Though the setup suggests that an integral is involved, using Gauss's Law as shown simplifies the solution process and the electric field component was derived simply without explicit integral computation. Computer software could be employed to simulate this field distribution over the entire surface to verify the uniformity and strength indicated by our calculation.

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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 principle of electromagnetism that provides a powerful way to calculate electric fields for symmetrical charge distributions. It states that the total electric flux through a closed surface, known as a Gaussian surface, is equal to the charge enclosed divided by the permittivity of free space, \[ \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{enc}}{\epsilon_0} \].This law is most useful when dealing with problems involving cylindrical, spherical, or planar symmetry. In these cases, the symmetry of the problem allows for the simplification of calculations, helping us avoid complex integrations. By smartly choosing our Gaussian surface so that the electric field is uniform over portions of the surface or has zero flux on other parts, we can solve for unknown electric fields straightforwardly. This exercise cleverly used Gauss's Law to find the electric field on the surface of a rectangular box, illustrating the concept beautifully.
Charge Density
Charge density (\(\rho\)) represents how much electric charge is distributed over a given volume, area, or length. In our exercise, we have a rectangular box with a uniform charge density, meaning that the charge is evenly spread throughout the entire volume of the box.
  • Volume charge density is used when the charge is spread over a 3D space and is measured in \(\frac{C}{m^3}\).
  • Surface charge density is for charges spread over a 2D area, measured in \(\frac{C}{m^2}\).
  • Linear charge density is used for a 1D line of charge, measured in \(\frac{C}{m}\).
Understanding charge density is crucial as it informs us about how strong an electric field we can expect in different parts of a charged object. A uniform charge density provides a straightforward scenario for applying Gauss’s Law as seen in the exercise.
Symmetry Considerations
When tackling problems in electric fields, considering symmetry can greatly simplify our approach and calculations. In the case of the rectangular box, we recognize a type of symmetry known as translational symmetry. This arises because the box is essentially infinite in length, behaving similarly to an infinite line of charge. Such symmetry tells us:
  • The electric field will primarily point perpendicular to the box's surface due to canceling effects of the horizontal field components.
  • By focusing on the symmetry, it allows us to devise a Gaussian surface that provides meaningful insights without laborious calculation.
Indeed, symmetry in this context ensures that the electric field is constant over certain faces of our Gaussian surface and zero over others, making use of Gauss's Law very effective.
Electric Flux
Electric flux is a measure of how much electric field passes through a given area. It helps quantify the electric field's effect over a surface and is defined as the integral of the electric field \[\Phi_E = \oint \mathbf{E} \cdot d\mathbf{A} \]. Electric flux provides insight into how field lines penetrate or skirt around an object. In scenarios with high symmetry, such as our infinite rectangular box, electric flux calculations become more intuitive. When using Gauss's Law, we translate the concept of electric flux into understanding the total closed surface area we choose, simplifying the steps to find electric fields when appropriate Gaussian surfaces are employed.
Rectangular Box
A rectangular box, in this exercise, is modeled with unique properties that make it resemble an infinite line when considering its length. This setup is crucial because it allows simplification using Gauss’s Law under symmetry considerations for infinite geometries.
  • The cross-section is square-shaped, which keeps the symmetry simple, crucial for manageable calculations.
  • Since the length is much larger than the cross-section dimensions, effects of edge charges are negligible, rendering this an ideal case for using Gauss's Law.
The choice of a rectangular box surveyed at the midpoint also means that any derived electric field results will apply uniformly across similar points, reinforcing the symmetry utility in these models. By considering the rectangular box with an infinite approach, standard complexity in field calculations can be sidestepped efficiently.

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Most popular questions from this chapter

Find every complex number \(z\) such that \(z^{3}=1\). (solution in the pdf version of the book)

In a certain region of space, the electric field is constant (i.e., the vector always has the same magnitude and direction). For simplicity, assume that the field points in the positive \(x\) direction. (a) Use Gauss's law to prove that there is no charge in this region of space. This is most easily done by considering a Gaussian surface consisting of a rectangular box, whose edges are parallel to the \(x\), \(y\), and \(z\) axes. (b) If there are no charges in this region of space, what could be making this electric field?

Calculate the quantity \(i^{i}\) (i.e., find its real and imaginary parts).(answer check available at lightandmatter.com)

(a) Use Gauss' law to find the fields inside and outside an infinite cylindrical surface with radius \(b\) and uniform surface charge density \(\sigma\).(answer check available at lightandmatter.com) (b) Show that there is a discontinuity in the electric field equal to \(4 \pi k \sigma\) between one side of the surface and the other, as there should be (see page 628). (c) Reexpress your result in terms of the charge per unit length, and compare with the field of a line of charge. (d) A coaxial cable has two conductors: a central conductor of radius \(a\), and an outer conductor of radius \(b\). These two conductors are separated by an insulator. Although such a cable is normally used for time-varying signals, assume throughout this problem that there is simply a DC voltage between the two conductors. The outer conductor is thin, as in part c. The inner 46. (a) Use Gauss' law to find the fields inside and outside an infinite cylindrical surface with radius \(b\) and uniform surface charge density \(\sigma\).(answer check available at lightandmatter.com) (b) Show that there is a discontinuity in the electric field equal to \(4 \pi k \sigma\) between one side of the surface and the other, as there should be (see page 628). (c) Reexpress your result in terms of the charge per unit length, and compare with the field of a line of charge. (d) A coaxial cable has two conductors: a central conductor of radius \(a\), and an outer conductor of radius \(b\). These two conductors are separated by an insulator. Although such a cable is normally used for time-varying signals, assume throughout this problem that there is simply a DC voltage between the two conductors. The outer conductor is thin, as in part c. The inner

This is a one-dimensional problem, with everything confined to the \(x\) axis. Dipole A consists of a \(-1.000\) C charge at \(x=0.000 \mathrm{~m}\) and a \(1.000 \mathrm{C}\) charge at \(x=1.000 \mathrm{~m}\). Dipole \(\mathrm{B}\) has a \(-2.000 \mathrm{C}\) charge at \(x=0.000 \mathrm{~m}\) and a \(2.000 \mathrm{C}\) charge at \(x=0.500 \mathrm{~m}\). (a) Compare the two dipole moments. (b) Calculate the field created by dipole A at \(x=10.000 \mathrm{~m}\), and compare with the field dipole B would make. Comment on the result.(answer check available at lightandmatter.com)

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