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Chapter 21: Problem 7

In Gauss's law, \(\oint \vec{E} \cdot d \vec{A}=q / \epsilon_{0},\) does the field \(\vec{E}\) necessarily arise only from charges within the closed surface?

Short Answer

Expert verified
No, the electric field in Gauss's law does not necessarily arise only from charges within the closed surface. It can be influenced by charges outside the surface as well. However, according to Gauss's law, the net electric flux through the surface is only contributed by the net charge enclosed by the surface.

Step by step solution

01

Gauss's Law Definition

Gauss's law, in its integral form, states that the total electric flux out of a closed surface is equal to the charge \( q \) enclosed by the surface, divided by the permittivity constant \( \epsilon_{0} \). Mathematically, it's expressed as \(\oint \vec{E} \cdot d \vec{A}=q / \epsilon_{0},\) where \( \vec{E} \) is the electric field, \( d \vec{A} \) is a differential area on the closed surface, and the dot product \( \vec{E} \cdot d \vec{A} \) represents the electric field flux.
02

Understanding the Meaning of Gauss's Law

Gauss's law accounts for all charges, whether they are inside or outside the closed surface. However, only the net charge enclosed by the Gaussian surface contributes to the net electric flux, according to the law. The electric fields from charges outside the closed surface enter and exit the surface an equal number of times, leading to a net flux of zero.
03

Clarifying the Source of Electric Field

The electric field \( \vec{E} \) in Gauss's law does not necessarily arise exclusively from charges within the closed surface. This is because the electric field at a particular point on the surface can be affected by charges outside the surface. However, as stated above, the net electric flux through the surface only depends on the enclosed charge.

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Most popular questions from this chapter

A spherical shell \(30 \mathrm{cm}\) in diameter carries \(85 \mu \mathrm{C}\) distributed uniformly over its surface. A \(1.0-\mu \mathrm{C}\) point charge is located at the shell's center. Find the electric field strength (a) \(5.0 \mathrm{cm}\) from the center and (b) \(45 \mathrm{cm}\) from the center. (c) How would your answers change if the charge on the shell were doubled?

A \(6.8-\mu \mathrm{C}\) charge and a \(-4.7-\mu \mathrm{C}\) charge are inside an uncharged sphere. What's the electric flux through the sphere?

What's the electric field strength in a region where the flux through a \(1.0 \mathrm{cm} \times 1.0 \mathrm{cm}\) flat surface is \(65 \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C},\) if the field is uniform and the surface is at right angles to the field?

An infinitely long rod of radius \(R\) carries a uniform volume charge density \(\rho .\) Show that the electric field strengths outside and inside the rod are given, respectively, by \(E=\rho R^{2} / 2 \epsilon_{0} r\) and \(E=\rho r / 2 \epsilon_{0},\) where \(r\) is the distance from the rod axis. (Although an infinite rod is an impossibility, your answer is a good approximation for the field of a finite rod whose length is much greater than its diameter.)

Eight field lines emerge from a closed surface surrounding an isolated point charge. Would the number of field lines change if a second identical charge were brought to a point just outside the surface? If not, would anything change? Explain.
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