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).

 

[diazona]

Really? Well if that is indeed the case I guess \vec{D} can be electric flux density. (I'm not able to preview the CRC in Google Books myself)

For reference, I'll continue to use the term to mean \vec{E} as that is the usage I'm familiar with.

 

[jtbell]

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?

If the charge is at the center of the sphere, then the electric field at the surface of the sphere is perpendicular to the surface, and has equal magnitude (strengh) everywhere on the surface. In this case, Gauss's Law reduces to

EA = \frac{Q}{\epsilon_0}

where A is the surface area of the sphere. Measure E at any convenient point on the sphere and you can easily calculate Q.

If the charge is not at the center of the sphere, this doesn't work because the field isn't perpendicular to the surface everywhere, and its magnitude varies with position. In this case, you have to divide the surface into many small pieces, small enough that the field is practically uniform over each individual piece. Then measure the field (both magnitude and direction) at each piece of surface, calculate the flux through each piece individually, using the dot product \vec E \cdot \vec A, and then add up all those little "bits" of flux to get the total flux, which equals Q / \epsilon_0 by Gauss's Law.

This gives you only an approximate value for the charge, but you can make it better by dividing the surface into smaller pieces.

 

[rockyshephear]

I think my analogy if flawed. It tends toward all the surfaces being impinged upon by a single source of radial arrows. But a lecturer on youtube was talking as if the electric field could approach the surace at ANY angle. My indians can only shoot normal to where they are standing and the hole in the surface is always normal to the arrows.

This lecturer drew incoming electric fields like perfectly parallel horizontal lines. How is that possible since any electric field comes from a point charge and spreads out radially?
At least if it's a single point charge in an empty universe.

So maybe for simplicity sake, my analogy should be qualified that nothing exists in the analogy universe except the one Earth and all it's indians and one ionospheric bubble. Then you have NOTHING but radially emminating arrows.

Now given my analogy and the new qualification, could someone explain
1. Charge
2. Electric Field
3. Electric Flux
4.dA
5 permittivity of free space

All in terms of my analogy. Then I will easily understand the concept.

 

[rockyshephear]

To prime the pump let me go first.

To reiterate the setup and conditions.

we have a totally smooth spherical world upon which stand an infinite tribe of indians all over the planet. They are only allowed to shoot normal to the infinitely small space they stand on. They are surrounded by a bubble located somewhere in the ionosphere. A shape is cut out of the bubble allowing whatever arrows are pointed at this opening to come out. We assume only one charge exists inside this bubble which is the perfect center of the planet. And it all happens in an otherwise empty universe. Now...

Charge would equal how far back the indians pull on their bows.

Electric field would be all indians shooting their arrows which travel radially to infinity.

Electric Flux would be a vector quantity depicting how many arrows escape a closed surface cut into the ionospheric bubble, summed into one vector arrow depicting the summed direction, which in this case is always normal to the cut out surface.

dA is the size of the area cut out, which is summed up to give the total flux, and which approaches zero as we approach the most accurate total flux.

permittivity of free space (since it is a constant) would state that forever and always, an exact percentage of bows will not function. Thus the reason why it is in the denominator because it tends to lower the flux amount. Q is in the numerator so since it's how far back the indians pull their bows, it gives a higher flux value because the lengths of the vectors are longer.

How's that? Once I nail it then I am going to do the same analogy using two point charges...that will be complicated. Then after that tackle magnetism using, you guessed, more indians shooting arrows.

note: In an effort not to appear as stereotyping, I should switch out indians for say...carthaginians, or the like. It only seems fair.
Squarkman

 

[diazona]

we have a totally smooth spherical world upon which stand an infinite tribe of indians all over the planet. They are only allowed to shoot normal to the infinitely small space they stand on. They are surrounded by a bubble located somewhere in the ionosphere. A shape is cut out of the bubble allowing whatever arrows are pointed at this opening to come out. We assume only one charge exists inside this bubble which is the perfect center of the planet. And it all happens in an otherwise empty universe. Now...

That's what I have been assuming all along.

Charge would equal how far back the indians pull on their bows.

Okay, if this is how you're thinking about it, then forget everything I posted earlier about the number of Indians, the number of arrows, the densities, etc. I was interpreting the analogy differently than you were.

