- #1

adamaero

- 109

- 1

- Homework Statement
- A co-axial 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

∇V =

∇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*8.854e-12*E

Work = 0.5*2.25*8.854e-12*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*8.854e-12*0.0012987^2

=

(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*8.854e-12*E

Work = 0.5*2.25*8.854e-12*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*8.854e-12*0.0012987^2

=

**1.68e-17**J/m^3