The Concept of Flux in Electromagnetism: Exploring Gauss' Law

[rockyshephear]

Gauss' Law for Electric field states:
The electric field equals Q (the charge inside small closed surface) / electric permittivity

Implications-permittivity is like a drain on the field, if it's high then you are dividing by a larger number giving a smaller field. I guess the permittivity is defined by what material the aforementioned small closed surface resides in. If it is highly insulative then the permittivity is higher and the electric field is smaller. If it is equal to 1 then you get Q. Hmmm Can and electric field be equal to Q? Do they have the same units even??

Gauss' Law for Magentism
Nabla dot producted with B = 0
or
The magnetic flux thru a small closed surface = 0

Implications-The rates of change of whatever in x hat, y hat and z hat unit vector directions (Nabla) of the magnetic flux (B) in the directions of x hat, y hat, and z hat = 0

So does that mean that the flux is not changing in any direction? I though flux propogated?

This seems contrary to one explanation of the amount of flux coming into a surface compared to that going out of a surface.

 

[cepheid]

Gauss' Law for Electric field states:
The electric field equals Q (the charge inside small closed surface) / electric permittivity

Not quite. Gauss' Law states that the FLUX of the electric field through a closed surface is equal to the electric charge enclosed by that surface (divided by the permittivity).

Implications-permittivity is like a drain on the field, if it's high then you are dividing by a larger number giving a smaller field. I guess the permittivity is defined by what material the aforementioned small closed surface resides in. If it is highly insulative then the permittivity is higher and the electric field is smaller. If it is equal to 1 then you get Q. Hmmm Can and electric field be equal to Q? Do they have the same units even??

The permittivity cannot be just 1, with NO units. This is because the permittivity is a constant of proportionality: it explains how the flux of the electric field is related to a corresponding amount of charge. Therefore, it has units that account for difference in physical dimensions between the two quantities (C²N^-1m^-2 in this case, I guess they would have to be). That is its job: to relate these two quantities. Now, it is conceivable that you could cook up a system of units in which the permittivity of free space, ε₀ was equal to 1, in THOSE units that you invented. However, in the SI units I mentioned, the permittivity of free space is given by

ε₀ = 8.854187817... × 10⁻¹² C²N^-1m^-2​

and this is the smallest value a permittivity can have. This is an important point. The specific numerical value of any dimensioned (as opposed to dimensionless) physical constant is arbitrary and therefore meaningless in and of itself, because it depends totally upon the system of units you choose. The same goes for other dimensioned physical constants (speed of light, Planck's constant, etc.).

Of course, the dielectric constant, which is the ratio of the permittivity of a medium to ε₀, can be just 1 (with no units) because this is a dimensionless constant.

 

[rockyshephear]

So the only difference is calling it FLUX of electric field vs electric field? So just saying electric field is not good enough? Otherwise my defiinition is spot on it seems. Yes? It's a simple equation E=Q/permittivity

I was just saying that IF the permittivity would be 1 regardless of units, then E =Q

 

[cepheid]

So the only difference is calling it FLUX of electric field vs electric field? So just saying electric field is not good enough? Otherwise my defiinition is spot on it seems. Yes? It's a simple equation E=Q/permittivity

I was just saying that IF the permittivity would be 1 regardless of units, then E =Q

No. The equation is not E = Q/permittivity. Gauss' Law in differential form is:

\nabla \cdot \mathbf{E} = \frac{\rho}{\epsilon_0}​

In integral form, Gauss' Law is:

\int_A \mathbf{E}\cdot d\mathbf{A} = \frac{Q_{\textrm{encl}}}{\epsilon_0}​

where A is the area enclosing the charge. The surface integral on the left hand side is the flux of the electric field vector. You can see that flux has dimensions of [electric field]*[area].

