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Coaxial Cable

by Winzer
Tags: cable, coaxial
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Winzer
#1
Sep20-07, 08:51 PM
P: 605
1. The problem statement, all variables and given/known data
The diagram below depicts a cross section of coaxial conductor with an inner wire of diameter and an outer conducting sheath of inside diameter , and some material placed in the space between the two wires. Suppose that you have a coaxial wire with di= 2.85 mm, do= 6.25 mm and mylar ( k= 3.10) is placed in the space between the two wires. If there is a potential of 1 kV between the wires, how much energy is stored in a 10 m piece of cable?

2. Relevant equations

[tex] U=\int V dQ[/tex]

3. The attempt at a solution
I perform the intergral and come up with a couple equations this one seem the best:
[tex] U= .5QV[/tex] Right?
Attached Thumbnails
InsulatedCoaxialCable.gif  
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learningphysics
#2
Sep21-07, 01:08 AM
HW Helper
P: 4,125
How did you get 0.5QV ?

The way I'd do it is to find the charge per unit length on the inner wire... the charge per unit length on the outer conductor is just - the inner charge per unit length...

You can get this using Gauss' law... and the voltage = -integral E.dr

When you find the charge per unit length x... then the total charge is 10*x. The energy stored is 10*x*V.
chaoseverlasting
#3
Sep21-07, 08:41 AM
P: 1,017
You need to know the charge right? Or how charge varies with voltage or something?

sterlingb06
#4
Apr22-10, 02:40 PM
P: 1
Coaxial Cable

The coax cable has capacitance per meter. You can look the equations up on wiki. Once you find the total capacitance (multiply by length), use E=(CV^2)/2.



Sterling
waters
#5
Jul14-12, 12:40 AM
P: 29
Quote Quote by learningphysics View Post
How did you get 0.5QV ?

The way I'd do it is to find the charge per unit length on the inner wire... the charge per unit length on the outer conductor is just - the inner charge per unit length...

You can get this using Gauss' law... and the voltage = -integral E.dr

When you find the charge per unit length x... then the total charge is 10*x. The energy stored is 10*x*V.
I do not know what I am doing wrong. Here is my work:
q/(кε) = EA, A = 2*pi*rx
E = q/(кεA) = q/(2кε*pi*rx)
V = -integral of Edr from a to b = -(q*ln(b/a))/(2кε*pi*x) = 1000 V
1000(2кε*pi)/ln(b/a) = q/x
(q/x)*10*1000 does not give me the answer. Where did I go wrong?
CWatters
#6
Jul14-12, 07:39 AM
P: 3,088
This page has the equation for the capacitance..

http://hyperphysics.phy-astr.gsu.edu...ic/capcyl.html

C/L=2*pi*k*ε0/ln(do/di)

So
C= L*2*pi*k*ε0/ln(do/di)

Then Energy is

U = 0.5*C*V2

Note it's V squared not V.
waters
#7
Jul14-12, 02:32 PM
P: 29
Thanks. It turns out my work is right, but the formula for energy is .5qV (i.e. C = q/V, so .5CV^2 = .5qV) , not qV.


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