Electric field would be all indians shooting their arrows which travel radially to infinity.

Not sure what you mean by that, but in this interpretation the electric field would be sort of like the pressure exerted by all these arrows. Very roughly. That's not a particularly great equivalence.

Electric Flux would be a vector quantity depicting how many arrows escape a closed surface cut into the ionospheric bubble, summed into one vector arrow depicting the summed direction, which in this case is always normal to the cut out surface.

Electric flux is a scalar quantity, not a vector quantity. In this analogy it would be basically the normal force exerted by all the arrows passing through the cut out surface. Sort of. That's not really precise, though, and I can't think of an ideal role for flux in the analogy.

dA is the size of the area cut out, which is summed up to give the total flux, and which approaches zero as we approach the most accurate total flux.

Well... technically A is the total amount of area cut out, but dA is just one little patch. (Imagine that when cutting out the area, you have to take out one square foot at a time; that's kind of like dA)

permittivity of free space (since it is a constant) would state that forever and always, an exact percentage of bows will not function. Thus the reason why it is in the denominator because it tends to lower the flux amount. Q is in the numerator so since it's how far back the indians pull their bows, it gives a higher flux value because the lengths of the vectors are longer.

Given that you are interpreting charge as how far the Indians pull back their bows, I would say that the permittivity would be like the mass of the arrow. Since the higher the mass of the arrow, the slower it will be traveling.

This arrow-shooting analogy can only take you so far. I suppose it more or less works for a single point charge, but once you try two point charges I anticipate you having to come up with some really messy hacks to make the analogy work, and who knows what is going to happen with magnetism. (Arrows spontaneously traveling in circles perhaps?) I'd really encourage you to take some time to get more comfortable with the math first, so you don't have to resort to analogies.

 

[rockyshephear]

Thanks. I believe I need to perfect this analogy so it doesn't stir debate in the engineering community, first. So I need the best description possible relating this analogy to the true nature of a point of charge.
Anyone care to help refine this analogy to as close a level of perfection possible. With it, I am sure I could easily teach an 8 year old this concept in a few hours.

So the previous poster refined my analogy as such----------------

We have a totally smooth spherical world upon which stand an infinite tribe of indians all over the planet. They are only allowed to shoot normal to the infinitely small space they stand on. They are surrounded by a bubble located somewhere in the ionosphere. A shape is cut out of the bubble allowing whatever arrows are pointed at this opening to come out. We assume only one charge exists inside this bubble which is the perfect center of the planet. And it all happens in an otherwise empty universe. Now...

Charge would equal how far back the indians pull on their bows (seems to conflict with the revised definition of electric field below)

Electric field would be the pressure exerted on the ionospheric bubble, which is related to how far the indians pull back their bows and depicted by the magnitude of the arrows.

Electric Flux would be a scalar quantity depicting how many arrows escape a closed surface cut into the ionospheric bubble, irrespective of direction since it's a scalar quantity. (Ironically it seems it would always be infinite since if you take the infinite field and divide it up, you still get infinity)

A is the total area of the ionospheric bubble

dA is the size of the area cut out of the bubble, which is summed up to give the total flux, and which approaches zero as we approach the most accurate total flux.

permittivity of free space (since it is a constant) would be equivalent to the mass of the arrows. (Not sure I buy this since electro-magnetic propogation is at the speed of light, precipitated by photons which are massless).


Are we getting closer or farther away?

 

[rockyshephear]

Something that bothers me about my analogy is that having the ionospheric bubble with an opening in it, every point on the surface that is cut out is normal to every arrow coming thru at that point. That's an idealized scenario since in reality many surfaces are at different angles to the incoming arrows.

If the closed surface and arrows have 0 angle between them, then there is 0 flux thru that dA surface. My analogy does not take this into account. But there's no way around having every point on the surface normal to each arrow if the ionspheric is a concentric sphere around the planet. I guess offsetting the planet inside the bubble would give arrows impinging on the cut away surface at DIFFERING angles other than normal. Is this correct?
Then the flux at an infinitely small location on the surface would then be Phi= EA cos q (it is even when normal).
So here's my new analogy based on the above consideration.
----------------------------------------------------------------------------

