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Gauss Law and Electric Field of a Dipole

  1. Jan 26, 2013 #1
    Hi,

    I don't understand this:
    You got a dipole, and a resulting electric field from it like this(hyperphysics):
    dipo2.gif
    As z approaches infinity, the field becomes zero.

    But if you draw a gaussian surface round the charges, then the net enclosed charge is zero.So the electric field must be zero, no matter how I draw the surface.
    So If i just draw a surface tiny enough to enclose both charges, my z is totally not going to infinity and the field is zero according to Gauss' LAw.

    How is this compatible with each other?What is the pitfall in my thinking.

    Thanks in advance.
     
  2. jcsd
  3. Jan 26, 2013 #2

    jtbell

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    Staff: Mentor

    No, the integral of the electric field over the entire surface (the flux of the electric field) must be zero.

    On some parts of the surface, the field points outwards, which contributes a positive amount to the flux. On other parts of the surface, the field points inwards, which contributes a negative amount to the flux. The positive and negative contributions add up to zero over the entire surface.
     
  4. Jan 26, 2013 #3

    vanhees71

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    2016 Award

    Of course, you have to use the full electric field due to your dipole and an electric field is a vector. By construction any partial wave corresponds to a charge distribution with total net charge 0 except the monopole term.

    The dipol field is given by the potential
    [tex]\Phi(\vec{x})=\frac{\vec{p} \cdot \vec{x}}{4 \pi r^3} \quad \text{with} \quad r=|\vec{x}|.[/tex]
    The field is thus given by
    [tex]\vec{E}=-\vec{\nabla} \Phi=\frac{1}{4 \pi r^5} [3(\vec{p} \cdot \vec{x}) \vec{x}-r^2 \vec{p}].[/tex]
    Here, the dipol sits at the origin of the coordinate system. The most simple surface to use is a sphere of radius [itex]a[/itex] around the origin. We use spherical coordinates with the polar axis in direction of the electric moment, [itex]\vec{p}[/itex]. Then we have
    [tex]\mathrm{d}^2 \vec{F}=\mathrm{d} \vartheta \mathrm{d} \varphi a^2 \sin \vartheta \vec{e}_r[/tex]
    and thus
    [tex]\int_{K_a} \mathrm{d}^2 \vec{F} \cdot \vec{E}=\int_0^{\pi} \mathrm{d} \vartheta \int_0^{2 \pi} \mathrm{d} \varphi \; \frac{p \sin \vartheta \cos \vartheta}{2 \pi a} =0.[/tex]
     
  5. Jan 27, 2013 #4

    stevendaryl

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    As others have pointed out, Gauss' Law tells us that if a surface encloses total charge 0, then the AVERAGE of the field over the surface is zero. That doesn't mean that it is zero at every point. In the case of a dipole, the part of surface nearest to the + charge will have a positive flux through it, and the part of the surface nearest to the - charge will have a negative flux through it.

    Gauss' Law doesn't really help much in figuring out what the electric field is for a distribution of charges unless there is some kind of symmetry involved. So for instance, a flat uniform sheet of charge, a line of charge, a spherical shell of charge.
     
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