For a coaxial cable (a cylindrical capacitor), use the differentialform of divergence to show that the divergence of the E-field between theconductor and the shielding is zero.
Question
For a coaxial cable (a cylindrical capacitor), use the differentialform of divergence to show that the divergence of the E-field between theconductor and the shielding is zero.
Solution
To show that the divergence of the E-field between the conductor and the shielding is zero, we need to use Gauss's law in differential form, which states that the divergence of the electric field E is equal to the charge density ρ divided by the permittivity of free space ε₀.
The differential form of Gauss's law is given by:
∇ • E = ρ / ε₀
In a coaxial cable, the electric field E is directed radially and only depends on the radial distance r from the center of the cable. Therefore, in cylindrical coordinates (r, φ, z), the electric field can be written as:
E = E(r) * e_r
where e_r is the radial unit vector.
The divergence of E in cylindrical coordinates is given by:
∇ • E = (1/r) * ∂(rE) / ∂r
Substituting E = E(r) * e_r into this equation gives:
∇ • E = (1/r) * ∂(rE(r)) / ∂r
Since the electric field E(r) is constant between the conductor and the shielding (i.e., ∂E/∂r = 0), the divergence of E simplifies to:
∇ • E = 0
This shows that the divergence of the E-field between the conductor and the shielding is zero, as required.
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