Confusion with the divergence of E fields

In summary, the conversation discusses how changing the direction of the electric field from ##\mathbf{E} = 3x\mathbf{i} + 3y\mathbf{j}## to ##\mathbf{E} = 3x\mathbf{i} - 3y\mathbf{j}## results in a change in the charge density from ##\rho = 6\epsilon_0## to zero. This is because the two fields, although geometrically similar, have different directions. The conversation also explores the concept of charge density in a 3D world, particularly in relation to a field of the form ##\mathbf{E} = 3x\hat{\mathbf{i
  • #1
maNoFchangE
116
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?
 
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  • #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}} ##?
 
  • #3
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Do these look "geometrically similar" to you?
 
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Likes Dale and maNoFchangE
  • #4
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.
 
  • #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.
 

FAQ: Confusion with the divergence of E fields

1. What is the divergence of an electric field?

The divergence of an electric field is a measure of how much the field is spreading out or converging at a particular point. It is represented by the symbol ∇ · E and is a vector quantity.

2. How is the divergence of an electric field related to charge?

The divergence of an electric field is directly proportional to the charge density at a point. This means that if there is a higher concentration of charge at a particular point, the divergence of the electric field will also be higher.

3. Why is confusion with the divergence of E fields common?

Confusion with the divergence of electric fields is common because it is a concept that is closely related to other fundamental concepts in electromagnetism, such as electric charge and electric flux. Additionally, the mathematical notation used to represent the divergence (∇ · E) can be confusing for those who are not familiar with vector calculus.

4. How does the divergence of an electric field affect the behavior of the field?

The divergence of an electric field determines the direction and strength of the electric field lines at a particular point. A positive divergence indicates that the field is spreading out from that point, while a negative divergence indicates that the field is converging towards that point.

5. What are some practical applications of understanding the divergence of E fields?

Understanding the divergence of electric fields is important in many practical applications, such as designing electrical circuits and predicting the behavior of charged particles in electromagnetic fields. It is also essential in the study of electromagnetism and its applications in various technologies, such as electric motors and generators.

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