# Flux and gauss law

1. May 5, 2010

I know that electric flux is defined as the number of electric field lines passing through an area but what kinda area are we talking about. Does it have to be perpendicular to the field lines like this

or could it be at an angle like this

does it have to be a flat area on 1 plane like the previous 2 examples or could it be a 3D area like this

2. May 5, 2010

### Staff: Mentor

All of your illustrations are perfectly OK examples of an electric flux through a surface.

3. May 5, 2010

### mooglue

A flux is defined as a dot product E dot A essentially, so your flux will be reduced if the surface is not perpendicular. Think about it like shining a flashlight at a friend. If you tilt the light away from him, he can still see the light, but it's not as bright. You can just imagine that less flux is reaching him because the light has been moved at an angle relative to his eyes.

4. May 5, 2010

### mooglue

Also, it doesn't have to be a nice shape. For example in your last picture, it is not, and in this case, you'd need to use the differential formula for the flux, which involves the integration over the surface. This just takes into account that the flux through different portions of your surface is not constant, and generally you'll need a function for the surface of your shape.

5. May 5, 2010

mooglue: So this surface function for the surface would take into account the angle between the field lines and the normal of the surface through the whole surface? This is the first physics application of integration I've run into so far.

Is Gauss's law then just the fact that the flux is proportional to the charge?

6. May 5, 2010

### LostConjugate

You should be able to reduce the problem of a complex blob like object to a simple plane though if it is a solid object or a membrane like a balloon.

http://ocw.mit.edu/OcwWeb/Physics/8...pring2002/VideoAndCaptions/detail/embed03.htm

"nd this is independent of the distance R.

And that's not so surprising because if you think of it as air flowing out then all the air has to come out somehow whether I make the sphere this big or whether I make the sphere this big.

So the flux being independent of the size of my sphere, the flux is given by the charge which is right here at the center divided by epsilon zero.

Now if I had chosen some other shape, not a sphere, but I have dented it like this, it's clear that the air that flows out would be exactly the same.

And so I don't have to take a sphere to find this result.

I could have taken any type of strange closed surface around this point charge and I would have found exactly the same result." -Walter Lewin

Last edited: May 5, 2010
7. May 5, 2010

### Staff: Mentor

Gauss's law states that the total flux through a closed surface is proportional to the charge enclosed. See: Gauss's Law

8. May 5, 2010

Also I'm having serious trouble understanding the difference between electric field magnitude and electric flux. Both are defined by the density of field lines per area aren't they?

9. May 5, 2010

### Staff: Mentor

No, they are not. If you want to think in terms of field lines, then the density of field lines gives you a measure of the field at some point. But you won't know the flux until an area is defined. Field is at a point; flux is over an area and depends on the orientation of the area with respect to the field.

10. May 5, 2010

### LostConjugate

I recommend watching Walter's videos, he is an expert advisor.

11. May 5, 2010

Thanks for the analogy that cleared up a lot of the confusion I had about $$\phi = \frac{q}{\epsilon_0}$$. As long as the thing is enclosed the amount of flux lines hitting the total surface area will be equal but the bigger the balloon the smaller the flux will be for small segments of the balloons surface area. I wasn't really thinking about a charge being a source of a finite amount of electric field lines but thinking about it like that I get the concept.