denniszhao
- 15
- 0
- Homework Statement
- Poynting vector
- Relevant Equations
- attached below in pic
I don't know which part gets wrong
Last edited by a moderator:
Poynting vector -- Calculate the EM power transmitted down a coax cable
- Thread starter denniszhao
- Start date
-
- Tags
- Cable Em Power Poynting vector Vector
4 replies · 2K views · last reply
I don't know which part gets wrong
Discussion
mitochan
- 294
- 134
Hi. How much W does your result show ?
denniszhao
- 15
- 0
I got 3.3uW which is same as the given answer but the derivation isn't correct thomitochan said:
mitochan
- 294
- 134
Hum.. I am so optimistic to say if you get the right value, the way of your derivation is right too.
Science Advisor
Homework Helper
Gold Member
- 14,830
- 4,812
@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.
---------------------------------------
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)##.
------------------
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.
---------------------------------------
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)##.
------------------
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.