# Parametric representation of a vector electric field

1. Oct 23, 2011

### jjand

1. The problem statement, all variables and given/known data

See Attachment

2. Relevant equations

None I can think of

3. The attempt at a solution

I'm fairly certain that phi_yx is zero
Also I tried factoring out the cos and splitting up the equation into it's respective components, but to no avail. Am I even going about this correctly?

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Last edited: Oct 23, 2011
2. Oct 24, 2011

### vela

Staff Emeritus
The last bit is a typo. It should be $\phi_{xy} = \phi_x$.

I'm not sure what you mean by "splitting up the equation into its respective components" since the equation you're supposed to show is true is a scalar equation.

3. Oct 25, 2011

### jjand

Thanks for the reply Vela! Actually I think I figured it out. Maybe you or someone else can check my work? I assumed that it the phi's were not typos, but they could very well be. I went ahead and attached my work.

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• ###### Scan 2.pdf
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4. Oct 25, 2011

### vela

Staff Emeritus
Your expressions for Ex and Ey are slightly wrong. For example,
$$E_x = A_x\cos(kz-\omega t+\phi_x) = A_x\cos(-r+\phi_x) = A_x\cos(r-\phi_x)$$since cosine is an even function. You have an extra negative signs floating around.

You must have made a mistake somewhere in the middle because the cross term has the wrong sign. The sign doesn't flip the way you did it because, again, $\cos (-\theta)=\cos\theta$.

You actually made it more complicated by keeping the E's and A's around. You know that
\begin{align*}
\frac{E_x}{A_x} &= \cos(\phi_x-r) \\
\frac{E_y}{A_y} &= \cos(r) \\
\end{align*}so you could have just written the lefthand side in terms of cosines and then shown it simplifies down to the righthand side.

5. Oct 30, 2011

### jjand

Ah that I forgot about that cosine property and I found a convenient sign mistake half way through the problem. Thanks Vela!