1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Why does a conducting slab double a uniform incident field?

  1. Sep 3, 2016 #1
    For a conducting sheet that is transverse to a constant uniform incident field, I believe the resulting total field would simply be the incident field doubled. I'm trying to figure out why this happens, preferrably avoiding the method of images since I am using this to figure out how the method of images works.

    What I have so far: The incident field will push positive charges in the sheet in the same direction that the incident field is pointing, and pull the sheet's negative charges opposite to the direction the incident field is pointing. It's clear from this that the sheet's presense will strengthen the field, but how do you know how much it strengthens the field? It seems to me that as long as the field due to the conducting sheet is transverse to the surface of the sheet, it is consistent with gauss's law, regardless of the magnitude.
  2. jcsd
  3. Sep 4, 2016 #2


    User Avatar
    Homework Helper
    Gold Member

    It's exactly the opposite.
    ...which makes the net field inside the conducting sheet zero . The field due to rearrangement of the charges on the sheet (called 'induced charges') is in direct opposition with the incident field. Hence, electric field inside a conductor is always zero in electrostatic equilibrium.
  4. Sep 4, 2016 #3
    Sorry I meant the field outside of the conductor. I will draw a picture for clarity.


    The yellow dots are negative charges and the purple dots are positive charges, and the field inside the conductor is zero. For an incident electric field pointing downward, how does the horizontal sheet affect the field outside of the conductor?
  5. Sep 5, 2016 #4
    It does not. The charge density σ on the surface of the conductor adjusts itself so that it is given by σ = E/ε0, where E is the electric field outside and ε0 is the usual constant.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted