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Electrostatic field at the square center

  1. Mar 10, 2015 #1
    I have just begun studying electrostatic and I'm trying to do this exercize:

    We have a square with charges +q , -2q, +2q, -q


    1)Compute the electrostatic field [tex]\vec{E}[/tex]at the center of the square.

    I did this way :

    I find [tex]\vec{E_A}=\frac{q}{2 \pi \epsilon_0} \vec{u}[/tex]
    [tex]{E_B}=\frac{-q}{ \pi \epsilon_0} \vec{u}[/tex]
    [tex]{E_C}=\frac{q}{ \pi \epsilon_0} \vec{u}[/tex]
    [tex]{E_D}=\frac{-q}{2 \pi \epsilon_0} \vec{u}[/tex]

    Then with projection :

    [tex]E_A=\frac{q}{2 \pi \epsilon_0}*cos(45)=\frac{\sqrt{2}q}{4\pi \epsilon_0}[/tex]

    [tex]E_B=\frac{-q}{ \pi \epsilon_0}*cos(45)=\frac{-\sqrt{2}q}{2\pi \epsilon_0}[/tex]

    [tex]E_C=\frac{q}{2 \pi \epsilon_0}*sin(-45)=\frac{-\sqrt{2}q}{2\pi \epsilon_0}[/tex]

    [tex]E_D=\frac{-q}{2 \pi \epsilon_0}*sin(45)=\frac{-\sqrt{2}q}{4\pi \epsilon_0}[/tex]

    Finally [tex]E_{total}=\frac{-\sqrt{2}q}{\pi \epsilon_0}[/tex]

    Is-it correct ? I'm not sure of my way of reasoning and the projection.

    Thank you
     
  2. jcsd
  3. Mar 10, 2015 #2

    rude man

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    You need to give the length of one of the sides; also the position of the four charges.
     
  4. Mar 10, 2015 #3
    Yes sorry the lenght of each side is 1 and this is a square with A (upper left), B(upper right) C(lower right) D(lower left) with respectively charges +q,-2q,+2q,-q
     
  5. Mar 10, 2015 #4

    rude man

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    What's the formula for the E field a distance d from a point source q?
     
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