Poynting vector -- Calculate the EM power transmitted down a coax cable

[denniszhao]

Homework Statement

Poynting vector

Relevant Equations

attached below in pic

I don't know which part gets wrong

 

[mitochan]

Hi. How much W does your result show ?

 

[denniszhao]

Hi. How much W does your result show ?

I got 3.3uW which is same as the given answer but the derivation isn't correct tho

 

[mitochan]

Hum.. I am so optimistic to say if you get the right value, the way of your derivation is right too.

 

[TSny]

@denniszhao, Welcome to PF! We ask that you try to type in your equations rather than post a picture of hand-written work. This allows the homework helpers to easily quote individual parts of your post.
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I'm not clear on the meaning of the statement:

"The distribution of charge on the inner wire is given by $q(z,t) = Q_0\cos \left( \omega( \frac{z}{c}-t) \right)$"

Does $q(z, t)$ represent the amount of charge on the wire at position $z$ at time $t$? That doesn't make sense to me. How can there be a finite amount of charge located at one value of $z$?

But, suppose we go ahead and assume that the charge per unit length $\lambda(z,t)$ on the wire is given by

$\lambda(z,t) =\frac{\partial q}{\partial z} = -\frac{\omega}{c} Q_0 \sin \left( \omega( \frac{z}{c}-t) \right)$

The current $I(z,t)$ is not given by $\large \frac{\partial q(z,t)}{\partial t}$. Rather, see if you can show that

$\large \frac {\partial I}{\partial z} = -\frac{\partial \lambda}{\partial t}$.

You can then use this relation to derive the expression for $I(z, t)$.

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Note that your result for $I(z,t)$ has the opposite sign of $\lambda(z, t)$. The directions of E and B will then be such that the Poynting vector propagates energy in the negative-z direction. But I think you should find that the energy propagates in the positive-z direction.