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Chapter 22: Problem 3

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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