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What does divergence of electric field = 0 mean?

  1. Apr 3, 2016 #1
    1. The problem statement, all variables and given/known data
    2hekqhk.png
    2d9sdg3.png
    I just want to focus on the divergence outside the cylinder (r >R)

    2. Relevant equations
    20l1e6b.png

    3. 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
    2ptrrkz.png

    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?
     
  2. jcsd
  3. Apr 3, 2016 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    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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