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A \(6.20-\mu \mathrm{C}\) 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.

Short Answer

Expert verified
The flux through one face is \(1.167 \times 10^5 \, \text{N}\cdot\text{m}^2/\text{C}\), unaffected by side length.

Step by step solution

01

Understanding the Problem

We have a point charge at the center of a cube, and we need to find the electric flux through one face of the cube. We also need to determine if the length of the cube's sides affects this flux.
02

Using Gauss's Law

Gauss's Law states that the electric flux \( \Phi \) through a closed surface is given by \( \Phi = \frac{Q}{\varepsilon_0} \), where \( Q \) is the charge inside the surface and \( \varepsilon_0 \) is the permittivity of free space \( (8.85 \times 10^{-12} \, \text{C}^2/\text{N}\cdot\text{m}^2) \).
03

Calculate Total Electric Flux for the Cube

First, calculate the total electric flux through the entire cube. The charge \( Q = 6.20 \times 10^{-6} \, \text{C} \). Using Gauss's Law: \[ \Phi_{\text{total}} = \frac{6.20 \times 10^{-6} \, \text{C}}{8.85 \times 10^{-12} \, \text{C}^2/\text{N}\cdot\text{m}^2} \approx 7.0045 \times 10^5 \, \text{N}\cdot\text{m}^2/\text{C} \]
04

Calculate Electric Flux Through One Face of the Cube

Since the cube has 6 faces, the flux through one face is one-sixth of the total flux: \[ \Phi_{\text{face}} = \frac{7.0045 \times 10^5 \, \text{N}\cdot\text{m}^2/\text{C}}{6} \approx 1.1674 \times 10^5 \, \text{N}\cdot\text{m}^2/\text{C} \]
05

Assess Side Length Impact

Gauss's Law implies that the flux through the entire closed surface depends only on the charge enclosed, not on the size of the cube. This means changing the side length of the cube to 0.250 \( \text{m} \) does not change the electric flux through each face.

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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 in electromagnetics describing how electric charges produce an electric field. It's expressed by the equation \( \Phi = \frac{Q}{\varepsilon_0} \), where \( \Phi \) is the electric flux through a closed surface, \( Q \) is the total charge enclosed by that surface, and \( \varepsilon_0 \) is the permittivity of free space. This law highlights that the total electric flux through a closed surface is determined solely by the enclosed charge, regardless of the shape or size of that surface.
This is incredibly useful for calculating electric fields in symmetrical situations, like spheres, cylinders, or cubes, where the mathematical complexities can be significantly reduced. In our cube scenario, Gauss's Law allows us to easily determine the electric flux associated with a point charge buried inside, without getting bogged down by intricate calculations involving each facet of its geometry. If we know the point charge's magnitude, we have all we need to calculate the electric flux.
Point Charge
A point charge refers to a hypothetical charge located at a single point in space. It's a convenient concept used in physics to simplify the study of electric fields and potentials when dealing with charges. In real-world terms, think of a point charge as an idealized version of actual charged particles, like electrons or protons.
In our problem, the point charge is 6.20 microcoulombs (\( \mu \text{C} \)). This charge, being at the center of a cube, allows us to use symmetry when applying Gauss's Law. The symmetry ensures that the electric field has the same magnitude at any point on a given surface equidistant from the charge. As a result, it becomes straightforward to analyze complex geometrical situations, such as determining the electric flux through individual faces of a cube.
Permittivity of Free Space
Permittivity of free space, represented by \( \varepsilon_0 \), is a universal constant with a value of approximately \( 8.85 \times 10^{-12} \, \text{C}^2/\text{N}\cdot\text{m}^2 \). It defines how much resistance exists to forming an electric field in a vacuum. In simpler terms, it characterizes the influence that free space itself has on the electric field creation and propagation around a charge.
This parameter is pivotal when applying Gauss's Law to calculate the electric flux. It acts as a scaling factor between the enclosed charge and the resulting electric flux through a closed surface. In our exercise, \( \varepsilon_0 \) helps bridge the conceptual gap between the point charge's intrinsic properties and the effects observed in the surrounding space, allowing for accurate flux calculation.
Electric Field
An electric field is a region around a charged particle where other charges experience a force. The strength and direction of this field can be represented by electric field lines. A positive point charge, for instance, has electric field lines emanating outward.
In the context of a point charge inside a cube, the electric field is uniform along each face of the cube. Thanks to Gauss's Law, we understand that the number and configuration of these field lines through the entire cube sum up to the total electric flux, which is dependent solely on the charge enclosed, not on the size or shape of the enclosure.
In practice, analyzing the electric field involves calculating how it influences the surrounding environment, determining force vectors, and understanding how materials in the vicinity might alter these effects. For this problem, while the electric field magnitude is crucial for deriving insights about flux distribution, Gauss's Law simplifies the computation by focusing on symmetry and charge centrality.

