Is the Capacitance of a Twin Lead TV Antenna Cable Within Specifications?

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Homework Statement



Your home video studio will use a TV antenna cable often known as "twin lead." Two 0.50mm--diameter wires are spaced 12mm apart. Your equipment requires that the total capacitance of the cable not exceed 1000 pF. The cable is 30m long. A technical paper on the cable gives the potential difference between the two conductors as

V = lamda / (pi*epsilon_0 ) * ln ( (b-a) / a )

where a is the radius of the wire and b is the separation.

Is your cable within the specifications? ____ (my answer is NO), see below

Attempt :

C = epsilon_0 * A / D

A = 2piRL
A = 2 * pi * ( 0.5/2 * 10^-3 ) * (30 m)
D = 12mm * 10 ^ -3

C = epsilon_0 * 2 * pi * (0.25 * 10^-3) * (30) / (12 * 10^-3)

= 3.47 * 10 ^ -11

which is greater than 1000pF = 10^-9, thus it should be NO, I think. Is this correct?
 
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Any one have any Ideas ,

I also tried this :

C = Q/V

V is given , and Q is lamda * L, but I don't think its right
 
Use the general definition for capacitance, C=Q/V.
 
I did that as well.

C = Q/V

V is given, and I use Q = q*Lamda

But I think its not correct though.