Why doesn't changing the Gausssian sphere change the net flux?

[mr_coffee]

Hello everyone, I'm stuck on yet another check point There is a certian net flux through a Gaussian surface radius r enclosing an isolated charged particle. Supppose the enclosing Gausssian surface is changed to (a) a larger Gaussian sphere, (b) a Gaussian cube with edge length equal to r, and (c) a Gaussian cube with edge length equal to 2r. In each case, is the net flux through the new Gaussian surface greater than, less than, or equal to the orginal net flux. I first tried to visual the difference, so I thought it made sense, if you increase the size, it will capture more electrical fields but then I remembered if the same amount of electrical fields enter and leave the surface there is a net flux of 0. Is that the answer why it doesn't matter how large the G surface is?

 

[Doc Al]

What does Gauss's law tell you about the net flux through any closed surface? Answer: The net flux is proportional to the net charge enclosed within that surface. (So... if the charge doesn't change, the flux doesn't change.)

 

[mr_coffee]

thanks! my book just says "relates" not proportional, i like your wording better!

 

[Jelfish]

I suppose you're already convinced, but I thought about this for a minute and came up with a sort of round-about logic to perhaps convince someone intuitively.

Flux is determined by taking a surface integral over a vector field. That is, you want to find the total amount of whatever going through a surface, so you find the rate at which stuff moves through the surface at every infinitesimal point and add them all up.

First consider a sphere with the charged particle at the center. Since every point on the surface of the sphere is the same distance away from the particle, the field on the surface is the same throughout (E is only a function of radius). Therefore, the flux equals the electric field at the surface times the surface area.

(for a sphere) $$\Phi = (\frac{kQ}{R^2})(4\pi R^2)=4kQ\pi$$ which is a constant.

(4*Pi*R^2 is the surface area of a sphere).

This means that no matter what the radius is, the flux is the same.

That solves the sphere cases, but the cubes require a bit more reasoning (I'm trying not to get too caught up in the math).

Visualize this: The charge particle is at the center of a small sphere. There is a cube that surrounds that sphere. Then there is another sphere that surrounds the cube, centered at the charged particle. Now, you know intuitively that the field falls off with distance. Therefore, it makes no sense that the cube should have a greater flux than the small sphere that it encloses, because the smaller sphere has the same flux as the larger sphere (as shown from above). You can sort of use the same argument for why the cube can't have a smaller flux than the spheres. This only works, however, if the cube is closed so none of the field 'leaks' out.

If you want to understand this a bit more rigorously, you can read about conservative fields and the derivation of Gauss' law.

I hope that helps.

 

[mr_coffee]

thanks for the explanation!

 

[James R]

If you want a non-mathematical way of thinking about electric or magnetic flux, it is roughly proportional to the number of field lines passing through the surface in question. You can increase the number of field lines in one or more of three ways:

1. Make the field strength stronger, so the field lines become more dense (more lines in a given volume); or
2. Increase the size of the area intercepted by the field lines; or
3. Alter the orientation of the area so it is "more perpendicular" to the field direction at the surface.

 

[Chronos]

You must integrate the charge over the surface area. The big question should be... what defines the surface area?

 

[Dr.Brain]

In the expression E.S on the left hand side of Gaussian Law , E is the electric field due to all charges present inside or outside the Gaussian Law and S is the Gaussian Surface Surface area which remains constant if you move your sphere or change its position.The net flux remains fixed if you keep the area fixed.

BJ