The diagram above shows a coaxial cable. The inner conductor has radius a = 0.00
ID: 2145677 • Letter: T
Question
The diagram above shows a coaxial cable. The inner conductor has radius a = 0.0025 m. The outer conductor is a cylindrical shell with inner radius b = 0.0075 m, and outer radius c = 0.008 m from the center. Both conductors are coaxial. For every length L = 10 m of cable, there is a total charge q = 2.8e-08 C on the inner conductor and a total charge of Q = -5.6e-06 C on the outer conductor.Determine the electric potential difference between the labeled points A and B.
capital delta VA,B = VB-VA =
Explanation / Answer
First find the E-field between the conductors using Gauss' Law and a Gaussian cylindrical surface concentric with the inner conductor and of radius "r" and length "L"; EA = E2pirL = q/e(o) Only the charge "q" enclosed by the Gaussian cylinder is used on the right side. The charge Q on the outer conductor establishes the direction of E as "outward" and from the symmetry "radial" along "r". E = (q/2piLe(o))(1/r) The potential difference from any point on the inner conductor (since its all at the same potential) to any point on the outer conductor (since its all at the same potential) is the negative of integral Edr from inner to outer cylinder. Since the initial & final points on the two conductors doesn't matter choose the integration to be along "r" from a to b; (remember E=0 inside a conductor so integrating a to c is essentially a to b) Vb - Va = -INTEGRAL[Edr] = -(q/2piLe(o))INTEGRAL[dr/r] = -(q/2piLe(o))Ln(b/a) = (q/2piLe(o))Ln(a/b) You can put in the numbers. Choole L=10m so the corresponding "q" is known. Since Va is the potential anywhere on (or in) the inner conductor and Vb is the potential anywhere on (or in) the outer conductor you can the write; VB - VA = Vb - Va = (q/2piLe(o))Ln(a/b)