Adding Electric Fields - Find Simplification Solution

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
5 replies · 3K views
PhDorBust
Messages
141
Reaction score
0

Homework Statement


Have two electric fields.

[tex]\hat{x} E_1 e^{i(kz- \omega t)}[/tex]
[tex]\hat{x} E_2 e^{i(-kz- \omega t)}[/tex]

Where E_1, E_2 are real.

Sum them such that the result can be expressed as one magnitude and exponential, e.g., |E|e^(iq), Where E is real.

I have no clue how I would begin to simplify this. Any ideas? It's from an undergraduate text.
 
Physics news on Phys.org
Factor out the common factor of e-iωt. Use Euler's formula, e=cos θ + i sin θ.
 
Sure.. but that's trivial detail and not key step.

[tex]e^{-i \omega t}[(E_1 + E_2) cos(kz) + i(E_1 - E_2) sin(kz)][/tex]
 
Well, the rest is even more trivial, so I have no idea where you're getting stuck.
 
Perhaps wording is unclear. Here is answer. Looks little complex for name of trivial, so long that I attach rather than typeset in latex =].

Disregard the unit vector.
 

Attachments

  • untitled.GIF
    untitled.GIF
    10.5 KB · Views: 553
PhDorBust said:
Sure.. but that's trivial detail and not key step.

[tex]e^{-i \omega t}[(E_1 + E_2) cos(kz) + i(E_1 - E_2) sin(kz)][/tex]

So now, how do you calculate the absolute value of a complex number? And its argument? That's just converting from cartesian to polar coordinates in the complex plane (if we ignore the factor e^(-iωt) which you can set apart).

(The only non-trivial part apart from that conversion will be using the identity sin² φ + cos² φ = 1).