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Electric field of infinite plate

  1. Mar 21, 2009 #1
    2 infinite plates are placed one next to another in a V formation with an angle β between them, they have charge densities of σ1 and σ2. what is the magitude of the field betweein them.

    ϕ=[tex]\oint[/tex]EdA=EA
    ϕ=[tex]\frac{q}{ϵ}[/tex]=[tex]\frac{σA}{ϵ}[/tex]

    E1=[tex]\frac{σ1}{ϵ}[/tex]
    E2=[tex]\frac{σ2}{ϵ}[/tex]

    E=E1+E2

    lets make our x axis parallel to plate 1 giving E1 a 0 value on x axis
    Ex= 0 + [tex]\frac{σ2}{ϵ}[/tex]*sinβ
    Ey= [tex]\frac{σ1}{ϵ}[/tex] + [tex]\frac{σ2}{ϵ}[/tex]*cosβ

    |E|2=|Ex|2 + |Ey|2
    =([tex]\frac{σ2}{ϵ}[/tex]*sinβ)2 +([tex]\frac{σ1}{ϵ}[/tex] + [tex]\frac{σ2}{ϵ}[/tex]*cosβ)2
    =[tex]\frac{1}{ϵ}[/tex]*(σ22sin2β + σ12 + σ22cos2β + 2(σ1)(σ2)cosβ )
    =[tex]\frac{1}{ϵ}[/tex]*(σ22(sin2β + cos2β ) + σ12 + 2(σ1)(σ2)cosβ)

    |E|2=[tex]\frac{1}{ϵ}[/tex]*(σ12 + σ22 + 2(σ1)(σ2)cosβ)

    which is almost right except that in my answers there is a second possible solution that

    |E|2=[tex]\frac{1}{ϵ}[/tex]*(σ12 + σ22 - 2(σ1)(σ2)cosβ)

    where does this minus come from? originally i thought it might be because of a possible negative charge density but surely if that was so the sign would be part of σ and come automatically when i plug in values
     
  2. jcsd
  3. Mar 21, 2009 #2
    i see the latex codes came all messed up, meant to read:

    Ex= 0 + (σ2/ɛ)*sinβ
    Ey=(σ1/ɛ) + (σ2/ɛ)*cosβ

    |E|^2=|Ex|^2 + |Ey|^2

    and at the end after calculations come to

    |E|^2=(1/ɛ)*(σ1^2 + σ2^2 + 2(σ1)(σ2)cosβ)

    where there is meant to be a second option of

    |E|^2=(1/ɛ)*(σ1^2 + σ2^2 + 2(σ1)(σ2)cosβ)

    and i cant see where from. originally i thought it might be because of a possible negative charge density but surely if that was so the sign would be part of σ and come automatically when i plug in values
     
  4. Mar 21, 2009 #3

    LowlyPion

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    Depending on the angle cosβ has a range of +1 to -1 doesn't it?
     
  5. Mar 22, 2009 #4
    correct, so does that mean that if 90<β<270 then i will use the minus option? could i then also say |E|^2=(1/ɛ)*(σ1^2 + σ2^2 + 2(σ1)(σ2)|cosβ|)
     
  6. Mar 22, 2009 #5

    LowlyPion

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    I think all it means is that there are 2 solutions depending on the angle of β.

    Your representation above ignores the second solution where cosβ is (-).
     
  7. Mar 22, 2009 #6
    what are the 2 options? is it

    |E|^2=(1/ɛ)*(σ1^2 + σ2^2 + 2(σ1)(σ2)cosβ) when (cosβ>1)

    and

    |E|^2=(1/ɛ)*(σ1^2 + σ2^2 - 2(σ1)(σ2)cosβ) when (cosβ<1)

    ????????????????????

    if so than can i not just say that it is
    |E|^2=(1/ɛ)*(σ1^2 + σ2^2 + 2(σ1)(σ2)|cosβ|)

    and then cos will always be positive, in effect that is what i am doing anyway, taking minus the value of the negative cos
     
  8. Mar 22, 2009 #7
    looking back at the original question and the one i posted, i see that in the original the 2 plates formed an X like shape whereas i wrote in my post a V shape, originally i never thought anything of it but now looking back i think that the (- cosB) is for the area between the top and bottom points of the X on either side whereas the ( + cosB) is for the area between the top 2 points or bottom 2 points, (when B is an acute angle,) or alternatively the opposite(when B is obtuse).
    is this a correct presumption?
     
  9. Mar 22, 2009 #8

    LowlyPion

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    An X configuration does simultaneously create 2 regions (4 actually, but 2 sets of 2).
     
  10. Mar 23, 2009 #9
    yes but is one the -cos option and the other the +cod option?
     
  11. Mar 23, 2009 #10

    LowlyPion

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    One governs acute angles and the other obtuse doesn't it?
     
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