Answer to Question #134768 in Electrical Engineering for pavan

Question #134768

 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. 


1
Expert's answer
2020-09-24T10:24:01-0400

ds=adϕdzds = ad\phi dz

Q=PsadϕdzQ=\iint Psad\phi dz

=2πPsL= 2\pi PsL ---------------- 1

from gauss law,

Q=DsAQ= DsA

Change Q at point P located is

Q=Ds×2πPLQ = Ds \times 2\pi PL ------------ 2

from equation 1 and 2

Q=2πaPsLQ= 2\pi aPsL

2πPsL=2\pi PsL = Ds×2πPLDs \times 2\pi PL

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

(a<P<b)


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