Question

--Q (for every 10m of cable) \ tq (for every 10m of cable) Perspective View Front View 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 lengt 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. A VA,B = Vb-VA= O V Enter Aha!

Answer 1

Qugetitative Analysis The work to move a test charge un from A to B is defined in terms of the integral from A to B og the component of the Electric field in parallel to chosen path : Ww=ffd = Ed _Integral can be evaluated in 3 steps, WAB=Wam+WmN+WN, B WA, B = 0 + Wmn + 0. I I The Electric field be the radial can be calculated two cylinder is from Gauss law, E = 2 2 T6% 26 27 E = 1 - 2 G 28 so, DUA = - WAB qt Again. AV,0 = - Wag AVA, B = - WAR pa

(exspective view) front liew Given, inner conductor radius = a = 0.0025 m. outer Conductor Cylindrical Shell with inner radius = b = 0.0075m. and outer radius = = 0.008 length of the cable (2) = com. lotal charge on inner conductor q = 2.8x108c Total charge on outer conducts Q = ~5.6X166C alow, Electric field between the conductors using Gauss law and a Gaussian cylindrical surface concentric with inner conductor and radius & and length L is given by E.A 14 = E ( 27171) = ? E = q_ апа со Now potential difference bhw any two points, Vo - VA - / Edr

an Vo - VA -q Q TL to 2HLG va - Vos droit - VA - Vo = 2 in (6) QITLGO AVAB = VB VA = - 9 in (b) = DVAIR = - 2x 8.99xio x 2.8 X10 10 n (0.0075 10.0025) - BVAB = VB VA = - 55 3 v Herice, change in potential from 8 to 4 ü -55.3V.