 #1
adamaero
 106
 1
 Homework Statement:

A coaxial cable has an inner conductor of radius r_i = 0.00045 m and a thin outer conductor or radius r_o = 0.0018 m. The dielectric between the conductors has a relative permittivity εr = 2.25. The dielectric's radius is 0.00148 m.
Find the electric energy density in the dielectric.
 Relevant Equations:

V(ρ) = V_o*ln(ρ/ρ_o)/ln(ρ_i/p_o)
Work = 0.5∫∫∫D•E dv [J]
Work = 0.5D•E = 0.5*p_v*V
Maybe relevant:
∇V = E
V(ρ) = V_o*ln(ρ/0.0018)/ln(45/180)
(Attached picture is where the unit vector of r is really ρ.)
In cylindrical coordinates
∇V = ρ*dV/dρ + 0 + 0
∇V =derivative[V_o*ln(ρ/0.0018)/1.386]dρ
∇V = V_o*0.0018/(1.386*ρ)
E = V_o*0.0012987/ρ
Work = 0.5∫∫∫εE•E dv
Bounds: 0.0018 to 0.00045 m
D = εE = 2.25*8.854e12*E
Work = 0.5*2.25*8.854e12*0.0012987^2*∫(V_o^2/ρ^2)dρ
...or can I just stop, before the integral, and do 0.5*D•E:
0.5*εE•E
0.5*2.25*8.854e12*0.0012987^2
= 1.68e17 J/m^3
(Attached picture is where the unit vector of r is really ρ.)
In cylindrical coordinates
∇V = ρ*dV/dρ + 0 + 0
∇V =derivative[V_o*ln(ρ/0.0018)/1.386]dρ
∇V = V_o*0.0018/(1.386*ρ)
E = V_o*0.0012987/ρ
Work = 0.5∫∫∫εE•E dv
Bounds: 0.0018 to 0.00045 m
D = εE = 2.25*8.854e12*E
Work = 0.5*2.25*8.854e12*0.0012987^2*∫(V_o^2/ρ^2)dρ
...or can I just stop, before the integral, and do 0.5*D•E:
0.5*εE•E
0.5*2.25*8.854e12*0.0012987^2
= 1.68e17 J/m^3