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Electrostatic field in a coaxial line
Msc. Julio Cesar Salazar Hernandez
A coaxial cable is a transmission line composed of two conductors, one internal and one external (see
fig. 1.1). The external conductor is a metal tube or mesh shield, but the internal conductor is
supported by a dielectric that can be a solid, expanded plastic or semi-solid (polyethylene discs,
helical tapes, plastic strips, etc.).
This transmission line is characterized by being a shielded structure, that is, the electromagnetic field
is limited between the space of the inner and outer conductors.
Illustration 1.1.- Structure of a Coaxial Cable.
A coaxial has an external diameter b and an internal diameter a. The space between the conductors is
formed by a dielectric with a certain permittivity. A potential difference is applied between the
conductors where the external one has 0V and the internal one V0 (see Fig. 1.2).
Illustration 1.2.- Internal and external radius of the coaxial cable separated by a
dielectric with a certain permittivity.
Analytical calculation.
Calculation of the electric field. The electric field equation for electrostatic fields uses the Poisson
equation (1) and the Laplace equation (2) as a function of fields.
𝛻2𝐸 = 𝛻
𝜌
𝜀
𝛻2
𝐸 = 0
(1)
(2)
Now based on potentials we have:
𝛻2
𝑉 = −
𝜌
𝜀
𝛻2
𝑉 = 0
(1.1)
(2.1)
In the region of a<r<b the dielectric is found, so there is no free charge (ρ=0). Applying the Laplace
equation of potentials in equation 2.1 with initial conditions:
𝑉 𝜌 = 𝑎 = 𝑉0
𝑉 𝜌 = 𝑏 = 0
This problem is solved directly because the structure is cylindrical and symmetrical. Rotating the coaxial
about the z axis does not change the geometry at all. As a result we know that the electric potential field
is a function of ρ. Solving equations in cylindrical coordinates we have the following equation:
𝛻2
𝑓 𝑟, ∅, 𝑧 =
1
𝑟
𝜕
𝜕𝑟
𝑟
𝜕𝑓
𝜕𝑟
+
1
𝑟2
𝜕2
𝑓
𝜕∅2 +
𝜕2
𝑓
𝜕𝑧2
In this case in the z and ∅ direction there are no changes, that is, they are always constant. Therefore,
what interests us is knowing the electric field at each point from radius a to radius b. Changing the
equation with the previously used variables we have:
𝛻2
𝑉(𝜌) = 0
1
𝜌
𝜕
𝜕𝜌
𝜌
)
𝜕𝑉(𝜌
𝜕𝜌
= 0
𝜕
𝜕𝜌
𝜌
)
𝜕𝑉(𝜌
𝜕𝜌
= 0 (2.2)
By integrating the two sides of the previous equation.
𝜕
𝜕𝜌
𝜌
)
𝜕𝑉(𝜌
𝜕𝜌
𝑑𝜌 = 0𝑑𝑝
𝜌
)
𝜕𝑉(𝜌
𝜕𝜌
= 𝐶1
)
𝜕𝑉(𝜌
𝜕𝜌
=
𝐶1
𝜌
Integrating again we have the following:
)
𝜕𝑉(𝜌
𝜕𝜌
𝑑𝜌 =
𝐶1
𝜌
𝑑𝑝
𝑽 𝝆 = 𝑪1𝒍𝒏 𝝆 + 𝑪2
This equation satisfies equation 2.2
𝜕
𝜕𝜌
𝜌
)
𝜕𝑉(𝜌
𝜕𝜌
= 0
𝜕
𝜕𝜌
𝜌
𝜕
𝜕𝜌
(𝐶1𝑙𝑛 𝜌 + 𝐶2 = 0
𝜕
𝜕𝜌
𝜌(𝐶1
1
𝜌
= 0
0 = 0
Now the initial conditions are evaluated to know the values ​​of the two constants:
𝑉 𝜌 = 𝑎 = 𝐶1 ln 𝑎 + 𝐶2 = 𝑉0
𝑉 𝜌 = 𝑏 = 𝐶1 ln 𝑏 + 𝐶2 = 0
(A)
(B)
Solving the system of equations of two unknowns, solve for B:
)
𝐶2 = −𝐶1l n( 𝑏
Substituting in A
𝐶1 ln )
𝑎 − 𝐶1l n( 𝑏 = 𝑉0
𝐶1 ln )
𝑏 − 𝐶1l n( 𝑎 = −𝑉0
𝑪1 =
−𝑽0
𝒍𝒏 𝒃 𝒂
Replacing C1 in C2:
𝑪2 =
)
𝑽0𝐥 𝐧( 𝒃
𝒍𝒏 𝒃 𝒂
Therefore, the final solution is:
𝑽 𝝆 =
−𝑽0𝒍𝒏 𝝆
𝒍𝒏 𝒃 𝒂
+
𝑽0 𝐥𝐧 𝒃
𝒍𝒏 𝒃 𝒂
; 𝒂 < 𝝆 < 𝒃
Now that we have the electric potential we can find the electric field as follows:
𝐸 = −𝛻𝑉
𝐸 𝜌 = −
𝜕
𝜕𝜌
−𝑉0𝑙𝑛 𝜌
𝑙𝑛 𝑏 𝑎
+
𝑉0 ln 𝑏
𝑙𝑛 𝑏 𝑎
𝑬 𝝆 =
𝑽0
𝒍𝒏 𝒃 𝒂
1
𝝆
𝒂𝝆; 𝒂 < 𝝆 < 𝒃
You can also know the electric flux density:
𝐷 = 𝜀𝐸
)
𝐷 𝜌 = 𝜀𝐸(𝜌
𝑫 𝝆 =
𝜺𝑽0
𝒍𝒏 𝒃 𝒂
1
𝝆
𝒂𝝆; 𝒂 < 𝝆 < 𝒃
