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Electric Field/Charge

  1. Apr 24, 2009 #1

    Air

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    Hello, I will post questions on this thread on Electric Field/Charge and other topics related to it. I'm studying it and am very weak at it thus any help will be appreciated.

    For now, here's one question.
    • I've come across two equations for E which are E = (Kq)/(r) and E = (Kq)/(r^2). I don't understand why it is r and sometimes (r^2). Are they specific for some cases? Also, which is used more often?

    Thanks in advance.
     
    Last edited: Apr 24, 2009
  2. jcsd
  3. Apr 24, 2009 #2
    Well the first equation you have stated is for Potential, and the second equation is electric field strength for a radial field.
     
  4. Apr 24, 2009 #3

    Doc Al

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    Staff: Mentor

    The second equation, E = (Kq)/(r^2), gives the field from a point charge. The first equation, E = (Kq)/(r), is not dimensionally correct, so I suspect you are mistaking it for another equation. (Perhaps the potential from a point charge: U = (Kq)/(r).)

    [RoryP beat me to it. :smile:]
     
  5. Apr 24, 2009 #4

    Air

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    What is the potential used for? When is it used in context?

    I know that it's related to Work by W=Ua - Ub.Work is the energy required to move a charge.
    __________________
    Also, for coulombs law, there is a unit vector (r) in the equation. Isn't unit vector value always one?

    (Thanks for the help so far :biggrin:)
     
  6. Apr 24, 2009 #5
    Hmm not entirely sure, i know the units for Potential are Volts. And i also know that if you integrate field strength dr between the limits infinity and current position you end up with Potential. =]
    Convieniently the laws for universal gravitation run in parallel to the ones for electric fields, so if you have knowledge of the gravitational ones then it could be helpful =]
     
    Last edited: Apr 24, 2009
  7. Apr 24, 2009 #6

    Air

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    I only know the gravitational theory briefly but I understood what you have said. :smile:

    __________________
    Another question:

    I understand that for a charged conducting sphere, the charge is built up on the surface thus there is no charge enclosed inside (if we are considering Gauss surface inside). When they talk about charged surface sphere, what is the structure like? Is is hollow or solid and is there no charge enclosed for both situations?
     
  8. Apr 24, 2009 #7

    Doc Al

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    Read about electric potential and potential energy here: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/volcon.html#c1"

    The magnitude of a unit vector is always one, but its direction varies. A unit vector is used to specify the direction of the force (or other vector).
     
    Last edited by a moderator: Apr 24, 2017
  9. Apr 24, 2009 #8

    Doc Al

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    A conducting sphere can be solid or hollow. For the electrostatic case, there is never any net charge within the material of the conductor, only on its surface.
     
  10. Apr 25, 2009 #9

    Air

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    Thanks Doc_Al. Amazing help as always. :smile:

    __________________
    Another question:

    Isn't the formula for Gauss law (When considering angle), EAcosx, so If the Gaussian surface is perpendicular, there would be no field as cos(90)=0. Why is it that when considering infinite sheet of charge, we draw the cyclindrical Gaussian surface, we say that the E is perpendicular to the sheet?
     
  11. Apr 25, 2009 #10

    Doc Al

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    That's a formula for finding the electric flux through a surface. Realize that the angle is measured from the normal to the surface. So if the field is perpendicular to the surface, the angle it makes to the normal is θ = 0, not 90.
    E is perpendicular to the sheet, and thus makes an angle of θ = 0 with the normal to the surface of the flat ends of the Gaussian surfaces. This should make sense, since when the field is perpendicular to the surface the flux is maximum (cos0 = 1).

    Read: http://hyperphysics.phy-astr.gsu.edu/HBASE/electric/elesht.html#c1"
     
    Last edited by a moderator: Apr 24, 2017
  12. Apr 25, 2009 #11

    Air

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    Last edited: Apr 25, 2009
  13. Apr 25, 2009 #12

    Doc Al

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    Good. The two charge distributions are very different. (Yet can often have the same electric field when r > a.)
    If you are asking if a sphere of uniform charge density can have a hollowed out portion--sure, why not?
     
  14. Apr 25, 2009 #13

    Air

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    So you cannot have a shell of uniformly charged density. The electric field inside would be zero for such a case. Is that correct?
     
  15. Apr 25, 2009 #14

    Doc Al

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    Why not? (Did you mean can, not cannot?)
    Yes. If you have a spherically symmetric shell of charge, the field within the hollow part of the shell will be zero.
     
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