We have a totally smooth spherical world upon which stand an infinite number of indians, completely covering the planet. They are only allowed to shoot normal to the infinitely small space they stand on, on the surface of the planet. They are surrounded by a concentric bubble located somewhere in the ionosphere. A closed shape is drawn on the bubble and this closed shape is extruded away from the center of the bubble for infinity. The ideal goal of the indians is to have their arrow go thru the extruded shape and never hit the sides but go as far as they can travel thru the extruded shape. Only those in the correct position can do so, or even get their arrow in the drawn shape at all. We are only concerned with arrows entering and leaving this closed surface. We assume only one charge exists inside the planet, which is the perfect center of the planet but not neccessarily colinear with the center point of the bubble. Assume that the surface of the planet is no where near the inside of the bubble. And it all happens in an otherwise empty universe. Now...

Charge would equal how far back the indians pull on their bows. If they are lucky, they get the highest flux score by making the arrow go really far and not touching the inside of the extruded shape by having the perfect entry angle. If they do not go as far or do not have the perfect entry angle, they lose points. (Phi=EA cos theta)

Electric field would be tantamount to, all the arrows shot at once, over and over, and hitting the inside of the bubble, and the force with which they hit, based on how far back the bowstring is pulled. It would be evenly distributed if the center of the bubble and the center of the planet are colinear.

Electric Flux would be a scalar quantity depicting how many arrows escape the drawn closed surface on the ionospheric bubble. Flux is stronger (they get more points), the more the tendency for the arrow to travel thru an infinite extruded shape of the closed surface, projected along the arrows trajectory away from the bubble center, and not hit the inside of the extruded shape. The ideal is the arrow goes on for infinity and never hits the sides.

A is the total area of the ionospheric bubble

dA is the size of the area drawn on the bubble, which is summed up to give the total flux on the entire bubble, and when dA approaches zero, we approach the most accurate measurement of the total flux for the entire charge inside the bubble.

permittivity of free space (a constant) would be equivalent to the mass of the arrows. (Not sure I buy this since electro-magnetic propogation is at the speed of light, precipitated by photons which are massless). The higher this number of permittivity (a constant), the harder it is for the indians to make their arrows go very far. To go further you need more charge (pulling back harder on the bowstrings).

I think I'm almost there with this analogy.

 

[rockyshephear]

Correction:

We have a totally smooth spherical world upon which stand an infinite number of indians, completely covering the planet. They are only allowed to shoot normal to the infinitely small space they stand on, on the surface of the planet. They are surrounded by a bubble located somewhere in the ionosphere but not necessarily concentric to the center of the planet. A closed shape is drawn on the bubble and this closed shape is extruded away from the center of the bubble for infinity. The ideal goal of the indians is to have their arrow go thru the extruded shape and never hit the sides but go as far as they can travel thru the extruded shape. Only those in the correct position can do so, or even get their arrow in the drawn shape at all. We are only concerned with arrows entering and leaving this closed surface. We assume only one charge exists inside the planet, which is the perfect center of the planet but not neccessarily colinear with the center point of the bubble. Assume that the surface of the planet is no where near the inside of the bubble. And it all happens in an otherwise empty universe. Now...

Charge would equal how far back the indians pull on their bows. If they are lucky, they get the highest flux score by making the arrow go really far and not touching the inside of the extruded shape by having the perfect entry angle. If they do not go as far or do not have the perfect entry angle, they lose points. (Phi=EA cos theta)

Electric field would be tantamount to, all the arrows shot at once, over and over, and hitting the inside of the bubble, and the force with which they hit, based on how far back the bowstring is pulled. It would be evenly distributed if the center of the bubble and the center of the planet are colinear.

Electric Flux would be a scalar quantity depicting how many arrows escape the drawn closed surface on the ionospheric bubble. Flux is stronger (they get more points), the more the tendency for the arrow to travel thru an infinite extruded shape of the closed surface, projected along the arrows trajectory away from the bubble center, and not hit the inside of the extruded shape. The ideal is the arrow goes on for infinity and never hits the sides.

A is the total area of the ionospheric bubble

dA is the size of the area drawn on the bubble, which is summed up to give the total flux on the entire bubble, and when dA approaches zero, we approach the most accurate measurement of the total flux for the entire charge inside the bubble.

permittivity of free space (a constant) would be equivalent to the mass of the arrows. (Not sure I buy this since electro-magnetic propogation is at the speed of light, precipitated by photons which are massless). The higher this number of permittivity (a constant), the harder it is for the indians to make their arrows go very far. To go further you need more charge (pulling back harder on the bowstrings).

