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Gauss's Law (Ch. 24)
Transcript of Gauss's Law (Ch. 24)
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.
Find the net electric flux through the spherical closed surface shown in Figure. The two charges on the right are inside the spherical surface.
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
1. i) The field is cylindrically radical to the filament, and is nowhere near zero at any face of the Gaussian surface, therefore the answer is (a) = 0.
(ii) Through how many of the cube’s faces is the electric flux zero? Choose from the same possibilities as in part (i).
ii) The flux is zero through the two faces pierced by the filament because the field is parallel to those surfaces. Therefore the answer is (b) = 2sides.