# Coaxial Cable

1. Sep 20, 2007

### Winzer

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

$$U=\int V dQ$$

3. The attempt at a solution
I perform the intergral and come up with a couple equations this one seem the best:
$$U= .5QV$$ Right?

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• ###### InsulatedCoaxialCable.gif
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Last edited: Sep 21, 2007
2. Sep 21, 2007

### learningphysics

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. Sep 21, 2007

### chaoseverlasting

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

4. Apr 22, 2010

### sterlingb06

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. Jul 14, 2012

### waters

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?

6. Jul 14, 2012

### CWatters

7. Jul 14, 2012

### waters

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.