Once the electrostatic case was resolved, the equations were evaluated with values ​​between the
internal and external radius. The proposed coaxial cable data is as follows:
𝑟𝑖 = 0.022 𝑚
𝑟𝑒 = 0.045 𝑚
𝜀𝑟 = 2.3
𝑉0 = 50𝑘[𝑉
Tables 1, 2 and 3 show the values ​​obtained for the electric field density, the electric potential and the
electric field respectively.
𝝆 𝑫 𝝆
0.022 6.46x10-5
0.025 5.68x10-5
0.030 4.74x10-5
0.032 4.44x10-5
0.035 4.06x10-5
0.040 3.55x10-5
0.042 3.38x10-5
0.045 3.16x10-5
𝝆 𝑽 𝝆
0.022 50000.00
0.025 41068.34
0.030 28329.63
0.032 23820.36
0.035 17559.20
0.040 8229.43
0.042 4820.49
0.045 0
Table 1.- Electric field density in the coaxial cable. Table 2.-Electrical Potential in the coaxial cable.
Table 3.-Electrical Field in the coaxial cable.
𝝆 𝑬 𝝆
0.022 3175885.466
0.025 2794779.21
0.030 2328982.675
0.032 2183421.258
0.035 1996270.864
0.040 1746737.006
0.042 1663559.054
0.045 1552655.117
Simulation using COMSOL Multiphysics.
By solving the problem as an electrostatic case and giving the dimensions and physical characteristics of
the problem, the following graphs of electric field and electric potential were obtained. Figure 1.3 shows
the graph of the electric field norm that goes from a little more than 30M[V/m] to a little more than
1M[V/m]. But on the other hand, Figure 1.4 shows the electric potential that goes from 50k [V] to 0 [V].
Illustration 1.3.- Graph of the electric field in the coaxial cable.
Illustration 1.4.- Graph of the electrical potential in the coaxial
cable.
Analytical vs simulated calculation.
In order to compare the results obtained, the analytical equations were graphed in Excel. The electric
potential graphs (fig. 1.4 and 1.5) and the electric field graph (fig. 1.3 and 1.6) are very similar since an
extra fine mesh was used when simulating it.
Illustration 1.5.- Analytical electrical potential graph in the coaxial
cable.
0
500000
1000000
1500000
2000000
2500000
3000000
3500000
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
Campo Eléctrico (V/m)
Illustration 1.6.- Analytical electric field graph of the coaxial cable
Illustration 1.7.- Distribution of electrical potential in the coaxial cable.
Illustration 1.8.- Distribution of the electric field in the coaxial cable.
Conclusion.
In solving this electrostatic problem, the advantage was that the problem could be reduced to a simple
differential equation because only the radius of the coaxial cable varied. If the opposite were the case,
perhaps the analytical solution would become more complex and difficult to solve since, although the
differential equations that represent the physical behavior of the problem are obtained, it does not
guarantee that an analytical solution exists, which is why simulation software is chosen. which uses
numerical methods to give an approximate solution.
In the case of COMSOL Multiphysic, it uses a mesh to solve the problem. If it uses a very fine mesh,
the results are very close to the analytical one, but it takes more computing time. If a poor mesh is
used, the results are far from the analytical one and uses less computing time.
The distribution of the electric potential and the electric field is very similar since most of it is found on
the surface of the central conductor and its intensity decreases until it disappears the closer it gets to
the external conductor.