I was just saying that IF the permittivity would be 1 regardless of units, then E =Q

Because the permittivity is a dimensioned quantity, it is not possible for it to be always equal to 1 regardless of units. If you change your system of units, the value of the permittivity will, in general, change. If the permittivity happens to be equal to 1 in some specific system of units, then the numerical values of Φ_E and Q will be the same (in their respective units). However that's no different than saying that if the mass of an object is m = 1 kg, then due to F = ma, the numerical value of a force acting on it (IN Newtons) is equal to the numerical value of its acceleration (IN ms^-2). However, that fact does NOT give you license to write F = a. This is NOT correct. You cannot equate two different physical quantities -- it just does not make any sense. This situation is no different. Algebraically, you have to keep the epsilon naught in there, even if it is numercally equal to 1 in whatever units it is being measured in. These are all dimensioned quantities.

 

[rockyshephear]

I've got a web page that specifically states
PHI (electric)=Q/permittivity. PHI must be electric flux. This is right, right?

 

[diazona]

Yep. It's defined as
\Phi = \int \vec{E}\cdot\mathrm{d}\vec{A}

 

[rockyshephear]

Oh so the integral of the E (electric field) dot producted with dA (ever shrinking area on the surface in question =Q/permittivity??

 

[diazona]

Yeah, that's just what cepheid wrote - the integral form of Gauss's law.

 

[rockyshephear]

regarding dA, one must need to know something about the geometry of the surface that contains the area so is keeping it simple, like sphere or cube beneficial as opposed to a globby mass whose geometric definition would be very messy? At some point one has to know something about A to define dA. Yes?

 

[cepheid]

For the sake of honesty I just want to say that I edited my most recent post in the following ways (to correct mistakes on my part):

1. I took away the subscript "encl" from the charge density rho in Gauss' Law. An equation in differential form is one that is applicable at a specific point in space (as opposed to the integral form of the equation, which is valid over some region A). In differential form, Gauss' Law states that the divergence of the electric field at a point in space is given by the charge density at that point in space. Density "enclosed" makes no sense.

2. After making a big point of chastising the OP for mistaking flux for electric field in the equation, I did the same thing myself. Hence I have changed E to Phi in my second paragraph.

regarding dA, one must need to know something about the geometry of the surface that contains the area so is keeping it simple, like sphere or cube beneficial as opposed to a globby mass whose geometric definition would be very messy? At some point one has to know something about A to define dA. Yes?

Yes, well, this integral, although we have used only one integration symbol for brevity, is a surface integral (a double integral) in disguise. To figure out how to define your area element, you need to pick a 2D coordinate system. For instance, dA could be given by:

dA = dxdy​

Of course, for curved surfaces, Cartesian coordinates would not be so useful, and we might use spherical coordinates:

dA = r²sinθdφdθ​

where (r,φ,θ) are the usual 3D spherical coordinates.

Finally, it's the LIMITS of integration over each of these two variables that determines the boundaries and therefore the specific shape of your region. For example, if the region A were the surface of a sphere of radius R, then the integral would become:

\int_A \mathbf{E} \cdot d\mathbf{A} = \int_{0}^{2 \pi}\int_{0}^{\pi} \mathbf{E}(\theta, \phi) \cdot \mathbf{\hat{n}} R^2 \sin{\theta}\, d\theta \, d\phi​

where n is a unit normal vector to the surface. You are right that if the region were not as simple (or symmetric) as a sphere, then the limits of integration would be more complicated.

Side note: dA is defined as a vector because a surface is defined to have an *orientation* determined by a unit vector normal to the surface. So, at each point on the surface, the dot product of E with dA ranges anywhere from -1*EdA (the field is pointing directly inward through the surface) to +1*EdA (the field is pointing directly outward through the surface). So, using this convention allows us to calculate the NET flux through the surface (by taking direction into account).

 

[rockyshephear]

Lot's to digest. Thanks. Is there a way to correctly distinguish a multiplication dot from a dot product dot? Or must the context of the equation always be concidered?

 

[jtbell]

Is there a way to correctly distinguish a multiplication dot from a dot product dot?

If the things on both sides of the dot are vectors (usually printed in boldface, or written with arrows on top), it's a dot product.

 

[rockyshephear]

And if only ONE is a vector what do you have?

