What does divergence of electric field = 0 mean?

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SUMMARY

The divergence of the electric field outside a charged cylinder (for r > R) is zero, indicating that there is no net charge present in that region. This conclusion is derived from Gauss' Law, which states that the divergence of the electric field (∇ * E) is equal to the charge density (ρ) divided by the permittivity of free space (ε₀). Since there is no charge outside the cylinder, the divergence must equal zero. Understanding this concept is crucial for grasping the behavior of electric fields in electrostatics.

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  • Knowledge of divergence as a mathematical operation
  • Basic principles of charge distribution in cylindrical coordinates
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Homework Statement


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I just want to focus on the divergence outside the cylinder (r >R)

Homework Equations


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The Attempt at a Solution


For r > R, I said ∇ * E = p/ε

But that's wrong. The answer is ∇ * E = 0

I'm confused because there is definitely an electric field outside the cylinder (r > R). The electric field points radially outwards and gets smaller the farther you get from the cylinder because
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So I don't understand how the divergence of the electric field can be 0. I think the main part of my confusion is that I don't understand what the divergence is. I know how to mathematically compute the divergence but I don't understand it physically. Like when the divergence of the electric field is 0, what does that mean in terms of the physical electric field?
 
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Gauss' law says that the divergence of E evaluated at some point equals the charge density at that same point divided by ##\epsilon_0##.

These videos might help improve your conceptual understanding of divergence:
 

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