
#1
Sep2007, 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? 



#2
Sep2107, 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. 



#3
Sep2107, 08:41 AM

P: 1,017

You need to know the charge right? Or how charge varies with voltage or something?




#4
Apr2210, 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 



#5
Jul1412, 12:40 AM

P: 29

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? 



#6
Jul1412, 07:39 AM

P: 2,861

This page has the equation for the capacitance..
http://hyperphysics.phyastr.gsu.edu...ic/capcyl.html C/L=2*pi*k*ε_{0}/ln(do/di) So C= L*2*pi*k*ε_{0}/ln(do/di) Then Energy is U = 0.5*C*V^{2} Note it's V squared not V. 



#7
Jul1412, 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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