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Electric Flux dimensions

  1. Jan 12, 2005 #1
    In a singular dimension case, can you say that the Flux is equal to the charge enclosed, or does it have to be the charge enclosed divided by the permitivity.

    My lecturers notes claim that Flux = Q {in the electrostatic case} in one dimension.

    Is that true?

    EDIT sorry its not singular dimension, i bite my lip..act of idiot..but still, my questions below i still feel valid..please read on :smile:
    Last edited: Jan 12, 2005
  2. jcsd
  3. Jan 12, 2005 #2
    The flux of a vector, is defined by Gauss's Theorem, that says:

    [tex] \int_{V} \vec{\nabla} \cdot \vec{v} dV = \int_{S} \vec{v} \cdot \hat{n} dS[/tex]

    The first law of Maxwell says that [tex]\vec{\nabla} \cdot \vec{E} = \frac{\rho}{\epsilon_0}[/tex], then:

    [tex]\int_{V} \frac{\rho}{\epsilon_0} dV = \int_{S} \vec{E} \cdot \hat{n} dS[/tex]

    If there is no charge in the infinity, the integration of the density is the total charge, then:

    [tex]\frac{Q_{total}}{\epsilon_0} = \int_{S} \vec{E} \cdot \hat{n} dS = \Phi [/tex]
    Last edited: Jan 12, 2005
  4. Jan 12, 2005 #3
    Flux of Electric field?
    I never did the 1-D case, but I'm guessing Electric field flux from a point charge Q would be constant in 1-D.
    [tex]\phi = \frac {E}{A}[/tex]
    [tex]E = \frac {Q} {4 \pi \epsilon_0 r^2}[/tex]

    Edit: The above Maxwell equation's usage seems more relevant.
  5. Jan 12, 2005 #4
    In one dimension? How you can talk about flux in one dimension?
  6. Jan 12, 2005 #5
    That's what he's asking for.
    Puzzled me too.
  7. Jan 12, 2005 #6
    Yeah im pretty confused too... I've done alot of calculus and and im still a bit 'oo-er' about this one..

    I have the EXACT problem here...and i'll upload a diagram now. I always thought that the sum of the fluxes through a surface was equal to the charge enclosed divided by the permittivity.. But here we go anyway, maybe im thinking of the wrong thing although this is supposed to be an example of gauss's law :grumpy:


    there's the statement, let me know what you think
    Last edited: Jan 12, 2005
  8. Jan 12, 2005 #7
    Is the correct way to do this {to work out the potential difference between R2 and R1} (assuming that its not correct as it stands) is to say :

    [tex] E(r) = \frac{Q}{4\pi\epsilon r^2} = \frac{dV}{dr} [/tex]
    [tex] v = \frac{Q}{4\pi\epsilon}\int_{r_1}^{r_2} \frac{1}{r^2} dr = \frac{Q}{4\pi\epsilon}\left [-\frac{1}{r}\right]_{r_1}^{r_2} [/tex]
  9. Jan 12, 2005 #8
    onnnnn, this is not a one D case... the graph is the cross section of the capacitor...... you shouldn't use the point charge formulas in this case.... remember tis is a infinte cylinder..... the jpg file is trying to use guess law to derive the formulas...
  10. Jan 12, 2005 #9
    Its not meant to be an infinite cylinder, its a cylinder of length 1m. Once you know the PD then you can use the equation that [itex] C = \frac{\epsilon_{0}A}{d}[/itex] and use Q = CV should you wish to find the charge on the surface surely... I can see that he's trying to use gauss's law to get those equations, but just from looking at the equation for E, it doesnt have the same dimensions as [tex] E = \frac{Q}{4\pi\epsilon r^2} [/tex]...Am i missing something here?!?! Can you just say Flux = Q?!! I've never heard of that concept before. I understand Gauss's law, but where did this come from?!
  11. Jan 12, 2005 #10
    Well I'll be damned..he's right!

    There's an equation :

    [tex] \int_{s} \mathbf{D} \cdot d\mathbf{S} \equiv \int_{v} \rho_{f} d\tau[/tex] and in words this means "The flux D out of a closed surface S = Total free charge enclosed"..which is "electric displacement", and is used for calculations in di-electrics.

    If you follow through the motions it comes to the correct outcome.

    Man I suck...

    Laters, chaps
  12. Jan 13, 2005 #11
    It seems to me he is asking if it is acceptable to set epsilon zero = to one.

    This is called the gaussian system of units, and is convenient for a purely mathematical discussion. Epsilon zero is a product of the metric system.
  13. Jan 14, 2005 #12
    I'm aware of the Gaussian system, but upon reading into it, there's an "electric displacement vector" that is related to the flux and the polarisation of the dielectric. I didn't catch all of it but the equation i quoted in my previous post summed it up.. The permittivity is encorporated somewhere in that equation, but I'm taking it on face value at the moment til I have time to review it.

    Hmm :rofl:

    Peace out!
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