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# Gauss&#039;s Law (Ch. 24)

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## Andres Obregon

on 15 March 2016

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#### Transcript of Gauss&#039;s Law (Ch. 24)

Gauss's Law (Ch. 24)
Conceptual Question
#10.
On the basis of the repulsive nature of the force between like charges and the freedom of motion of charge within a conductor, explain why excess charge on the isolated conductor must reside in the surface.
In a conductor like metal, the structure of the atoms is in such a way that they are arranged in a lattice within a “sea” of electrons. The charge is carried by electrons which are the particles that freely move within that conductor. If an isolated conductor is charged, more electrons are added to this already existing “sea” of electrons. Since the induced electrons carry the same charge as the existing electrons, they therefore experience a repulsive force, and are pushed away towards the surface of the conductor and since the conductor is isolated, the repelled electrons have nowhere to go hence they will reside on the surface of the conductor.

Problem
#8
Find the net electric flux through the spherical closed surface shown in Figure. The two charges on the right are inside the spherical surface.
Objective Question
#1
1. A cubical Gaussian surface surrounds a long, straight, charged filament that passes perpendicularly through two opposite faces. No other charges are nearby.
(i) Over how many of the cube’s faces is the electric field zero? (a) 0 (b) 2 (c) 4 (d) 6

Charge 1: 1.00x〖10〗^(-9) C
Charge 2: -3.00x〖10〗^(-9) C

Total charge enclosed: -2.00x〖10〗^(-9) C

Eo=8.85x〖10〗^(-12 ) C^2/N m^2
EF=q/Eo = (-2.00x〖10〗^(-9))/(8.85x〖10〗^(-12) ) =-2.26x〖10〗^2 N.m^2/C