EEI

Exercise 4: Worked solutions

Electrical Engineering I · Module 92315 · Exercise 4

Try each part before opening the answer. We apply the interface conditions to the measured field, then use cylindrical Gauss surfaces and voltage integration for the layered coaxial geometry.

Take the solutions offline

Download clean PDFDownload LaTeX source (.tex)

Task 9: Field at a flat dielectric boundary

An infinite line charge of density λ0 lies along the z-axis. The lower half-space y < h is air with permittivity ε0, and the upper half-space y ≥ h has permittivity ε1. In the air, measurements give the field below.

E0(x,y) = λ0π(ε0 + ε1)(x2 + y2) [xex + ε1ε0yey]

Connected topic: Material boundary conditions (§1.11) keep tangential E continuous and normal D continuous when no free surface charge is present.

Resolve the measured field into interface-parallel and interface-normal parts. The x-component stays continuous; the y-component changes with permittivity.

9: Electric field in the dielectric at y = h

Try first: evaluate the measured air field at the interface, then apply the tangential-E and normal-D boundary conditions component by component.

Attempted it? Open the worked steps

9 · Apply both interface conditions

  1. The interface is horizontal, so its normal is along y and its tangent is along x. There is no free charge sheet in the problem. Therefore:

    E1x = E0x;   D1y = D0y,   equivalently ε1E1y = ε0E0y

  2. First evaluate the given air field at y = h. Its tangential component is:

    E0x(x,h) = λ0xπ(ε0 + ε1)(x2 + h2)

  3. Keep that x-component unchanged across the boundary:

    E1x(x,h) = E0x(x,h)

  4. The measured normal component contains the factor ε1/ε0. Multiplying it by ε0/ε1 to preserve D cancels that factor:

    E1y(x,h) = ε0ε1E0y(x,h) = λ0hπ(ε0 + ε1)(x2 + h2)

  5. Result: combine the two components into the dielectric-side vector:

    E1(x,h) = λ0π(ε0 + ε1)(x2 + h2) [xex + hey]

    Check: the tangential component is unchanged; the normal flux density matches on both sides. With SI inputs, E is in V/m (equivalently N/C).

Task 10: Two-dielectric coaxial capacitor

Two infinitely long concentric metal cylinders share the z-axis. The inner metal’s outer radius is a0; the outer metal’s inner radius is a2 > a0. Between them, permittivity is ε1 for a0 < ρ < a1 and ε2 for a1 < ρ < a2. The conductor walls each have thickness d; outside the conductors is material ε0. The only charges are σ0 at the inner conductor’s outer surface and σ2 at the outer conductor’s inner surface.

Connected topics: Gauss’s law (§1.10) finds the radial flux; capacitance (§1.13) follows from charge per length divided by voltage.

Track each cylindrical boundary by radius. The field runs radially through ε1 then ε2; free surface charges sit on the two metal faces bordering the dielectric gap.

In the dielectric gap, symmetry gives D = Dρ(ρ)eρ. The sheet fixes the voltage direction: U20 = ∫ from P2 (outer) to P0 (inner) of E · ds, matching the figure’s arrow from the outer to the inner terminal and the course convention U20 = φ2 − φ0. The path therefore runs inward.

10a: Flux density, field, and outer surface charge

Try first: apply cylindrical Gauss surfaces in each dielectric, then use continuity of normal D at ρ = a1 and the conductor boundary at ρ = a2.

Attempted it? Open the worked steps

10a · Use enclosed charge per unit length

  1. Choose a coaxial Gaussian cylinder of radius ρ and length Δz, with a0 < ρ < a2. Radial D is constant over its curved side; its end caps contribute no flux. The enclosed free charge is the inner surface charge on length Δz:

    Dρ(ρ) 2πρ Δz = σ0 2πa0 Δz

  2. Cancel the common factors. No free charge lies at the dielectric boundary ρ = a1, so normal D has the same value in both layers:

    D(ρ) = σ0a0ρ eρ,   a0 < ρ < a2

  3. Use D = εE separately in each material:

    E(ρ) = σ0a0ε1ρeρ,   a0 < ρ < a1

    E(ρ) = σ0a0ε2ρeρ,   a1 < ρ < a2

  4. At the inner face of the outer metal cylinder, the normal from the conductor into the gap points inward, −eρ. Thus σ2 = D · (−eρ) at ρ = a2:

    σ2 = −σ0a0a2

    Result: D is radial and continuous through the dielectric interface; E changes magnitude inversely with permittivity. The two metal-face charges have opposite signs.