 

[rockyshephear]

This will freak some of you mathemticians out but I am approaching my question with a crazy analogy. So prepare yourselves.
I am trying to understand electric flux.
Imagine the Earth completely covered with indians. They all have bow and arrows. They are only allowed shoot arrows along a lines that flow from the center of the earth, thru them (normal to this assumed perfect sphere of earth). Imagine that the ionosphere is a very large bubble surrounding the Earth covered indians. The indians all shoot and the arrows fly out symmetrically. The hit the inside of the bubble and all fall to earth.
Now someone cuts a one mile in diameter hole in this plastic bubble. The indians shoot again and some of the arrows escape the bubble because of the opening.
Is this a fair analogy to the principle of flux? The opening in the bubble is A, the intensity of the electric field relates to, say, if every other indian fired or every fourth indian fired, etc. - the higher the intensity of the field, the more arrows escape, and therefore the more flux, the permittivity would maybe relate to the friction in the air on the arrows. Not sure what Q would be in this analogy.
Can anyone answer these questions in terms of my analogy only?
I can see some of you rolling your eyes already. lol
Thanks,
Squarkman

 

[rockyshephear]

Oh, and I forgot to throw in the the electric field is the sum of all indians shooting their arrows over and over and over again.

 

[diazona]

I think it is actually a fair analogy to the principle of flux. It would be basically legitimate to talk about the flux of arrows escaping the hole, or the flux of arrows fired. In either case "flux" would mean the number of arrows. Q would be, I guess, the number of Indians shooting arrows. But I think a better analogy for permittivity would be the failure rate of the bows. For a given number of Indians shooting, the higher the permittivity (i.e. failure rate), the lower the flux of arrows.

Of course, with an electric field, there is no number of anything to count, since it's a continuous field "smoothed" out over space. That's why we invented "field lines", because when you draw field lines, there is a number of something that you can count to determine the flux.

Trivia: electric field is also sometimes called "electric flux density," because it is the density of electric flux (per unit cross-sectional area, though, not per unit volume).

Regarding your previous question, if you have a vector (dot) a number, it's regular multiplication, not a dot product. You can only have a dot product when there are two vectors.

 

[jtbell]

And if only ONE is a vector what do you have?

Then you have "scalar multiplication" of the vector, which multiplies the length of the vector by the other (scalar) quantity. Or, in rectangular coordinates, multiply each component of the vector by the scalar:

\vec B = k \cdot \vec A (usually written without the dot as \vec B = k \vec A) means

B_x = kA_x

B_y = kA_y

B_z = kA_z

I leave it as an exercise to prove that the length (magnitude) of \vec B is k times the length of \vec A.

Hmmm, the vector arrows don't seem to come out as well as they used to...

 

[rockyshephear]

Now we are getting somewhere with this analogy. Cool. Thanks.
Summary
1. flux = number of arrows
2.Q = number of indians or how densely they are represented across the planet
3.permittivity = failure of bows (which limits the number of arrows getting thru the opening.
The electric field will have to be defined in terms of this analogy. Since it's continuous, yet described in a non-continuous manner, is it fair to say that the field equates to the total of all indians shooting with no bow failure and all arrows get thru since we are not blocking their flight with a ionospheric bubble at all?
lol
I know, I'm laughing, myself.
Squarkman

 

[rockyshephear]

Trivia: electric field is also sometimes called "electric flux density," because it is the density of electric flux (per unit cross-sectional area, though, not per unit volume).

Hmmm this confuses me. The flux we said was the number of arrows getting thru the A opening. But electric field density sounds precisely the same. Could you clear that up in terms of my analogy.
Thanks

 

[rockyshephear]

Here's a statement from Wikipedia

In electromagnetism, electric flux is the flux of the electric field. Electric flux is proportional to the number of electric field lines going through a virtual surface. The electric flux d\Phi_E\, through a small area d\mathbf{A} is given by
d\Phi_E = \mathbf{E} \cdot d\mathbf{A} ---------------------------------------

electric flux is proportional to number of field lines going thru a surface?? Well, we know that electric field is continuous, no discrete distance between lines, and field lines are just a convention used by man since you cannot show infinite vectors in illustrations.
I don't see how the number of field lines makes any difference since it's not an accurate depiction of a true field. Plus if they are infinite, then there's an infinite number of flux lines coming through any surface. I just don't get this contradiction.
If you can't see electric field lines, how does anyone even know how many are going thru a surface, TO make the electric flux proportional to?
I just don't think mathematicians really are counting flux lines. ie
"Oh, there's 12 lines coming out of this small surface so the electric flux is proportional to 12!"

