Electric energy density in the dielectric of a coaxial cable

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SUMMARY

The discussion focuses on calculating the electric energy density in the dielectric of a coaxial cable with an inner radius of 0.00045 m and an outer radius of 0.0018 m. The dielectric material has a relative permittivity (εr) of 2.25. The derived electric field (E) is expressed as E = V_o*0.0012987/ρ, and the energy density is calculated using the formula Work = 0.5*εE•E, resulting in a value of 1.68e-17 J/m³. The conversation also addresses assumptions regarding the applied voltage and the interpretation of energy density as linear energy density.

PREREQUISITES
  • Cylindrical coordinates and vector calculus
  • Understanding of electric fields and energy density
  • Knowledge of coaxial cable structure and properties
  • Familiarity with dielectric materials and their permittivity
NEXT STEPS
  • Study the derivation of electric fields in cylindrical capacitors
  • Learn about the relationship between capacitance and energy in capacitors
  • Explore numerical integration techniques for calculating energy density
  • Investigate the effects of varying dielectric materials on energy density
USEFUL FOR

Electrical engineers, physics students, and professionals working with coaxial cables and dielectric materials will benefit from this discussion, particularly those involved in energy density calculations and capacitor design.

adamaero
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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 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
 

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adamaero said:
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.
The problem is poorly phrased.
1. I assume the applied voltage is V_o?
2. I assume by "energy density" is meant the linear energy density, i.e. energy per unit length of cable?
If so, you might proceed by finding the charge q per unit length on what you will recognize is a cylindrical capacitor.
To do this, consider -V_o = ## \int_a^c E \, dr ##
and ## \iint_S D \cdot dA ## = q.
That would give you E(r) and D(r) so you could volume-integrate 1/2 E(r)D(r) over the distance a to b and again b to c, the differential volume being ##2\pi r dr ## per unit length.
 
Alternatively, if you previously solved for, or otherwise obtained, the capacitance of a cylindrical capacitor, it would be easier to compute the total capacitance per unit length, then use the familiar expression relating energy to C and V.
 

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