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A point charge, \(+Q\), is located on the \(x\) -axis at \(x=a\), and a second point charge, \(-Q\), is located on the \(x\) -axis at \(x=-a\). A Gaussian surface with radius \(r=2 a\) is centered at the origin. The flux through this Gaussian surface is a) zero. b) greater than zero. c) less than zero. d) none of the above.

Short Answer

Expert verified
Answer: The electric flux through the Gaussian surface is (a) zero.

Step by step solution

01

Identify the given variables

We are given the following information: 1. The charge for the point charge is +Q and is located at x=a. 2. The charge for the second point charge is -Q and is located at x=-a. 3. The Gaussian surface radius is r=2a and is centered at the origin.
02

Apply Gauss's Law

Gauss's Law states that the total electric flux through a closed surface is equal to the total charge enclosed by the surface, divided by the permittivity of free space ε₀. \(\Phi_E = \frac{Q_{enclosed}}{\epsilon_0}\)
03

Calculate the total charge enclosed by the Gaussian surface

Since the Gaussian surface has radius 2a and is centered at the origin, it encloses both charges +Q and -Q. To find the total charge enclosed by the Gaussian surface, add these two charges: \(Q_{enclosed} = Q + (-Q) = 0\)
04

Calculate the electric flux through the Gaussian surface

Now we can plug the total charge enclosed by the Gaussian surface into Gauss's Law: \(\Phi_E = \frac{0}{\epsilon_0} = 0\) The electric flux through the Gaussian surface is zero. Therefore, the correct answer is (a) zero.

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Key Concepts

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

Electric Flux
Electric flux helps us quantify the number of electric field lines passing through a given surface. Think of it as a measure of how much electric field is 'flowing' through a surface.
Essentially, electric flux gives us an idea of the strength of the electric field across a particular area. We calculate electric flux using Gauss's Law, where the electric flux, \( \Phi_E \), is \( \Phi_E = \oint \mathbf{E} \cdot d\mathbf{A} \), or simplified in some cases to \( \Phi_E = \frac{Q_{enclosed}}{\epsilon_0} \).
Here, \( \mathbf{E} \) is the electric field, \( d\mathbf{A} \) is a small area on the surface, and \( Q_{enclosed} \) is the charge enclosed by the surface.Key points to remember about electric flux:
  • Flux is directly proportional to the number of electric field lines passing through the surface.
  • If the field lines enter and exit the surface equally, the net flux is zero.
  • Flux can be positive, negative, or zero, depending on the direction of the electric field and its enclosed charges.
Understanding electric flux helps in analyzing charges in different configurations, especially when employing Gauss's Law.
Point Charge
A point charge is a hypothetical charge that is assumed to be concentrated at a single point in space. Although real charges occupy some space, imagining them as point charges simplifies calculations and helps in studying complex electrostatics.
By using point charges, we can easily define the electric field strength and direction produced by them.Here are a few properties of point charges:
  • The electric field produced by a point charge is inversely proportional to the square of the distance from the charge: \( E = \frac{kQ}{r^2} \), where \( k \) is Coulomb's constant, \( Q \) is the charge, and \( r \) is the distance.
  • The point charge produces a radial electric field, which means the field lines radiate outward from a positive charge and inward to a negative charge.
  • They are used in thought experiments to help understand charge interactions and distributions.
In this exercise, we encounter two point charges, \(+Q\) and \(-Q\), placed symmetrically along the x-axis. Their equal and opposite nature plays a crucial role in understanding the net electric field and enclosed charge within the Gaussian surface.
Gaussian Surface
A Gaussian surface is an imaginary, closed surface used in Gauss's law to help calculate electric flux. Depending on the symmetry of the problem, Gaussian surfaces can take various shapes like spheres, cylinders, or planes.
Choosing the right Gaussian surface can significantly simplify the process of calculating electric fields and flux.Here are some important characteristics of Gaussian surfaces:
  • The surface must always be closed to fully enclose the charge distribution.
  • Spherical Gaussian surfaces are chosen for point charges because of their symmetry.
  • Cylindrical or planar surfaces are used for infinite lines or plane charges, respectively.
In this exercise, the Gaussian surface is a sphere of radius \( r = 2a \), centered at the origin. By enclosing both the positive and negative point charges, the surface shows that the net charge enclosed is zero. This explains why the flux is zero through this spherical Gaussian surface, as predicted by Gauss's Law.
Permittivity of Free Space
Permittivity of free space, denoted as \( \epsilon_0 \), is a fundamental constant that describes how electric fields interact with the vacuum of free space. It serves as a proportionality factor in Coulomb's law and Gauss's law, making it a pivotal concept in electrostatics.
With a value of approximately \( 8.854 \times 10^{-12} \text{C}^2/\text{N} \, \text{m}^2 \), it is often used to determine electric field strength.Key facts about permittivity include:
  • It quantifies the ability of a vacuum to allow electric field lines to pass through it.
  • Permittivity affects the force between charges; lower permittivity means stronger interaction.
  • It provides consistency across different systems of units in electromagnetic equations.
Understanding the permittivity of free space is essential, especially when employing Gauss's Law or studying electromagnetic phenomena in a vacuum or near-vacuum scenarios.

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Most popular questions from this chapter

Suppose you have a large spherical balloon and you are able to measure the component \(E_{n}\) of the electric field normal to its surface. If you sum \(E_{n} d A\) over the whole surface area of the balloon and obtain a magnitude of \(10 \mathrm{~N} \mathrm{~m}^{2} / \mathrm{C}\) what is the electric charge enclosed by the balloon?

Two parallel, infinite, nonconducting plates are \(10.0 \mathrm{~cm}\) apart and have charge distributions of \(+1.00 \mu \mathrm{C} / \mathrm{m}^{2}\) and \(-1.00 \mu \mathrm{C} / \mathrm{m}^{2} .\) What is the force on an electron in the space between the plates? What is the force on an electron located outside the two plates near the surface of one of the two plates?

A uniformly charged rod of length \(L\) with total charge \(Q\) lies along the \(y\) -axis, from \(y=0\) to \(y=L\). Find an expression for the electric field at the point \((d, 0)\) (that is, the point at \(x=d\) on the \(x\) -axis).

A spherical aluminized Mylar balloon carries a charge \(Q\) on its surface. You are measuring the electric field at a distance \(R\) from the balloon's center. The balloon is slowly inflated, and its radius approaches but never reaches R. What happens to the electric field you measure as the balloon increases in radius. Explain.

Which of the following statements is (are) true? a) There will be no change in the charge on the inner surface of a hollow conducting sphere if additional charge is placed on the outer surface. b) There will be some change in the charge on the inner surface of a hollow conducting sphere if additional charge is placed on the outer surface. c) There will be no change in the charge on the inner surface of a hollow conducting sphere if additional charge is placed at the center of the sphere. d) There will be some change in the charge on the inner surface of a hollow conducting sphere if additional charge is placed at the center of the sphere.

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