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!

Charge induction

  1. Jun 14, 2012 #1
    1. The problem statement, all variables and given/known data
    This is no homework question, i am only trying to clear my concepts.
    Lets say i have a conducting rod placed vertically. I give it a charge Q. The charge would stay on the outer surface. The charge will distribute equally over the surface. What would happen if i place the charged rod in an electric field of magnitude E?
    (see attachment)

    2. Relevant equations



    3. The attempt at a solution
    I think the negative charge will get concentrated on the left side of rod and the positive charge on the right side. But is it possible to find the amount of charge which will shift to the left and right faces or rather, the final charges on both the faces?

    Any help is appreciated! :smile:
     

    Attached Files:

    • ci.jpg
      ci.jpg
      File size:
      17.1 KB
      Views:
      54
  2. jcsd
  3. Jun 14, 2012 #2
    There is no 'net negative' charge. The only excess charge is the +Q you put on the conductor. Now this conductor will always have a zero field inside it. So in presence of an electric field, the charges in the conductor(including the charge that it naturally possesses) will be rearranged in whatever way, so that net field inside remains zero.

    I'm not sure how you can calculate the final charges on both faces, as charges of the native conductor also participate in the distribution.
     
  4. Jun 14, 2012 #3

    I like Serena

    User Avatar
    Homework Helper

    Hi Pranav! :smile:

    You make it a bit difficult when you use a rod...

    But suppose you use a conducting plate that is perpendicular to your electric field E, with some surface area A.

    Then inside the plate the electric field generated by the charge in the plate will neutralize your electric field E.
    Suppose ΔQ is the difference in charge on the left and right sides of the plate.
    Then the electric field inside is ΔQ/A which will be equal to E.
    So ΔQ=EA.
     
  5. Jun 14, 2012 #4
    Hi ILS!

    This is where I was unsure of exact distribution. But still, for understanding, can you please describe how the distribution can be calculated if it were a rod? A brief idea will do :smile:
     
  6. Jun 14, 2012 #5
    Thank you both for the replies! :smile:

    @ILS: I think the electric field inside the plate should be ΔQ/Aεo, you seem to have miss the εo.
     
  7. Jun 14, 2012 #6

    ehild

    User Avatar
    Homework Helper
    Gold Member

    It is not an easy problem. Putting a conducting body in an electric field, it redistributes the charge on the conductor, and the charge distribution on the body - either having a net charge or not - influences the field near itself. The external electric field produces such surface charge distribution that the surface of the conductor becomes equipotential. This also means that the lines of the resultant field are perpendicular to the surface. And the relation of the surface charge density σ and the resultant electric field Er at a place is σ = ε0Er
    And of course, the electric field inside the conductor is zero.

    ehild
     
  8. Jun 15, 2012 #7

    I like Serena

    User Avatar
    Homework Helper

    Yep. Well spotted! :)


    Gee! I don't really know.
    Usually we work from objects that are nicely symmetrical in some way.

    The rod should have a surface distribution that would only depend on the angle. And the summation of the electric fields of all infinitesimal charges should come out as expected.
    But that's not easy to work through.
     
  9. Jun 15, 2012 #8

    I was also thinking along similar lines. Assuming the field perpendicular to the surface of the rod, I believe it would vary linearly with the thickness of the rod... I have no mathematical proof for this, though...
     
  10. Jun 15, 2012 #9

    ehild

    User Avatar
    Homework Helper
    Gold Member

    I think it would be interesting to solve the simpler problem first: A plate of large area A and small thickness d, with Q excess charge, put into a homogeneous electric field E, perpendicular to the plate. What is the surface charge density on both sides of the plate?

    ehild
     
  11. Jun 15, 2012 #10
    I am not sure about it but i guess we can solve it by the hints provided by ILS.
     
  12. Jun 15, 2012 #11

    I like Serena

    User Avatar
    Homework Helper

    Since you're not sure, care to give it a try? ;)
     
  13. Jun 15, 2012 #12
    Since we have kept it in a uniform field E, an equivalent but opposite in direction, a field gets induced in the plate. As ILS said, the field inside inside is equal to ΔQ/Aεo, therefore ΔQ=AEεo. I am not sure how this charge will appear on the surface, i mean i know the charge will go to outer surface but how will this affect the outer surface charge distribution, what will be the final charge on outer surface and what's the use of the small thickness "d" that ehild has provided?
     
  14. Jun 15, 2012 #13

    ehild

    User Avatar
    Homework Helper
    Gold Member

    That d is small means only that the extension of the plate can be considered very big - infinite- with respect to thickness. Otherwise you need to consider the "edge effects". The value of the thickness does not influence the result if it is very small with respect to the lateral sizes.

    When the plate is placed in the electric field, the field appears inside the plate at the first instant. The electric field inside moves the free electrons of the conductor till they reach the surface. Also free electrons will flow from the opposite surface towards the inside of the conductor to make the inside neutral. We say that the applied electric field induces excess surface charge; positive on one side of the plate and negative on the other side. The resultant surface charges cause such an electric field that is opposite to the applied field and cancels it inside the plate.

    You need to find the amount of surface charge which appears on both sides of the plate when an electric field E is applied.

    When the conductor has a net charge Q, it will change the outer field. Find out the resultant field on both sides of the plate.

    ehild
     
    Last edited: Jun 15, 2012
  15. Jun 16, 2012 #14
    Assume the electric field outside also enters the conductor. This would have a field E to the right(inside the conductor), going by your diagram. Now the conductor plate needs a field E to the left, to counterbalance the external field, and hence redistributes charge. Taking a charge Q1 for one side of the plate, equate the field sum to E. :smile:
     
  16. Jun 16, 2012 #15
    The initial charge on both the surfaces were Q/2, when we place the plate in electric field, the final charge will be Q/2-AEεo and Q/2+AEεo, am i right?
     
  17. Jun 16, 2012 #16

    I like Serena

    User Avatar
    Homework Helper

    Close! :)

    What is your resulting ΔQ between the charge on the left and the charge on the right?
     
  18. Jun 16, 2012 #17
    2AEεo?
     
  19. Jun 16, 2012 #18

    ehild

    User Avatar
    Homework Helper
    Gold Member

    It is correct. :smile:

    ehild
     
  20. Jun 16, 2012 #19
    Ah, thanks ehild! But i am still confused on which surface, Q/2-AEεo or Q/2+AEεo will reside? Any clarification on that will help. :smile:
     
  21. Jun 16, 2012 #20

    ehild

    User Avatar
    Homework Helper
    Gold Member

    The charge will reside on both surfaces of the plate. If the applied electric field points to the right, Q/2+AEε0 charge is on the right-hand side surface of the plate and Q/2-AEεo is on the left-hand side.

    ehild
     

    Attached Files:

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Charge induction
  1. Charging by Induction (Replies: 1)

  2. Charging by induction (Replies: 3)

  3. Charging by Induction (Replies: 3)

  4. Induction charge (Replies: 1)

Loading...