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Electric field.

  1. Aug 6, 2013 #1
    What will be the electric field at the points p3 and p4 in the attached image.

    I think at p3 it will be zero as it is inside the conducting surface
    But what about p4??
    Will it be zero too? Due to negative and positive charges or will it be sigma/E(sub)o(/sub).. And why? ImageUploadedByTapatalk1375775117.593771.jpg
     
  2. jcsd
  3. Aug 6, 2013 #2
    Well as I believe p3 is inside a conductor which looks more like a capacitor well anyway the field strength is above zero only outside of a conducting or charged surface because that's where all the particles with similar charges are located...

    But maybe you just had to do a better forums search...

    https://www.physicsforums.com/showthread.php?t=109422
     
  4. Aug 6, 2013 #3
    I do not understand..
     
  5. Aug 6, 2013 #4
    Okay let me go like this , there are electrons (one out of the list of fundamental elementary particles)
    Electrons have charge , which is negative , now as you have probably heard charges with the same polarity tend to repel from one another.
    Now a conductor or any metal into which current flows or which is just set at a given potential even without current flow has these electrons aligned so that they go and reside themselves on the outer part of that metal or wire or plate as in a capacitor.

    That's why the field has a value only outside of the conductor and at the surface of it because that's where " all the fun is" or all the charge.

    as in your case it looks like a capacitor as there are two separate but close conductors with opposite charges on them and opposite charges tend to get together so there is no other way but for them to be located nearest one another which happens to be the boundary of the conductor because that's where they are located.
    It's like with prisoners or inmates whenever there is something interesting going on they stick to the doors and bars and watch , now they would like to go further but they can't because they are restricted.
     
  6. Aug 6, 2013 #5
    Thank you very much..
     
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