Confusion with the divergence of E fields

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
4 replies · 1K views
maNoFchangE
Messages
115
Reaction score
4
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?
 
Physics news on Phys.org
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}} ##?
 
Untitled-1.png
Untitled-2.png

Do these look "geometrically similar" to you?
 
  • Like
Likes   Reactions: Dale and maNoFchangE
theodoros.mihos said:
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}} ##?
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.
 
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.