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Electrostatic field in a coaxial transmission line

  • 1. Electrostatic field in a coaxial line Msc. Julio Cesar Salazar Hernandez
  • 2. A coaxial cable is a transmission line composed of two conductors, one internal and one external (see fig. 1.1). The external conductor is a metal tube or mesh shield, but the internal conductor is supported by a dielectric that can be a solid, expanded plastic or semi-solid (polyethylene discs, helical tapes, plastic strips, etc.). This transmission line is characterized by being a shielded structure, that is, the electromagnetic field is limited between the space of the inner and outer conductors. Illustration 1.1.- Structure of a Coaxial Cable.
  • 3. A coaxial has an external diameter b and an internal diameter a. The space between the conductors is formed by a dielectric with a certain permittivity. A potential difference is applied between the conductors where the external one has 0V and the internal one V0 (see Fig. 1.2). Illustration 1.2.- Internal and external radius of the coaxial cable separated by a dielectric with a certain permittivity.
  • 4. Analytical calculation. Calculation of the electric field. The electric field equation for electrostatic fields uses the Poisson equation (1) and the Laplace equation (2) as a function of fields. 𝛻2𝐸 = 𝛻 𝜌 𝜀 𝛻2 𝐸 = 0 (1) (2) Now based on potentials we have: 𝛻2 𝑉 = − 𝜌 𝜀 𝛻2 𝑉 = 0 (1.1) (2.1)
  • 5. In the region of a<r<b the dielectric is found, so there is no free charge (ρ=0). Applying the Laplace equation of potentials in equation 2.1 with initial conditions: 𝑉 𝜌 = 𝑎 = 𝑉0 𝑉 𝜌 = 𝑏 = 0 This problem is solved directly because the structure is cylindrical and symmetrical. Rotating the coaxial about the z axis does not change the geometry at all. As a result we know that the electric potential field is a function of ρ. Solving equations in cylindrical coordinates we have the following equation: 𝛻2 𝑓 𝑟, ∅, 𝑧 = 1 𝑟 𝜕 𝜕𝑟 𝑟 𝜕𝑓 𝜕𝑟 + 1 𝑟2 𝜕2 𝑓 𝜕∅2 + 𝜕2 𝑓 𝜕𝑧2
  • 6. In this case in the z and ∅ direction there are no changes, that is, they are always constant. Therefore, what interests us is knowing the electric field at each point from radius a to radius b. Changing the equation with the previously used variables we have: 𝛻2 𝑉(𝜌) = 0 1 𝜌 𝜕 𝜕𝜌 𝜌 ) 𝜕𝑉(𝜌 𝜕𝜌 = 0 𝜕 𝜕𝜌 𝜌 ) 𝜕𝑉(𝜌 𝜕𝜌 = 0 (2.2) By integrating the two sides of the previous equation. 𝜕 𝜕𝜌 𝜌 ) 𝜕𝑉(𝜌 𝜕𝜌 𝑑𝜌 = 0𝑑𝑝 𝜌 ) 𝜕𝑉(𝜌 𝜕𝜌 = 𝐶1 ) 𝜕𝑉(𝜌 𝜕𝜌 = 𝐶1 𝜌
  • 7. Integrating again we have the following: ) 𝜕𝑉(𝜌 𝜕𝜌 𝑑𝜌 = 𝐶1 𝜌 𝑑𝑝 𝑽 𝝆 = 𝑪1𝒍𝒏 𝝆 + 𝑪2 This equation satisfies equation 2.2 𝜕 𝜕𝜌 𝜌 ) 𝜕𝑉(𝜌 𝜕𝜌 = 0 𝜕 𝜕𝜌 𝜌 𝜕 𝜕𝜌 (𝐶1𝑙𝑛 𝜌 + 𝐶2 = 0 𝜕 𝜕𝜌 𝜌(𝐶1 1 𝜌 = 0 0 = 0
  • 8. Now the initial conditions are evaluated to know the values ​​of the two constants: 𝑉 𝜌 = 𝑎 = 𝐶1 ln 𝑎 + 𝐶2 = 𝑉0 𝑉 𝜌 = 𝑏 = 𝐶1 ln 𝑏 + 𝐶2 = 0 (A) (B) Solving the system of equations of two unknowns, solve for B: ) 𝐶2 = −𝐶1l n( 𝑏 Substituting in A 𝐶1 ln ) 𝑎 − 𝐶1l n( 𝑏 = 𝑉0 𝐶1 ln ) 𝑏 − 𝐶1l n( 𝑎 = −𝑉0 𝑪1 = −𝑽0 𝒍𝒏 𝒃 𝒂
  • 9. Replacing C1 in C2: 𝑪2 = ) 𝑽0𝐥 𝐧( 𝒃 𝒍𝒏 𝒃 𝒂 Therefore, the final solution is: 𝑽 𝝆 = −𝑽0𝒍𝒏 𝝆 𝒍𝒏 𝒃 𝒂 + 𝑽0 𝐥𝐧 𝒃 𝒍𝒏 𝒃 𝒂 ; 𝒂 < 𝝆 < 𝒃 Now that we have the electric potential we can find the electric field as follows: 𝐸 = −𝛻𝑉 𝐸 𝜌 = − 𝜕 𝜕𝜌 −𝑉0𝑙𝑛 𝜌 𝑙𝑛 𝑏 𝑎 + 𝑉0 ln 𝑏 𝑙𝑛 𝑏 𝑎 𝑬 𝝆 = 𝑽0 𝒍𝒏 𝒃 𝒂 1 𝝆 𝒂𝝆; 𝒂 < 𝝆 < 𝒃
