# Compute electric field

1. Jul 12, 2012

### ggeo1

Hello,

I have this equation that computes the electric field from electric currents.

E=-j*k_n (J*C_1- (J*R|R*C_2) e^ -j*k*R/4πR )

The data are:

k_n=377 , k=20.93 , R=Sqrt(3) ,J_x=0.72576821 ,J_y=0.03763659 , J_z=0.86104220 ,
C_1=0.999239077741 + 1 / 36.2518234024j
C_2=0.997717233223+ 3 / 36.25182344024j

R_x=R_y=R_z=0.57735026919

I have 3 questions.

1) From the above equation i must compute E_x ,E_y,E_z.
When i compute E_x i will compute J_x*C_1 - (J*R|R*C_2) ?
And specific: I will compute the J*R dot product as : J*R=J_x*R_x+J_y*R_y+J_z*R_z ?
Or only the J*R=J_x*R_x ?

2) The computation J*R|R*C_2 . How will i do the computation for the symbol ' | ' ?
I will compute the dot product , then the R_x*C_2 and then what?How to compute the whole
J*R|R*C_2 ?

3) How to deal with the C_1 and C_2? I mean they have the imaginary part in the denominator.What is the best approach?

Thank you!

2. Jul 12, 2012

### Staff: Mentor

What does the symbol "|" represent? It does not look like standard notation.

C_2 is a scalar? In this case, both J*R and C_2 are scalars, and I would assume that you can just perform a scalar multiplication with your vector.

Don't forget the exponential which is multiplied with the latter term.

You should do this, right.

1/10 does not look high, but I would perform the whole calculation with complex numbers. If the result is not real, try to find an interpretation - maybe you can simply drop the imaginary part.

3. Jul 12, 2012

### ggeo1

Hello and thanks for the help.

C_1 and C_2 are scalars.

So, ok J*R=..=scalar.

But i can't understand how to do the (J*R|R*C_2) .
I will compute the J*R and then the R*C_2 , and then what?

4. Jul 12, 2012

### ggeo1

I think it's

((J*R) *C_2 )*R , where R i will put R_x?

Then for E_y i will put R_y?

5. Jul 12, 2012

### Staff: Mentor

I would try to just multiply them as scalar*vector.

$$(\vec{J}*\vec{R}|\vec{R}*C_2) \stackrel{?}{=} \left(C_2 (\vec{J}*\vec{R})\right) \vec{R}$$

Edit:
Should work.
It is speculative, and maybe you should check where the formula comes from.

6. Jul 12, 2012

Ok , thanks!