Answer to Question #134768 in Electrical Engineering for pavan

Question

 Consider a coaxial cable formed by two concentric cylindrical conductors with the radius of the inner conductor being “a” and the outer conductor being “b”. Let L be the length of the cable. Assume the surface charge distribution on the inner conductor be defined by ρs. Determine the flux density D at a point ρ located radially between a and b, i.e. a < ρ < b. 

Answer 1

$ds = ad\phi dz$

$Q=\iint Psad\phi dz$

$= 2\pi PsL$ ---------------- 1

from gauss law,

$Q= DsA$

Change Q at point P located is

$Q = Ds \times 2\pi PL$ ------------ 2

from equation 1 and 2

$Q= 2\pi aPsL$

$2\pi PsL =$ $Ds \times 2\pi PL$

$Ds = a\frac{Ps}{p}$

(a<P<b)