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Flux through a closed surface

by guitarphysics
Tags: electricity, flux, gauss
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guitarphysics
#1
Apr7-13, 02:25 PM
P: 180
Hi, I'm trying to teach myself electricity and magnetism (and it's not easy!) and I'm not sure I understand flux...
For one thing, why is the flux through a closed surface zero if there is no charge inside of the surface (but there IS one outside)?
Another thing I'm not really sure about this either; why the flux through a closed surface is equal to 1/epsilon times ∫ ρ dv. That is, why the flux through any closed surface is independent of that surface's shape or size! One way I have of thinking about this (which may be completely wrong, I don't know), is this: The electric field decreases over distance as 1/r2, but the surface area will also be increasing (this only works if it increases by the same proportion), so when a bigger surface surrounds a charge, the electric field will decrease while the surface area will increase, so the flux will stay the same. Is this a good way of thinking about it, or is it completely wrong?
Thank you
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mathskier
#2
Apr7-13, 02:33 PM
P: 30
Flux through a closed surface is 0 by the divergence theorem... But only for the case of an inverse square field. In that case, the divergence of the field is 0 unless you have a point charge (a source or a sink) within the surface. So that relation is specific to electrostatics, gravity, and other inverse square laws. If the field fell off exponentially or something, it wouldn't be true that closed surfaces have 0 net flux.
guitarphysics
#3
Apr7-13, 02:38 PM
P: 180
Quote Quote by mathskier View Post
Flux through a closed surface is 0 by the divergence theorem... But only for the case of an inverse square field. In that case, the divergence of the field is 0 unless you have a point charge (a source or a sink) within the surface. So that relation is specific to electrostatics, gravity, and other inverse square laws. If the field fell off exponentially or something, it wouldn't be true that closed surfaces have 0 net flux.
Sorry, I didn't understand much of that.. I don't know what div is yet. Could you explain it in a different way? (Sorry to make this so difficult). Also, is my intuition on why the flux through closed surfaces containing a charge is independent of their shape and size correct?

mikeph
#4
Apr7-13, 02:59 PM
P: 1,212
Flux through a closed surface

Vector calculus is pretty integral to the study of electromagnetism. The best introduction I've ever read is by H.M.Schey, I recommend finding a copy of that book to get started.
guitarphysics
#5
Apr7-13, 03:05 PM
P: 180
Well, I'm reading Electricity and Magnetism by Purcell and Morin, which introduces vector calculus from scratch, so I don't think I need any extra resources (although we'll see).
EDIT: I actually thought of getting that book before, and if it's actually good, I think I'll get it.
WannabeNewton
#6
Apr7-13, 03:32 PM
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I assume you mean closed and bounded surface (i.e. compact surface) such as a sphere. Anyways so assume we have an arbitrary such surface and further assume it has no charge inside it. Remember that field lines of electric fields must start with a charge and end with some charge or go off to infinity. If there is no charge inside the surface then the field lines of any electric field generated by charge distributions outside the surface passing through this surface must also exit the surface because there is no charge inside the surface for the field lines to terminate on. This means the number of fields lines entering the surface cannot be greater than the number exiting so the net flux through the surface is zero.

Also, flux through a surface is not independent of the geometry and orientation of the surface. The ##dV## is the volume element which depends on the geometry of the surface (the metric). The volume element for a sphere is not the same as the volume element for a plane.
DrZoidberg
#7
Apr7-13, 04:11 PM
P: 389
Quote Quote by guitarphysics View Post
Another thing I'm not really sure about this either; why the flux through a closed surface is equal to 1/epsilon times ∫ ρ dv. That is, why the flux through any closed surface is independent of that surface's shape or size!
The flux through a closed surface is always equal to the charge contained inside or alternatively the charge divided by e0 depending on what system of units you use.
One way to visualise that is to imagine each proton and each electron have a field line emerging from them or ending at them. Each field line has to and at a charge. The shape of the surface doesn't change the charge inside and so it doesn't change the number of field lines.
In this model the electric flux is simply the number of field lines that pass through the surface. And that number can't change if you change the shape or size of the surface. It only changes if you change the charge.
BruceW
#8
Apr7-13, 04:17 PM
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Yeah, I think the best way to 'bridge the gap' to learning Gauss' law is by thinking about what the field lines are doing. It probably will still feel slightly 'mystical'. I know that Gauss' law felt mystical for me, even a long time after first learning about it. I only really felt comfortable with it after learning about divergence, and the differential form of Gauss' law.

Edit: and continuum mechanics in general also made me feel like I understood it better. For example, there is a Gauss' law for gravity, too.

final edit: so, what I'm trying to say, is don't worry that the concept feels a bit alien. Doing problems and derivations as practice will help (as it does for most physics).
guitarphysics
#9
Apr7-13, 04:57 PM
P: 180
Thanks, you were all very helpful. WannabeNewton, you cleared everything up very nicely (as always :D). BruceW, you're right! Mystical is a very good word to describe my thought's about Gauss' Law (and fields).


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