# Electric energy density in the dielectric of a coaxial cable

In summary, the problem involves finding the electric energy density in a co-axial cable with inner and outer conductors of different radii and a dielectric with a relative permittivity. It can be solved by first finding the charge per unit length on the cylindrical capacitor formed by the cable, and then using the capacitance and applied voltage to calculate the energy density. Alternatively, one can directly calculate the total capacitance per unit length and use this to find the energy density.
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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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.

## 1. What is electric energy density in the dielectric of a coaxial cable?

Electric energy density refers to the amount of electric energy that is stored in the dielectric material of a coaxial cable. It is measured in joules per cubic meter (J/m3).

## 2. How does the electric energy density affect the performance of a coaxial cable?

The higher the electric energy density in the dielectric, the more efficient the coaxial cable will be in transmitting electrical signals. This is because a higher density allows for a greater amount of energy to be stored and therefore a stronger signal can be transmitted.

## 3. What factors can affect the electric energy density in the dielectric of a coaxial cable?

The electric energy density in the dielectric of a coaxial cable can be affected by the type of dielectric material used, the thickness of the dielectric layer, and the voltage applied to the cable.

## 4. Can the electric energy density in the dielectric of a coaxial cable be changed?

Yes, the electric energy density can be changed by altering the properties of the dielectric material or by adjusting the voltage applied to the cable. This is often done in order to optimize the performance of the cable for specific applications.

## 5. Is there an ideal electric energy density for a coaxial cable?

The ideal electric energy density for a coaxial cable will depend on the specific application and the desired performance. In general, a higher density will result in a stronger signal, but it is important to balance this with other factors such as cost and practicality.

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