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Thread: Algebra: Coulomb's law & electric field

  1. #1
    Junior Member froodles01's Avatar
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    Algebra: Coulomb's law & electric field

    Quite a simple thing, but can't work out one step.
    Charge √3Q is at cartesian point (3a, a, 2a). Calculate the electric field at the point (2a, 2a, 3a) due to this charge.

    using the formula;
    E(r) = q / 4 ∏ ε0 |r - r0|3 (r - r0)
    Here
    r = (2a ex + 2a ey + 3a ez)
    r0 = (3a ex + a ey + 2a ez)

    r - r0 = (-a ex + a ey + a ez)
    |r - r0| = √-1a2 +1a2 +1a2

    so
    E(r) =
    √3 Q / 4 ∏ ε0 (√3a)3 * (-a ex + a ey + a ez) ------1
    which goes to

    E(r) = Q / 4 ∏ ε0 a2 * (-0.33 ex + 0.33 ey + 0.33 ez) -------2

    I need to know how to get to the last part.
    i.e. the algebra bit from 1 to 2
    where does the √3 bits get transformed into the 0.33 in the vector at the end.
    Just feeling dumb when I REALLY shouldn't.

    Thank you
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  2. #2
    MHF Contributor
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    Re: Algebra: Coulomb's law & electric field

    Quote Originally Posted by froodles01 View Post
    E(r) = [/SUP]√3 Q / 4 ∏ ε0 (√3a)3 * (-a ex + a ey + a ez) ------1
    which goes to

    E(r) = Q / 4 ∏ ε0 a2 * (-0.33 ex + 0.33 ey + 0.33 ez) -------2

    I need to know how to get to the last part.
    i.e. the algebra bit from 1 to 2
    $\displaystyle \frac{\sqrt{3}Q}{4\pi\varepsilon_0 \left(\sqrt{3}a\right)^3} \cdot a(-1,1,1) = \frac{\sqrt{3}Q}{4\pi\varepsilon_0 3\sqrt{3}a^3} \cdot a(-1,1,1) = \frac{Q}{4\pi\varepsilon_0 a^2}\cdot (-1/3,1/3,1/3)$.

    You need to insert more parentheses. For example, it is not clear where the denominator ends in q / 4 ∏ ε0 |r - r0|3 (r - r0). This should be written as (q / (4 ∏ ε0 |r - r0|3)) (r - r0). Similarly, √-1a2 +1a2 +1a2 can be written as sqrt(-1a2 +1a2 +1a2).
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