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Let's see if I have it now.

  1. Jul 10, 2009 #1
    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
    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.
  2. jcsd
  3. Jul 10, 2009 #2


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    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).

    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.
  4. Jul 10, 2009 #3
    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
  5. Jul 10, 2009 #4


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

    [tex] \nabla \cdot \mathbf{E} = \frac{\rho}{\epsilon_0} [/tex] ​

    In integral form, Gauss' Law is:

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

    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].

    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 ΦE 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 licence 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.
    Last edited: Jul 11, 2009
  6. Jul 10, 2009 #5
    I've got a web page that specifically states
    PHI (electric)=Q/permittivity. PHI must be electric flux. This is right, right?
  7. Jul 10, 2009 #6


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    Yep. It's defined as
    [tex]\Phi = \int \vec{E}\cdot\mathrm{d}\vec{A}[/tex]
  8. Jul 10, 2009 #7
    Oh so the integral of the E (electric field) dot producted with dA (ever shrinking area on the surface in question =Q/permittivity??
  9. Jul 10, 2009 #8


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    Yeah, that's just what cepheid wrote - the integral form of Gauss's law.
  10. Jul 10, 2009 #9
    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?
  11. Jul 11, 2009 #10


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    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.

    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:

    [tex] \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 [/tex] ​

    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).
  12. Jul 11, 2009 #11
    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?
  13. Jul 11, 2009 #12


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    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.
  14. Jul 11, 2009 #13
    And if only ONE is a vector what do you have?
  15. Jul 11, 2009 #14
    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
  16. Jul 11, 2009 #15
    Oh, and I forgot to throw in the the electric field is the sum of all indians shooting thier arrows over and over and over again.
  17. Jul 11, 2009 #16


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    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.
  18. Jul 11, 2009 #17


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    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:

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

    [tex]B_x = kA_x[/tex]

    [tex]B_y = kA_y[/tex]

    [tex]B_z = kA_z[/tex]

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

    Hmmm, the vector arrows don't seem to come out as well as they used to...
  19. Jul 11, 2009 #18
    Now we are getting somewhere with this analogy. Cool. Thanks.
    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?
    I know, I'm laughing, myself.
  20. Jul 11, 2009 #19
    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.
  21. Jul 12, 2009 #20
    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 any one 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.
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