  • 10. You can also know the electric flux density: 𝐷 = 𝜀𝐸 ) 𝐷 𝜌 = 𝜀𝐸(𝜌 𝑫 𝝆 = 𝜺𝑽0 𝒍𝒏 𝒃 𝒂 1 𝝆 𝒂𝝆; 𝒂 < 𝝆 < 𝒃 Once the electrostatic case was resolved, the equations were evaluated with values ​​between the internal and external radius. The proposed coaxial cable data is as follows: 𝑟𝑖 = 0.022 𝑚 𝑟𝑒 = 0.045 𝑚 𝜀𝑟 = 2.3 𝑉0 = 50𝑘[𝑉
  • 11. Tables 1, 2 and 3 show the values ​​obtained for the electric field density, the electric potential and the electric field respectively. 𝝆 𝑫 𝝆 0.022 6.46x10-5 0.025 5.68x10-5 0.030 4.74x10-5 0.032 4.44x10-5 0.035 4.06x10-5 0.040 3.55x10-5 0.042 3.38x10-5 0.045 3.16x10-5 𝝆 𝑽 𝝆 0.022 50000.00 0.025 41068.34 0.030 28329.63 0.032 23820.36 0.035 17559.20 0.040 8229.43 0.042 4820.49 0.045 0 Table 1.- Electric field density in the coaxial cable. Table 2.-Electrical Potential in the coaxial cable.
  • 12. Table 3.-Electrical Field in the coaxial cable. 𝝆 𝑬 𝝆 0.022 3175885.466 0.025 2794779.21 0.030 2328982.675 0.032 2183421.258 0.035 1996270.864 0.040 1746737.006 0.042 1663559.054 0.045 1552655.117
  • 13. Simulation using COMSOL Multiphysics. By solving the problem as an electrostatic case and giving the dimensions and physical characteristics of the problem, the following graphs of electric field and electric potential were obtained. Figure 1.3 shows the graph of the electric field norm that goes from a little more than 30M[V/m] to a little more than 1M[V/m]. But on the other hand, Figure 1.4 shows the electric potential that goes from 50k [V] to 0 [V]. Illustration 1.3.- Graph of the electric field in the coaxial cable.
  • 14. Illustration 1.4.- Graph of the electrical potential in the coaxial cable.
  • 15. Analytical vs simulated calculation. In order to compare the results obtained, the analytical equations were graphed in Excel. The electric potential graphs (fig. 1.4 and 1.5) and the electric field graph (fig. 1.3 and 1.6) are very similar since an extra fine mesh was used when simulating it. Illustration 1.5.- Analytical electrical potential graph in the coaxial cable.
  • 16. 0 500000 1000000 1500000 2000000 2500000 3000000 3500000 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 Campo Eléctrico (V/m) Illustration 1.6.- Analytical electric field graph of the coaxial cable
  • 17. Illustration 1.7.- Distribution of electrical potential in the coaxial cable.
  • 18. Illustration 1.8.- Distribution of the electric field in the coaxial cable.
  • 19. Conclusion. In solving this electrostatic problem, the advantage was that the problem could be reduced to a simple differential equation because only the radius of the coaxial cable varied. If the opposite were the case, perhaps the analytical solution would become more complex and difficult to solve since, although the differential equations that represent the physical behavior of the problem are obtained, it does not guarantee that an analytical solution exists, which is why simulation software is chosen. which uses numerical methods to give an approximate solution. In the case of COMSOL Multiphysic, it uses a mesh to solve the problem. If it uses a very fine mesh, the results are very close to the analytical one, but it takes more computing time. If a poor mesh is used, the results are far from the analytical one and uses less computing time. The distribution of the electric potential and the electric field is very similar since most of it is found on the surface of the central conductor and its intensity decreases until it disappears the closer it gets to the external conductor.