 

[cepheid]

Just a minor point. I am Indian (racially or ethnically), because my ancestors once lived in India. This is the only type of Indian I tend to generally accept in modern parlance. Here in Canada, I believe that the people who used to live in the region before the arrival of Europeans are officially referred to as either First Nations or Aboriginals, and are sometimes called Natives. In the U.S. I believe they are known as Native Americans. I'm not claiming that the term has disappeared entirely. I believe that our federal government still has a Minister of Indian Affairs (and no, his job is not to deal with the influx of brown people immigrating from the subcontinent, LOL), but my point is that these are generally relics of the 19th century. I'm not taking any offence to your statement, but I feel obliged to warn you that some people might.

I certainly can't claim to have read all of this, but let's examine some parts of your analogy.

Now...

Charge would equal how far back the indians pull on their bows (seems to conflict with the revised definition of electric field below)

An electric field is a vector field (a function that assigns a vector to every point in space). In other words, the electric field is a function of position: E = E(x,y,z).

For a static field, the vectors don't move, so your analogy of firing speed doesn't really have any bearing here.

What the charge determines is the strength of the electric field at any given point in space. The strength of the electric field is typically visualized in terms of the length of the electric field vector at that point in space. However, this also has no bearing to your analogy.

As an alternative to mapping out the electric field vectors in space, we can visualize the electric field using field lines, which are curves to which the electric field vectors are tangent at every point. Now, sometimes, the density of the electric field lines (how closely spaced they are) is used to indicate the strength of the field (so the field line density would be higher for a higher charge). In this case, the charge would be analogous to the number of people firing arrows. However, I want to emphasize that this is an arbitrary convention. The field lines are not actually more or less dense with more or less charge. Since the field has a value and direction at EVERY point in space, we would actually have to draw an infinite number of field lines (in every scenario) in order to be "accurate." This, of course, is useless. Therefore, the number of field lines drawn in practice is entirely a matter of convenience.

Electric field would be the pressure exerted on the ionospheric bubble, which is related to how far the indians pull back their bows and depicted by the magnitude of the arrows.

The direction of the electric field would be the direction of the arrows. The magnitude of the electric field, as I have stated already, would not be analogous to anything at all (at least, not anything obvious). Again, using the field line analogy, the closest thing we might say is that the the stronger the field, the more closely-spaced the arrows are.

Electric Flux would be a scalar quantity depicting how many arrows escape a closed surface cut into the ionospheric bubble, irrespective of direction since it's a scalar quantity. (Ironically it seems it would always be infinite since if you take the infinite field and divide it up, you still get infinity)

It does depend on direction: a scalar can be positive or negative, and as I explained several posts back, the dot product of the electric field vector with the unit normal vector ensures that what is calculated is the NET flux. Field vectors whose normal component points inward from the surface contribute to the flux in a negative way, and field vectors whose normal component point outward from the surface contribute in a positive way. In this simplified scenario (a single positive charge at the origin), obviously all of the field vectors point radially outward (radially because you have no sources that are offset from the centre of the sphere, and outward because all the sources are inside the sphere and none are outside).

I would say that a reasonable analogy for flux in this scenario is the number of arrows passing through the surface per unit area, per unit time. EDIT: No. Scratch that. That's a different use of the word "flux" that I was thinking of. As diazona pointed out, there is no good analogy for the flux here.

permittivity of free space (since it is a constant) would be equivalent to the mass of the arrows. (Not sure I buy this since electro-magnetic propogation is at the speed of light, precipitated by photons which are massless).

You are confusing two different issues. It is true that if I have some charges "over here", and I shake them around a bit, then I ask myself, how long will it take before the people "over there" a distance x away will realize that the electric field at their location has changed due to the time variation of the source charge distribution, the answer is that they will not realize that anything has happened until a time t = x/c later. This is because c is the speed at which information about the change propagates.

