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Confusion with the divergence of E fields

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  1. Dec 28, 2015 #1
    Suppose I have electric field of the form ##\mathbf{E} = 3x\mathbf{i} + 3y\mathbf{j}##. Calculating the charge density gives me ##\rho = \epsilon_0 \nabla\cdot\mathbf{E} = 6\epsilon_0##.
    But now if I turn one of the components of the field in the opposite direction, for example ##\mathbf{E} = 3x\mathbf{i} - 3y\mathbf{j}##, then the charge density vanishes. I am confused with this because the only difference between the first and the second fields is just the direction, geometrically they are similar. Where do I go wrong?
     
  2. jcsd
  3. Dec 28, 2015 #2
    You are on a 2D world. What must be ##\rho## on 3D world to make a field like ##\mathbf{E} = 3x\hat{\mathbf{i}}+3y\hat{\mathbf{j}} + 0\hat{\mathbf{k}} ##?
     
  4. Dec 28, 2015 #3
    Untitled-1.png Untitled-2.png
    Do these look "geometrically similar" to you?
     
  5. Dec 28, 2015 #4
    The divergence of such a field is ##6\epsilon##, so this kind of charge distribution may generate that field.
    @Fightfish ah I see so they are actually quite different.
     
  6. Dec 28, 2015 #5
    A constant surface charge density ##\rho## make a field ##\mathbf{E} = c\mathbf{k}##, for infinity surface. Flux by point sources relates by ##1/r^2## for 3D, by ##1/r## for 2D and are constants for 1D. Just trust your calculation.
     
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