What am I missing here? I posted somewhere else on this site an analogy of electric flux which I am refining so I can understand flux better.
It's planet Earth covered with indians who shoot arrows normal to where they stand. They are inside a larger sphere who has an opening "A". The arrows that get thru the opening represent electric flux.
The bigger the opening obviously the more arrows can get thru and the more flux. Right?
But the number seems to be totally dependent on the size of the opening. Nothing else since the field is infinite and the flux is what part of that infinite field gets thru the opening.
Thanks for commenting.

 

[rockyshephear]

Maybe this is a better way to phrase my question.
The equation Electric Flux =Q/permittivity of free space
The electric field lines are proprtional to Electric Flux so the electric field lines are proportional to Q/permittivity of free space. Which means how many (if you COULD count them) is directly proportional to the charge and inversely proportional to permittivity.
Given that, how does charge 'a' of 1 C look compared to charge 'b' of 1000 coulombs?
They both have infinite arrows coming out of the charge points normal to the charge point. So they look exactly alike? What are the differentiating characteristics?
Thx

 

[rockyshephear]

Here's what I'd like to see someone do. A real world problem. I'll set up the conditions and someone shows the math to get to my answer.

There is a single charge of 1 C inside a sphere of 1 inch in diamter. That's all you know. Show me how, if you didn't know the charge was 1C, you would find out using Gauss' Law of Electrostatics. Or is there not enough information to calculate it?

 

[Born2bwire]

Trivia: electric field is also sometimes called "electric flux density," because it is the density of electric flux (per unit cross-sectional area, though, not per unit volume).

Actually the two are different. The flux densities are related to the fields by the appropriate permittivity or (I know I am going to get in trouble for this, see below) permeability. The electric flux density (or also called the electric displacement field) and magnetic flux densities, \mathbf{D} and \mathbf{B} respectively, are related to the electric and magnetic fields, \mathbf{E} and \mathbf{H}, by

\mathbf{D}=\epsilon\mathbf{E}
\mathbf{B}=\mu\mathbf{H}

Unfortunately, when it comes to the magnetic fields names both B and H will be called the magnetic field among many other names. The references I was raised on set H to be the magnetic field and B to be the magnetic flux density. This makes nice sense to me because it keeps a symmetry with the naming of the electric flux density and electric field. In addition, this allows us to relate only the vectors D and B to the enclosed charges directly as opposed to using the intermediary permittivity and permeability constants. The names for D and E do not seem to have any ambiguity to them like H and B do.

 

[diazona]

Now we are getting somewhere with this analogy. Cool. Thanks.
Summary
1. flux = number of arrows
2.Q = number of indians or how densely they are represented across the planet
3.permittivity = failure of bows (which limits the number of arrows getting thru the opening.
The electric field will have to be defined in terms of this analogy. Since it's continuous, yet described in a non-continuous manner, is it fair to say that the field equates to the total of all indians shooting with no bow failure and all arrows get thru since we are not blocking their flight with a ionospheric bubble at all?
lol
I know, I'm laughing, myself.
Squarkman

Q = number of Indians. I guess that is related to density, but making the planet bigger would decrease the density, and that is not the way charge works. Making a ball of charge bigger does not decrease the charge. (\rho would be the equivalent of the density of Indians, or actually I guess \sigma since it's a surface density... but I digress)

Trivia: electric field is also sometimes called "electric flux density," because it is the density of electric flux (per unit cross-sectional area, though, not per unit volume).

