Thread: 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 e_x + 2a e_y + 3a e_z)
r₀ = (3a e_x + a e_y + 2a e_z)

r - r₀ = (-a e_x + a e_y + a e_z)
|r - r₀| = √-1a² +1a² +1a<sup>2

so
E(r) =</sup> √3 Q / 4 ∏ ε_{0 (}√3a)³ * (-a e_x + a e_y + a e_z) ------1
which goes to

^{E(r) =} Q / 4 ∏ ε₀ a² * (-0.33 e_x + 0.33 e_y + 0.33 e_z) -------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

Originally Posted by froodles01

E(r) = [/SUP]√3 Q / 4 ∏ ε_{0 (}√3a)³ * (-a e_x + a e_y + a e_z) ------1
which goes to

^{E(r) =} Q / 4 ∏ ε₀ a² * (-0.33 e_x + 0.33 e_y + 0.33 e_z) -------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).