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Most popular questions from this chapter

A Sphere in a Sphere. A solid conducting sphere carrying charge \(q\) has radius \(a\) It is inside a concentric hollow conducting sphere with inner radius \(b\) and outer radius \(c .\) The hollow sphere has no net charge. (a) Derive expressions for the electricfield magnitude in terms of the distance \(r\) from the center for the regions \(r < a, a < r < b, b < r < c,\) and \(r > c .\) (b) Graph the magnitude of the electric field as a function of \(r\) from \(r=0\) to \(r=2 c .\) (c) What is the charge on the inner surface of the hollow sphere? (d) On the outer surface? (e) Represent the charge of the small sphere by four plus signs. Sketch the field lines of the system within a spherical volume of radius 2\(c .\)

A very long, solid cylinder with radius \(R\) has positive charge uniformly distributed throughout it, with charge per unit volume \(\rho .\) (a) Derive the expression for the electric field inside the volume at a distance \(r\) from the axis of the cylinder in terms of the charge density \(\rho\) . What is the electric field at a point outside the volume in terms of the charge per unit length \(\lambda\) in the cylinder? (c) Compare the answers to parts (a) and (b) for \(r=R .\) (d) Graph the electric-field magnitude as a function of \(r\) from \(r=0\) to \(r=3 R .\)

Thomson's Model of the Atom. In the early years of the 20 th century, a leading model of the structure of the atom was that of the English physicist J. J. Thomson (the discoverer of the electron). In Thomson's model, an atom consisted of a sphere of positively charged material in which were embedded negatively charged electrons, like chocolate chips in a ball of cookie dough. Consider such an atom consisting of one electron with mass \(m\) and charge \(-e,\) which may be regarded as a point charge, and a uniformly charged sphere of charge \(+e\) and radius \(R\) (a) Explain why the equilibrium position of the electron is at the center of the nucleus. (b) In Thomson's model, it was assumed that the positive material provided little or no resistance to the motion of the electron. If the electron is displaced from equilibrium by a distance less than \(R,\) show that the resulting motion of the electron will be simple harmonic, and calculate the frequency of oscillation. Hint: Review the definition of simple harmonic motion in Section \(14.2 .\) If it can be shown that the net force on the electron is of this form, then it follows that the motion is simple harmonic. Conversely, if the net force on the electron does not follow this form, the motion is not simple harmonic.) (c) By Thomson's time, it was known that excited atoms emit light waves of only certain frequencies. In his model, the frequency of emitted light is the same as the oscillation frequency of the electron or electrons in the atom. What would the radius of a Thomson-model atom have to be for it to produce red light of frequency \(4.57 \times 10^{14}\) Hz? Compare your answer to the radii of real atoms, which are of the order of \(10^{-10} \mathrm{m}\) (see Appendix For data about the electron). (d) If the electron were displaced from equilibrium by a distance greater than \(R,\) would the electron oscillate? Would its motion be simple harmonic? Explain your reasoning. (Historical note: In \(1910,\) the atomic nucleus was discovered, proving the Thomson model to be incorrect. An atom's positive charge is not spread over its volume as Thomson supposed, but is concentrated in the tiny nucleus of radius \(10^{-14}\) to \(10^{-15} \mathrm{m} . )\)

A very long conducting tube (hollow cylinder) has inner radius \(a\) and outer radius \(b\) . It carries charge per unit length \(+\alpha\) where \(\alpha\) is a positive constant with units of \(\mathrm{C} / \mathrm{m} .\) A line of charge lies along the axis of the tube. The line of charge has charge per unit length \(+\alpha .\) (a) Calculate the electric field in terms of \(\alpha\) and the distance \(r\) from the axis of the tube for (i) \(r < a ;\) (ii) \(a < r < b ;\) (iii) \(r > b .\) Show your results in a graph of \(E\) as a function of \(r .\) (b) What is the charge per unit length on (i) the inner surface of the tube and (ii) the outer surface of the tube?

You measure an electric field of \(1.25 \times 10^{6} \mathrm{N} / \mathrm{C}\) at a distance of 0.150 \(\mathrm{m}\) from a point charge. There is no other source of electric field in the region other than this point charge. (a) What is the electric flux through the surface of a sphere that has this charge at its center and that has radius 0.150 m? (b) What is the magnitude of this charge?

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