Hmmm this confuses me. The flux we said was the number of arrows getting thru the A opening. But electric field density sounds precisely the same. Could you clear that up in terms of my analogy.
Thanks

The flux is the number of arrows; the field would be the density of arrows. So for instance, if you were to double the size of the A opening, you would have twice the number of arrows, so twice the flux, even though the density of arrows (the field) is the same. Be careful not to mix up those two terms (flux and field).

electric flux is proportional to number of field lines going thru a surface?? Well, we know that electric field is continuous, no discrete distance between lines, and field lines are just a convention used by man since you cannot show infinite vectors in illustrations.
I don't see how the number of field lines makes any difference since it's not an accurate depiction of a true field. Plus if they are infinite, then there's an infinite number of flux lines coming through any surface. I just don't get this contradiction.
If you can't see electric field lines, how does anyone even know how many are going thru a surface, TO make the electric flux proportional to?
I just don't think mathematicians really are counting flux lines. ie
"Oh, there's 12 lines coming out of this small surface so the electric flux is proportional to 12!"

What am I missing here? I posted somewhere else on this site an analogy of electric flux which I am refining so I can understand flux better.
It's planet Earth covered with indians who shoot arrows normal to where they stand. They are inside a larger sphere who has an opening "A". The arrows that get thru the opening represent electric flux.
The bigger the opening obviously the more arrows can get thru and the more flux. Right?
But the number seems to be totally dependent on the size of the opening. Nothing else since the field is infinite and the flux is what part of that infinite field gets thru the opening.
Thanks for commenting.

Your confusion about the field lines is understandable because Wikipedia, like most sources, talks about field lines as if they are something real, but of course they're not. What really happens is that the person making the drawing chooses where to draw field lines, in such a way that (1) the diagram is clearly readable and (2) equal field strengths correspond to equally spaced field lines. That way, someone else reading the diagram could, in principle, say that if 12 field lines pass through one part of the diagram, and 6 field lines pass through another part of the same diagram, then the field in the first part is twice as strong as the field in the second part. That's really all it is; field lines are just a device to show the relative field intensity between different parts of the same drawing. (Also, nobody actually counts field lines. The most any real physicist would use them for is to get a rough sense of where the field is relatively strong and where it is relatively weak, in a given illustration.)

In the analogy, the flux is dependent not only on the size of the opening, but also on the density of the arrows. As I said before, if the opening were twice as big, there would be twice as many arrows passing through - but then if you remove half the Indians, you would be back to the original number of arrows because the density of arrows would have dropped by 50%.

Maybe this is a better way to phrase my question.
The equation Electric Flux =Q/permittivity of free space
The electric field lines are proprtional to Electric Flux so the electric field lines are proportional to Q/permittivity of free space. Which means how many (if you COULD count them) is directly proportional to the charge and inversely proportional to permittivity.
Given that, how does charge 'a' of 1 C look compared to charge 'b' of 1000 coulombs?
They both have infinite arrows coming out of the charge points normal to the charge point. So they look exactly alike? What are the differentiating characteristics?
Thx

Charge 'b' would have 1000 times as many field lines (arrows) coming out. Even though the number of field lines is technically infinite in both cases, still one can be 1000x the other. In a drawing, the number of field lines drawn would be finite, but it would be 1000 times as many around charge 'b' as around charge 'a'. This is where the idea of lines is not very useful.

Here's what I'd like to see someone do. A real world problem. I'll set up the conditions and someone shows the math to get to my answer.

There is a single charge of 1 C inside a sphere of 1 inch in diamter. That's all you know. Show me how, if you didn't know the charge was 1C, you would find out using Gauss' Law of Electrostatics. Or is there not enough information to calculate it?

Well, if we don't know that the charge is 1 C, all we do know is that the sphere is 1 inch in diameter. And that by itself is not enough information to calculate anything. If all you know is that a sphere is 1 inch in diameter, you have no idea whether it has any charge at all. We would need to know something else, like (for example) the electric field strength on the sphere's surface.

