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Math Help - Electric fields

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    Electric fields

    Charges in the corner are 10*10^(-9) C.
    What is strenght of electric field at red point.
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    Quote Originally Posted by totalnewbie View Post
    Charges in the corner are 10*10^(-9) C.
    What is strenght of electric field at red point.
    You need to find the electric field at that point due to each charge then add them vectorally.

    Call q = 10 nanoCoulomb and r = 0.05 m
    I am setting up a cooredinate system with +x to the right and +y upward.

    So:
    E(A) = k*q(A)/r^2(A) = k*q/(r/2)^2 = 4kq/r^2
    (If you aren't using "k" just substitute k = 1/(4*pi*epsilon0) )

    This field will be pointing away from A, so this E component is pointing in the -y direction.

    E(D) = k*q(D)/r^2(D) = k*q/(r/2)^2 = 4kq/r^2

    This field will be pointing toward D, so this E component is pointing in the -y direction.

    E(B) = k*q(B)/r^2(B) = k*q/(r^2 + (r/2)^2) <-- By the Pythagorean theorem

    E(B) = 4kq/5r^2

    This field will be pointing toward B, so
    E(B)x = E(B)*cos(t)
    E(B)y = E(B)*sin(t)
    where
    sin(t) = (r/2)/sqrt{5r^2/4} = 1/sqrt{5}
    cos(t) = r/sqrt{5r^2/4} = 2/sqrt{5}

    So
    E(B)x = 2E(B)/sqrt{5} = 8kq/(5sqrt{5}*r^2)
    E(B)y = E(B)*sin(t) = 4kq/(5sqrt{5}*r^2)

    Similarly:
    E(C)x = -8kq/(5sqrt{5}*r^2)
    E(C)y = 4kq/(5sqrt{5}*r^2)

    So the overall electric field E has components:
    Ex = 0 + 8kq/(5sqrt{5}*r^2) + -8kq/(5sqrt{5}*r^2) + 0 = 0

    Ey = -4kq/r^2 + 4kq/(5sqrt{5}*r^2) + 4kq/(5sqrt{5}*r^2) + -4kq/r^2 = - 8kq/r^2 + 8kq/(5sqrt{5}*r^2)

    The magnitude of E at this point is then:
    E = sqrt{Ex^2 + Ey^2} = sqrt{[0]^2 + [- 8kq/r^2 + 8kq/(5sqrt{5}*r^2)]^2} = - 8kq/r^2 + 8kq/(5sqrt{5}*r^2)

    -Dan
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