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Electric flux density a direct measuer of E?

  1. May 10, 2009 #1
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    This is what a book says -

    “The number of lines per unit area through a surface perpendicular to the lines is
    proportional to the magnitude of the electric field in that region. Thus, the field
    lines are close together where the electric field is strong and far apart where the
    field is weak.”
    +
    “These properties are illustrated in Figure 23.20. The density of lines through
    surface A is greater than the density of lines through surface B. Therefore, the magni-
    tude of the electric field is larger on surface A than on surface B. Furthermore, the fact
    that the lines at different locations point in different directions indicates that the field
    is nonuniform.”

    What I concluded from the above -

    That means the total intensity of E.F passing though an area is independent of the number of lines of force, depends more its density and is directly proportional to it.

    So even if we are comparing 2 areas having areas A and 2A and suppose x lines of forces passes through them so the total intensity of the field at A should be greater than 2A, if the areas are charged partially, then the force on A will be more than 2A despite the fact that the number of lines of forces passing through both of them are the same the E.F on A will be higher cause the density of the lines are A is higher.
    Talking about insane analogies, Suppose we have 2 area A and 500000*A, and if the density of the lines in A is more than 500000*A, the force on A will be more than on 500000*A. Suppose the the flux density on A is 1000 lines of forces/m2 and on 500000*A is 999 lines of forces/m2........then also by what the book says, A will experience more force/field.

    Considering the above, suppose we have 2 areas -

    http://img223.imageshack.us/img223/5449/2areas.jpg [Broken]

    Then even if the small area is one trillionth of the larger area, it will experience more force simply cause the flux density is more on it.

    Edit:Consider the charge on both the areas as the same.

    Well...sounds insane to me.

    And what do you mean when one line of force passes through an area?...I mean the field at this state should have a minimum possible value, but still needs to decrease with an inverse square relation to distance...what will happen then?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. May 10, 2009 #2

    dx

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    The fact that the electric field on A is greater doesn't mean the force on A is greater. Assuming E points in the same direction everywhere on the plate, the force on A will be equal to σAE where σ is the surface charge density.
     
  4. May 10, 2009 #3
    Ok...I've changed the question a bit, I've included the fact that the total charge on each of the areas are the same.
     
  5. May 10, 2009 #4

    dx

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    Let's say the total charge on each plate is Q, distributed uniformly.

    If the area of plate one is A, then the charge density on it will be σ1 = Q/A. If the area of plate two is 2A, then the charge density on it will be σ2 = Q/(2A).

    If E is the same on both plates, then the force on both plates will be the same because EAQ/A = E⋅2A⋅Q/2A = EQ. If E is different on the two plates, say E1 and E2, then the forces will be E1Q and E2Q respectively.
     
  6. May 10, 2009 #5
    Ok, I thought about it -

    Suppose we have 2 areas...A and xA...where x is a positive integer greater than 1.
    On xA and flux density of y falls, while on A a flux density of ay falls...where a is a positive integer greater than 1.

    The net charge on both these areas is the same...q.

    What I mean to say here, is that suppose we have 2 areas one larger, one smaller, the flux density on A is greater than xA and the charges on both the areas is the same.
    So according to the fact that flux density is a direct measure of strength of the field, what will the force on A and xA?

    The charge (and so force) on both these areas will be by virtue of certain charged particles uniformly distributed on both these area, since the amount of charged particles on both these areas is the same, the final thing that will matter is the flux density, since that will define force falling on each particle...and the net force will be the summation of all these forces, which's the same for both the areas.

    We can replace the area with these charges and get the same effect, force is directly proportional to the flux density.
     
  7. May 10, 2009 #6
    Thanks for telling me about that charge on the areas.
     
  8. May 10, 2009 #7

    dx

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    I already answered this question. If E on plate one is E1 and E on plate two is E2, then the forces are QE1 and QE2.
     
  9. May 10, 2009 #8
    Yes, I didn't see that, I was making that...but you posted it before.

    Thanks!
     
  10. May 11, 2009 #9
    Is this true for magnetic fields also?
     
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