Gauss's law intuitive explanation?by lord_james Tags: electric field, electricity, enclosed charge, gauss's law, physics 

#1
May1212, 03:33 PM

P: 1

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? 



#2
May1312, 01:53 PM

P: 6





#3
May1312, 06:10 PM

P: 6

In reference to your statement about the enclosed charge equaling zero telling you nothing about the [itex]\vec E [/itex] field, that's false. It tells you that [itex]\vec E = 0[/itex]. 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.




#4
May1312, 07:40 PM

Sci Advisor
P: 5,468

Gauss's law intuitive explanation?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. 


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