HOWEVER, in this case, the field we are talking about is static. It is not propagating anywhere. It merely permeates space. This is just one example of where the arrow analogy is leading you astray. What permittivity measures is the tendency of a medium to respond to an applied electric field by becoming "polarized" (meaning that the applied field induces a spatial separation of + and - charges in the atoms or molecules of that medium, which is known as an electric dipole). If this medium is polarized (has a bunch of electric dipoles in it), then those dipoles will set up a field of their own, one that tends to oppose the applied field. The net result is a shielding effect -- the net field within the medium is weakened. That's what permittivity is, period.

Are we getting closer or farther away?

To be honest, I do not think that this analogy is a useful intellectual exercise that will lead you to greater insights about electrostatics. Analogies by their very nature are imprecise, so trying to "perfect" this one makes little sense. Please please take the advice of diazona in his/her last post (#35 -- the very last sentence) to heart. Your intellectual energies would be better spent trying to get a handle on vector calculus. Physical theories are formulated mathematically, and vector calculus is the most natural language for formulating electromagnetism (at least at this level of understanding).

 

[rockyshephear]

Well, thanks for that. I don't feel like I'm making progress with this analogy thing. I am sensing engineering antipathy for all but using math and vector calc to describe this. I feel that's somewhat narrow thinking since people all learn in differing ways. But your critique leaves me not knowing how to proceed. I can't be satisfied with an analogy full of holes.
I can't believe the concept is so complex it defies a concise analogy that could be memorized as a narrative that would enable a higher degree of understanding to the young and those who are far more visually minded that analytical.

 

[diazona]

This "engineering antipathy" you think you're sensing is - well, first of all, not engineering, since I at least am a physicist, not an engineer But seriously, it's not that we have some ignorant disrespect for every learning method that isn't math. I know I've tried to use many analogies for these sorts of concepts, and inevitably they all fall apart at some point - the only description that I know to be completely accurate and not misleading is the mathematical one. So by telling you to forget about analogies and learn to use the math, we're trying to give you the benefit of experience.

(P.S. @cepheid: I'm a "him" )

 

[rockyshephear]

Thanks but I'm a dyed in the wool optomist and I KNOW that an amazingly accurat analogy can be devised. I just know it. And no, I don't believe in Santa anymore and I'm agnostic. So for me to say this indicates a true sincerity on my part. There is nothing that cannot be explained with analogy. Sorry, physicist, not engineer. You've been wrongly categorized twice today. lol
So, I challenge all on this forum to come up with an excellent analogy for electrostatics.

So without this, my next question would be...

You said..."The strength of the electric field is typically visualized in terms of the length of the electric field vector at that point in space."

So considering a sphere with a point charge at it's center, around the center will have the greatest magnitude vector and they will radially fall off as the inverse of R^2.

So I see two things alone that can be differentiated at the surface dA. One of them is not how many lines of flux pass thru, since it is infinite. The only two things are the angle that the vector makes at the surface dA and the magnitude of that vector. These, to me are the only two pieces of information to work with in describing flux.
So this agrees with finding the total flux by Phi(total)=q/epsilon(0).
Uses:
If you know Phi you can find q

I

 

[rockyshephear]

I only see two pieces of data that are useful at the dA surface. The angle that the vector makes to the surface and the magnitude of the vector. Since flux vectors are infinite, they are infinite thru dA.

So are these equations correct?

Phi (total)= Surf Integral of E dot producted with dA = Surf Integral |E| times |dA | times cos theta (scalar)

Phi at dA is
|E| times |dA | time cos (scalar)

Phi (total)=q/epsilon (0) (scalar)

Oh, and there is not such thing as an electric field that is not radially spherical. Right?

Thx

 

[cepheid]

Hi rockyshephear,

I'm sorry if I sounded overly critical. I flipped through a few pages of Electrodynamics by David J. Griffiths. It was my undergraduate text for both of my third-year E & M (Electricity and Magnetism) courses. It's a pretty standard text for undergraduate physics, although I am sure that some people here feel that it is no good, preferring a more rigourous text like Jackson (which I'm told is the de facto standard for Classical E & M). In any case, I don't think that Griffiths is used in Electrical Engineering, in which the course that introduces Maxwell's equations tends to be called "Electromagnetics", and has a slightly different emphasis (the goal being to get right to applications like transmission line theory and waveguides, from what I gather). Nevertheless, because my education was in Engineering Physics, I got the best of both worlds. In particular, I had the good fortune (in my opinion) of being able to learn from Griffiths, which I felt to be really excellently written, and highly entertaining to boot. One of my friends said that an acquaintance of his once commented that the book, "reads like a novel," and was among his favourite books! I would highly recommend borrowing a copy of this book from the library if you get the chance. I took vector calculus before E & M, and I didn't really have much of an intuitive feel for divergence, gradient, and curl (just the definitions). Seeing them applied in Griffiths really helped. I think it was the right order in which to do things, however, because knowing the math and then seeing it applied meant that I had no impediments to learning the physics. I would not recommend doing things the other way around (tackling E & M first, and learning vector calculus as you go along).