    Check: σ2 2πa2Δz = −σ0 2πa0Δz, so the two facing surfaces carry equal and opposite charge.

10b: Field and σ0 as functions of U20

Try first: integrate E along the path the sheet gives, inward from P2 at a2 to P0 at a0, splitting at a1, then solve the voltage relation for σ0.

Attempted it? Open the worked steps

10b · Integrate through both layers

  1. The path runs inward, from ρ = a2 to ρ = a0. Along a radial path ds = eρ dρ, and the limits carry the direction, so E · ds = Eρ dρ with ρ decreasing. Insert the two E expressions from 10a:

    U20 = ∫a2a1 σ0a0ε2ρ dρ + ∫a1a0 σ0a0ε1ρ dρ

  2. Each integral of 1/ρ gives a logarithm; running inward makes each one negative, for example ∫ from a1 to a0 of dρ/ρ = ln(a0/a1) = −ln(a1/a0):

    U20 = −σ0a0[1ε1 lna1a0 + 1ε2 lna2a1]

  3. Define the positive geometry/material factor:

    S = 1ε1 lna1a0 + 1ε2 lna2a1

  4. Rearrange for the inner surface charge density, then substitute it back into D and E:

    σ0 = −U20a0S;   D(ρ) = −U20Sρeρ

    E(ρ) = −U20ε1Sρeρ,   a0 < ρ < a1

    E(ρ) = −U20ε2Sρeρ,   a1 < ρ < a2

  5. Result: for positive U20 the outer cylinder is at the higher potential, so σ0 < 0, σ2 = −σ0a0/a2 > 0, and the field points inward, from the outer cylinder to the inner one.

    Check: the field points from high to low potential, and the outer cylinder (node 2) is high when U20 > 0, so an inward field is right. Integrating the resulting E inward from a2 to a0 returns +U20. Likely mistake: integrating outward from P0 to P2, which flips every sign. The field has units V/m and σ0 has units C/m².

10c: Capacitance for length Δz

Try first: use the inner conductor’s charge over length Δz and divide by the voltage from 10b.

Attempted it? Open the worked steps

10c · Capacitance per unit length

  1. Capacitance is the charge on the electrode at the higher potential divided by the voltage. For U20 > 0 that is the outer cylinder, whose facing surface carries Q = σ2 2πa2Δz = −σ0 2πa0Δz:

    C = QU20 = −σ02πa0ΔzU20

  2. Substitute σ0 = −U20/(a0S). The signs, the voltage and a0 cancel:

    C = 2πΔzS;   CB = CΔz = 2πS

  3. Result: C is the capacitance over length Δz; CB is the capacitance loading per unit length.

    Check: S has units m/F, so C is in farads and CB in F/m. If ε1 = ε2 = ε, then S = ln(a2/a0)/ε, recovering the homogeneous coaxial result.

10d: Field inside the inner conductor and outside the outer one

Try first: use Gauss’s law and the conductor condition D = 0, remembering that the two facing surface charges cancel per unit length.

Attempted it? Open the worked steps

10d · Zero field outside the charged gap

  1. For 0 ≤ ρ < a0, a coaxial Gaussian cylinder encloses no free charge: σ0 lies at ρ = a0, outside this Gaussian surface. Thus its radial flux and field are zero. Within the inner metal wall D is also zero by the conductor condition.

  2. For ρ > a2, a Gaussian cylinder outside both facing surfaces encloses their total charge per length:

    qencΔz = 2πa0σ0 + 2πa2σ2 = 0

    The cancellation uses σ2 = −σ0a0/a2. Therefore D and E vanish outside the outer conductor; they also vanish within its metal wall.

  3. Result:

    E = 0,   0 ≤ ρ < a0

    E = 0,   ρ > a2

    Check: the applied voltage and field are confined to the annular dielectric gap in this ideal infinitely long, concentric arrangement.

Personal worked-solution guide based on the supplied EEI Exercise 4 sheet, written in my own words; no exercise text or figures are reproduced. Not an official FAU publication; check course details against current material. Made by Thanate Amornjiraporn (Focus).