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E-field greater at sharp edges?

  1. Dec 5, 2013 #1
    Hello everyone,

    Could someone explain or direct me to some detailed explanations as to why is the electric field at a conductor greatest where the curvature is sharpest?

    I just can't seem to grasp this concept. I understand sharp edges do have this property so is this that there is more charge density at those points? And if so, what would prevent another point close by (on a rounded surface) to have the same charge density given that both points are at the same KV potential?

    In another words, how is a sharp edge (high curvature) affecting the local charge distribution at an atomic level?

    Thanks for enlightening me :)

    Regards
     
  2. jcsd
  3. Dec 5, 2013 #2
    The charge density is indeed higher at sharper regions.

    I think this site is quite good at explaining why.
    http://www.physicsclassroom.com/class/estatics/u8l4d.cfm

    Just scroll down to the "Electric Fields and Surface Curvature" section and it'll explain how the sharper regions results in a smaller component of the electron-electron electrostatic repulsive force being directed parallel to the conductor surface.
     
  4. Dec 5, 2013 #3
    Where there is a flat surface, the charge distribution is even because the exposed surface area of the object per unit volume of space surrounding it is constant. Where you have a sharp edge, or a pointed protrusion on an object, there is more surface area exposed per unit volume of space immediately surrounding it - and so a higher charge density in that region of 3D space.

    The link posted by the other poster is very good, and should explain the rest.
     
  5. Dec 7, 2013 #4
    Let's consider this charged object:

    O-------o

    Two metal balls and a thin metal rod between them.

    Let's say the small ball has a radius that is half of the big one's radius. So its capacitance is half of the other one's capacitance. So its charge is half of the big ball's charge.


    I'm lazy, so I leave this question as an exersice for the reader:
    Why is the electric field near the small ball's surface twice as large as the electric field near the large ball's surface?
     
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