So, like I said, I was flipping through the section that introduces Gauss' Law, and I realized that I may have been too quick to pooh-pooh the density of field lines as an indicator of electric field strength. Even if you haven't drawn the radial vectors decreasing in length, and have only drawn the field lines themselves, you have still not thrown out the information about the strength of the electric field. The reason is because for these field lines for a single point charge, which extend radially outward like spokes on a wheel, the lines (spokes) really are denser (more closely-spaced) near the centre of the wheel than they are farther out. Now, what I said is true. The number of lines you draw is arbitrary. However, so long as you choose a consistent way of sampling the infinite number of field lines (i.e you decide that a charge of q will have x number of lines emanating from it, and a charge of 2q will have 2x lines), then the density of field lines is indeed an indicator of the electric field strength.

Based on these notions, Griffiths goes on to state that the flux through a surface can simply be thought of as the number of lines passing through it. From this idea, he immediately concludes that the flux through a closed surface is proportional to the amount of electric charge enclosed by it. His reasoning is entirely intuitive (not mathematical). He points out that there are only two possible fates for the field lines of any positive charge enclosed by the surface: these lines must either pass through the surface, or terminate on another negative charge within the surface. Therefore, the total number of lines passing through the surface (i.e. the flux of the electric field) is proportional to the net charge within that surface. From this argument, two conclusions also immediately follow:

1. Charges outside the surface do NOT contribute to the NET flux, because their field lines will pass into the surface through one side, and out of it through the other side. Hence it is only the charge *enclosed* that matters.

2. The shape and size of the closed surface doesn't matter, because any closed surface will capture exactly the same number of field lines as any other closed surface. (This is supported by the math, for although the strength of the electric field and resulting density of lines falls off as 1/r², the surface area available to catch them increases as r², and the two dependencies cancel out).

I think that these ideas might help you out in understanding some questions in another thread you posted here in General Physics. Griffiths does not go through a formal mathematical derivation of Gauss' Law until after he has outlined these intuitive ideas. I sincerely hope that this helps you in your understanding!

diazona: Sorry for my gender uncertainty. As you can see, I've been around PF for quite some time, but I wouldn't say I really know too many of the members or interact with them too much. I definitely don't know too much about people's backgrounds or personal info. Also, I am just a lowly grad student, so if I say anything that is wrong, be sure to step in and put me in my place.

 

[diazona]

diazona: Sorry for my gender uncertainty. As you can see, I've been around PF for quite some time, but I wouldn't say I really know too many of the members or interact with them too much. I definitely don't know too much about people's backgrounds or personal info. Also, I am just a lowly grad student, so if I say anything that is wrong, be sure to step in and put me in my place.

No apology necessary I certainly didn't take any offense at your uncertainty, I just figured it couldn't hurt to clarify. And actually I'm just a lowly grad student myself! (Sorry if I suggested otherwise... I'm either embarrassed or flattered ) So it's certainly not my place to put you in your place (though if I saw you posting something that seemed egregiously wrong, I wouldn't hesitate to bring it up).

I do agree with you about Griffiths, by the way. I also used it as the primary textbook for my 3rd-year E&M course and I thought it was great as an introductory textbook, at least compared to the (admittedly few) other E&M books I've looked at in any detail.

rockyshephear, if you'd like a recommendation for a book at a somewhat less technical level, a standard is http://he-cda.wiley.com/WileyCDA/HigherEdTitle/productCd-0471758019.html by Halliday, Resnick, and Walker (unfortunately Griffiths hasn't written a book at this level ) Part 3 covers electromagnetism, including electric flux and Gauss's law. I think that if you're not already familiar with those concepts, the HRW book may be more useful to you than Griffiths right now. Just a suggestion, of course.