 

[rockyshephear]

Sorry. I don't follow that.
What is the physical difference between a point charge of 1 C vs a point charge of 1000 C?
They both radiate radially from the point charge. Does one have different magnitudes for the vectors and that's the difference?

 

[diazona]

Actually the two are different. The flux densities are related to the fields by the appropriate permittivity or (I know I am going to get in trouble for this, see below) permeability. The electric flux density (or also called the electric displacement field) and magnetic flux densities, \mathbf{D} and \mathbf{B} respectively, are related to the electric and magnetic fields, \mathbf{E} and \mathbf{H}, by

\mathbf{D}=\epsilon\mathbf{E}
\mathbf{B}=\mu\mathbf{H}

Unfortunately, when it comes to the magnetic fields names both B and H will be called the magnetic field among many other names. The references I was raised on set H to be the magnetic field and B to be the magnetic flux density. This makes nice sense to me because it keeps a symmetry with the naming of the electric flux density and electric field. In addition, this allows us to relate only the vectors D and B to the enclosed charges directly as opposed to using the intermediary permittivity and permeability constants. The names for D and E do not seem to have any ambiguity to them like H and B do.

Well, the nomenclature is kind of ambiguous, true, and that can lead to confusion... but my usage is that \vec{E} = "electric field" = "electric flux density" and \vec{B} = "magnetic field" = "magnetic flux density". Using that convention, field and flux density are the same. Unless you disagree that in
\Phi = \iint \vec{E}\cdot\mathrm{d}\vec{A}
\vec{E} is the area density of \Phi? (And in that case I'm not sure what to say except, "well, it is" )

 

[diazona]

Sorry. I don't follow that.
What is the physical difference between a point charge of 1 C vs a point charge of 1000 C?
They both radiate radially from the point charge. Does one have different magnitudes for the vectors and that's the difference?

Yes. That is the real difference.

An unfortunate thing about field lines is that they hide the magnitudes of the vectors, and so to retain some information about them, we have to "translate" the vector magnitudes into spacing of the field lines.

 

[Born2bwire]

Well, the nomenclature is kind of ambiguous, true, and that can lead to confusion... but my usage is that \vec{E} = "electric field" = "electric flux density" and \vec{B} = "magnetic field" = "magnetic flux density". Using that convention, field and flux density are the same. Unless you disagree that in
\Phi = \iint \vec{E}\cdot\mathrm{d}\vec{A}
\vec{E} is the area density of \Phi? (And in that case I'm not sure what to say except, "well, it is" )

No, but I would argue that there is no ambiguity in that the D field is called the electric flux density. I have seen plenty of back and forth in terms of the naming of B and H but not E and D. The flux density usage of the terms seems to be more of an engineering nomenclature. Open pretty much any EE textbook and you should find D as the electric flux density. Just a quick search on Amazon reveals this is the case in "Elements of Electromagnetics."

 

[diazona]

This is the first I'm hearing of \vec{D} being called the electric flux density. Maybe it is just an EE thing :-/ IIRC I first saw electric field (\vec{E}) being called electric flux density in the CRC Handbook of Chemistry and Physics, although I think I've seen it in other places since then...

Anyway, bottom line: unless you can convince me that
\Phi = \iint \vec{D}\cdot\mathrm{d}\vec{A}
is a true equation, I don't accept "electric flux density" as a name for \vec{D}.

 

[Born2bwire]

This is the first I'm hearing of \vec{D} being called the electric flux density. Maybe it is just an EE thing :-/ IIRC I first saw electric field (\vec{E}) being called electric flux density in the CRC Handbook of Chemistry and Physics, although I think I've seen it in other places since then...

Anyway, bottom line: unless you can convince me that
\Phi = \iint \vec{D}\cdot\mathrm{d}\vec{A}
is a true equation, I don't accept "electric flux density" as a name for \vec{D}.

The CRC calls E the electric field and D the displacement field. However, the CRC does define the electric flux as

\Psi = \int \vec{D}\cdot\mathrm{d}\vec{A}

and in this case it does make sense that D is the electric flux density (page 2-3 via Google Books).