1. Not finding help here? Sign up for a free 30min 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!

Help with an electrostatics problem

  1. Oct 18, 2013 #1
    question: A positive charge exists in free space.Now, the charge is enclosed by a spherical shell of finite thickness.What effect does it have on the field lines emanating from the charge.

    answer:The field lines emanate outwards from the charge till they reach the inner surface of the shell after which they continue from the outer surface. The field lines do not exist over the thickness of the shell.

    query:Why should the field lines not exist over the thickness of the shell?
     
  2. jcsd
  3. Oct 18, 2013 #2

    UltrafastPED

    User Avatar
    Science Advisor
    Gold Member

    If the shell is a conducting sphere:
    (1) electrons move within the conductor to the inner surface so as to cancel the enclosed positive charge.
    Every field line has been terminated.
    (2) the electrons that moved to the inner surface left a net positive charge at the outer surface. These will be evenly spaced (because electrons move so as to minimize local charges).
    This creates new field lines (to infinity) that are spherically symmetrical ... even if the original positive charge was not.

    And there are no field lines inside the conducting shell.
     
  4. Oct 18, 2013 #3
    I understand your point. If i draw a Gaussian surface arbitrarily close to the outer surface of the shell ( in between the the two surfaces), it will enclose no net charge hence no flux and no field lines either. However, is it not true that their is a positive charge at the center of the shell and induced negative charge on the inner surface? So shouldn't there be field lines originating from positive charge and terminating on the negative charge just like that between the plates of a parallel plates capacitor which are oppositely charged?
     
  5. Oct 19, 2013 #4

    ehild

    User Avatar
    Homework Helper
    Gold Member

    It was not specified if the shell is a conductor or a dielectrics (insulator).

    In case of a dielectric shell, the electric field is not zero inside the shell, but different from that inside the cavity, as the inner charge polarizes the dielectrics and some of the field lines continue in dipole chains.

    If it is a metal shell, you know that there can not exist static electric field in the metal. If there was some field, it would drive the free electrons till they reach the surface. There is surface charge at both sides of a metal layer, and the net charge on both surfaces are of equal magnitude and opposite sign.

    ehild
     
  6. Oct 19, 2013 #5

    ehild

    User Avatar
    Homework Helper
    Gold Member

    The induced negative charge is equal in magnitude to the positive charge at the centre. The number of field lines emerging from a positive charge or ending in a negative charge q is q/ε. So all field lines emerging from the central charge end at the surface charges on the inner surface on the shell.
    As the shell was neutral, a positive surface charge appear on the outer surface, equal in magnitude with the inner charge.
    In case of a capacitor, there are two plates, and opposite surface charges q and -q on the inner surfaces. There is air or some dielectrics between the plates, and the number of the electric field lines is q/ε.
     
  7. Oct 19, 2013 #6
    You said,"all field lines emerging from the central charge end at the surface charges on the inner surface on the shell". Does that mean the electric field vanishes with the termination of field lines?
     
  8. Oct 19, 2013 #7

    ehild

    User Avatar
    Homework Helper
    Gold Member

    The field lines represent the electric field, and the electric field is proportional with the density of the field lines.
    You can say that the electric field lines terminate at the surface charges, as the electric field is zero inside the metal.

    ehild
     
  9. Oct 19, 2013 #8
    OK i seem to be getting it. But following the same argument can't we conclude that there is no electric field in between the capacitor plates?(since there too the lines terminate at the negative plate). Or is it that the dielectric makes a difference? If so, how exactly? Help will be appreciated..
     
  10. Oct 19, 2013 #9

    ehild

    User Avatar
    Homework Helper
    Gold Member

    In case of a capacitor, the field lines are between the plates. The surface charges are on the inner surfaces. The field lines emerge from the positive charges and terminate on the negative ones. No field lines are present inside the metal plates or outside the capacitor.

    ehild
     

    Attached Files:

    Last edited: Oct 19, 2013
  11. Oct 19, 2013 #10
    OK! thank you.
     
  12. Oct 20, 2013 #11

    ehild

    User Avatar
    Homework Helper
    Gold Member

    You are welcome.

    ehild
     
  13. Nov 30, 2013 #12
    Dear All

    I planned to give electrostatic induction to the abrasive grain (Di electric) to get attracted by the positive charged plate, where i have kept sticky pad, so that the abrasive grains are attracted & deposited.

    The size of abrasive grains are : 60 grit size. ( Kept at Negative Plate)
    The Size of sticky pad is: 1 mtr x 1 mtr. (Kept at Postive Plate)

    Can you please tell me the following:

    1 . Amount of Electrical energy has to be applied in Positive & negative plates.

    2. Nature of Metal which i have to use to design positive & negative plates and its thickness?

    Your guidance will help me a lot
     
  14. Nov 30, 2013 #13

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Hi Kiran, welcome to PF.

    I can not answer your question. And this is a forum for homework problems. Try to send your question to one of the Engineering forums, like Materials and Chemical Engineering.

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

Have something to add?
Draft saved Draft deleted



Similar Discussions: Help with an electrostatics problem
Loading...