Gauss's law intuitive explanation?

[lord_james]

I'm not sure what Gauss's law really means. "The electric flux through any closed surface is proportional to the enclosed electric charge." How does this apply to finding the electric field?
apcentral.collegeboard. com/apc/public/repository/ap11_frq_physics_cem.pdf
Look at parts 1 a and b. Part 1a is easy enough to do, but I want to really understand why Gauss's law applies here. collegeboard. com/apc/public/repository/ap11_physics_c_electricity_magnetism_scoring_guidelines.pdf
Here are their solutions. (Remove the space before com in both links). The fact that the enclosed charge is zero does not tell you anything about the electric field, though, as evidenced by part (b). What if I draw a Gaussian surface next to, but not enclosing, a point charge? There is no enclosed charge, and no net flux, but there is still obviously an electric field. So why do they want Gauss's law used in these situations?

 

[wkassis]

I'm not sure what Gauss's law really means. "The electric flux through any closed surface is proportional to the enclosed electric charge." How does this apply to finding the electric field?
apcentral.collegeboard. com/apc/public/repository/ap11_frq_physics_cem.pdf
Look at parts 1 a and b. Part 1a is easy enough to do, but I want to really understand why Gauss's law applies here. collegeboard. com/apc/public/repository/ap11_physics_c_electricity_magnetism_scoring_guidelines.pdf
Here are their solutions. (Remove the space before com in both links). The fact that the enclosed charge is zero does not tell you anything about the electric field, though, as evidenced by part (b). What if I draw a Gaussian surface next to, but not enclosing, a point charge? There is no enclosed charge, and no net flux, but there is still obviously an electric field. So why do they want Gauss's law used in these situations?

Gauss' Law in it's mathematical form is \oint\vec E \cdot \vec {da} = \dfrac{1}{\epsilon _o}Q_{enc}. If we draw a Gaussian surface encompassing the region in which we seek to determine the electric field, the left side of the equation 'picks out' the all sources in this region, and says that the electric field is proportional to the charge enclosed only in this region we've defined with our Gaussian surface, which is the right side of the equation. Notice that in regions that no charge is enclosed, the left side of the equation is zero because all flux entering a Gaussian surface leaves the surface as well.

 

[wkassis]

In reference to your statement about the enclosed charge equaling zero telling you nothing about the \vec E field, that's false. It tells you that \vec E = 0. Gauss' Law allows you to determine electric fields for regions. If you want to know the field in a particular region, the Gaussian surface must enclose that region, and the charge generating the field.

 

[Andy Resnick]

I'm not sure what Gauss's law really means. <snip>

Gauss's law is a form of conservation of energy for a field. For example, we say that a charge generates an electric field. If you enclose the charge with a spherical surface, the electric field at the surface is Q/r^2. No matter what radius you choose, the total field 4πr^2*Q/r^2 through the surface is constant.

Conceptually, Gauss's law in electrostatics states that electric charges create electric fields, magnetic charges create magnetic fields (and since there are no magnetic charges, div(B) = 0). Gauss's law (in other contexts) means the intensity of light from a point source falls off quadratically with distance, the gravitational field of a point source falls off quadratically with distance, etc. etc.