1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Check work on capacitance.

  1. Oct 12, 2009 #1
    1. The problem statement, all variables and given/known data

    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?
     
    Last edited: Oct 12, 2009
  2. jcsd
  3. Oct 13, 2009 #2
    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
     
  4. Oct 13, 2009 #3

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Use the general definition for capacitance, C=Q/V.
     
  5. Oct 13, 2009 #4
    I did that as well.

    C = Q/V

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

    But I think its not correct though.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Check work on capacitance.
  1. Check work (Replies: 1)

Loading...