# Math Help - Algebra: Coulomb's law & electric field

1. ## 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

2. ## Re: Algebra: Coulomb's law & electric field

Originally Posted by froodles01
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
$\frac{\sqrt{3}Q}{4\pi\varepsilon_0 \left(\sqrt{3}a\right)^3} \cdot a(-1,1,1) = \frac{\sqrt{3}Q}{4\pi\varepsilon_0 3\sqrt{3}a^3} \cdot a(-1,1,1) = \frac{Q}{4\pi\varepsilon_0 a^2}\cdot (-1/3,1/3,1/3)$.

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).