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Homework Help: What is the physical significance of an electric field divergance of zero?

  1. Jan 10, 2010 #1
    This is not a homework problem, I'm going over some notes and seem to have a discrepancy. In a local region with no charge, del.E=0. Since the divergence is zero, there are no sources and sinks (ie charges).

    In my notes I have;
    "In cartesian co-ords, a solution of del.E=0 is a uniform Electric field."

    Yet later in the notes I seem to have;
    "del.E=0 does not imply a uniform electric field".

    I'm assuming I've misheard something, but can't find anything in any of my textbooks on it. I can't see what difference changing the co-ordinate system should have on del.E.

    It is that since del.E=O suggests no charges, the field CAN be uniform but doesn't have to be (and del.E at a small point in a uniform field must be zero since on a small scale the net flux in and out of an infintesimally small volume are identical)? But on an infintesimaly small volume any E field through the volume will be uniform because on such a scale, all of the field lines will point in the same direction and hence on a small scale any del.E=0 result will be a uniform field since it contains no charge?

    Thanks
     
  2. jcsd
  3. Jan 10, 2010 #2
    del.E = 0

    implies there is no charge inside the closed curve you are operating on.
    aber there could be field inside the curve due to the charges present outside it.
    the field doesnt necessarily have to be a uniform field.
     
  4. Jan 10, 2010 #3

    diazona

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    That's it.

    But... if [itex]\vec{\nabla}\cdot\vec{E}[/itex] is zero everywhere, at all points in space, all the way out to infinity, then you could conclude that there are no charges anywhere, and that the electric field is uniform (zero, in fact, given the usual boundary conditions).
     
  5. Jan 10, 2010 #4

    vela

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    Yes.

    No. Think of applying the integral formulation of Gauss's law to an infinitesimal cube of size dx by dy by dz. Looking at the x-component of E, you can calculate the flux through two sides of this cube:

    [tex]d\Phi_x = E_x(x+dx,y,z)dydz - E_x(x,y,z)_xdydz = \frac{\partial E_x}{\partial x}dxdydz[/tex]

    Note that this takes into account that E may vary with x. Doing the same thing for the y and z directions, you find the total flux through the cube is

    [tex]d\Phi = (\frac{\partial E_x}{\partial x}+\frac{\partial E_y}{\partial y}+\frac{\partial E_z}{\partial z})dxdydz[/tex]

    If there is no charge in the volume, you get [itex]\nabla\cdot \vec{E}=0[/itex] even though the field is not necessarily uniform.
     
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