Tutorial 5: Worked solutions

9a · Qualitative field

1. The metal cylinders are equipotentials. Away from the finite ends, the field meets each cylindrical conductor normally, so its direction is radial, along e_ρ.

2. The voltage is U₁₀ = φ_e,outer − φ_e,inner. If U₁₀ > 0, the outer conductor is at higher potential, so E points inward (−e_ρ) from the outer cylinder toward the inner one. Reverse the arrows if U₁₀ < 0.

3. Draw radial lines from radius b to radius a, with denser spacing near the inner cylinder because the field magnitude varies as 1/ρ. Neglect end fringing as the ideal long-coax model requires.

9b · Boundary conditions

1. A constant-φ interface has normal n = ±e_φ. The radial field is perpendicular to that normal, so it lies tangent to the interface:

   E_n,1 = E_n,2 = 0,    E_t,1 = E_t,2

2. With no free charge on the dielectric boundary, the normal component of displacement is continuous; it is zero on both sides here. Tangential displacement need not be continuous and scales with each material permittivity:

   D_t,1D_t,2 = ε₁ε₂

3. Thus E crosses the angular boundary without a change in direction or magnitude. D changes magnitude at the boundary in proportion to ε. The radial field lines run along the interface, rather than crossing it.

9c · Radial solution

1. The coaxial geometry and constant conductor potentials allow a z- and φ-independent potential in each sector. The cylindrical Laplace equation reduces to:

   1ρ · ddρ(ρ dφ_edρ) = 0

   φ_e(ρ) = A ln ρ + B

2. Apply φ_e(a) = φ_e,0 and φ_e(b) = φ_e,1. The same solution works in both sectors because their conductor boundary values are identical and the radial field is tangent to the interface:

   φ_e(ρ) = φ_e,0 + U₁₀ln(b/a) ln(ρa)

3. The electric field is the negative potential gradient:

   E(ρ) = −U₁₀ρ ln(b/a) e_ρ

   The 1/ρ dependence is required by cylindrical symmetry. Positive U₁₀ gives the inward direction.

9d · Piecewise permittivity

1. For a linear, isotropic dielectric, D = εE. The permittivity depends on which half-annulus contains the point:

   D(ρ, φ) = ε₁ E(ρ)     for 90° < φ < 270°

   D(ρ, φ) = ε₂ E(ρ)     for −90° < φ < 90°

2. Because E is the same on both sides of the radial interface, the displacement magnitudes differ by ε₁/ε₂. At φ = ±90° the normal displacement remains zero and continuous.

9e · Outer-conductor charge sign

1. For U₁₀ > 0, the outer cylinder’s potential exceeds the inner cylinder’s. The field points from the outer conductor toward the inner conductor.

2. The normal from the metal into the dielectric at the outer cylinder’s inner wall is −e_ρ, parallel to E. The conductor surface charge density there is D · (−e_ρ) > 0. Therefore the net charge on the outer cylinder is positive.

9f · Charge fixes the field scale

1. Let the inward field have magnitude K/ρ, so D in sector i has magnitude ε_iK/ρ. The normal from the outer metal into the gap is −e_ρ; its facing area is half the cylinder, πb l, in each sector:

   Q = ε₁ Kb(πbl) + ε₂ Kb(πbl)

2. Solve for the field constant and substitute into the radial field:

   K = Qπl(ε₁ + ε₂)

   E(ρ) = −Qπl(ε₁ + ε₂) ρ e_ρ

   This is uniform in φ although D is piecewise. The sign of Q sets the field direction.

9g · Charge–voltage relation

1. Because the field points inward, the outer-minus-inner voltage is U₁₀ = −∫_a^b E · e_ρ dρ:

   U₁₀ = K ∫_a^b dρρ = K ln(b/a)

2. Insert K from 9f:

   U₁₀ = Q ln(b/a)πl(ε₁ + ε₂)

   Positive Q gives positive U₁₀, consistent with 9e.

9h · Coaxial capacitance

1. For this linear arrangement, capacitance is the charge magnitude per applied voltage. Cancel Q using 9g:

   C = πl(ε₁ + ε₂)ln(b/a)

2. With ε₁ = ε₀, ε₂ = 3ε₀, a = 0.020 m, b = 0.021 m and l = 0.10 m:

   C = 4πε₀ · 0.10 mln(0.021/0.020)

   C ≈ 2.28 × 10⁻¹⁰ F = 228 pF

   Check: a uniform ε₀ gap would have half this capacitance, since the mean permittivity here is 2ε₀. The result is also positive and has units of farads.