Home exercise 6.4.4: Parallel polarisation on a perfect conductor

Home exercise 6.4.4

Parallel polarisation on a perfect conductor

In the geometry of Problem 6.4.4 (perfect conductor at $z = 0$ , air in $z < 0$ ) the incident wave now has the magnetic field $H^{i} (x, z) = \frac{1}{6 π} e^{- j (6 x + 8 z)} a_{y} A/m$ .

Find $E^{i}$ and state the polarisation.
Find the reflected fields and the total fields in air.
Find the surface charge and current densities, and check $\nabla \cdot J_{S} = - j ω ρ_{S}$ .
Where could a second conducting plane be placed without disturbing the field?

Adapted from: D. K. Cheng, Field and Wave Electromagnetics, 2nd ed., Problem P.8-23.

Try it yourself first.

Strategy

The direction is that of Problem 6.4.4, $a_{n}^{i} = 0.6 a_{x} + 0.8 a_{z}$ , $β_{1} = 10 rad/m$ , $ω = 3 \times 1 0^{9} rad/s$ . Now $H$ is normal to the plane of incidence: P polarisation. Get $E$ from $E = η_{0} H \times a_{n}$ , choose the reflected wave so that the tangential $E_{x}$ vanishes on the wall, and read the surface sources from the boundary conditions $ρ_{S} = a_{n} \cdot ε_{0} E_{1}$ and $J_{S} = a_{n} \times H_{1}$ with $a_{n} = - a_{z}$ .

E^{i} = 120 π \times \frac{1}{6 π} a_{y} \times (0.6 a_{x} + 0.8 a_{z}) \times e^{- j (6 x + 8 z)} = 20 (0.8 a_{x} - 0.6 a_{z}) e^{- j (6 x + 8 z)},

E^{i} = (16 a_{x} - 12 a_{z}) e^{- j (6 x + 8 z)} V/m,

parallel (P) polarisation, $∣ E^{i} ∣ = 20 V/m$ .

The reflected wave travels along $a_{n}^{r} = 0.6 a_{x} - 0.8 a_{z}$ . $E_{x}^{r} (x, 0) = - 16 e^{- j 6 x}$ cancels the incident $E_{x}$ , and $E^{r} ⊥ a_{n}^{r}$ fixes its $z$ -component:

E^{r} H^{r} = - (16 a_{x} + 12 a_{z}) e^{- j (6 x - 8 z)} V/m, = \frac{1}{η _{0}} a_{n}^{r} \times E^{r} = \frac{1}{6 π} e^{- j (6 x - 8 z)} a_{y} A/m .

Adding,

E_{1} H_{1} = (- j 32 sin 8 z a_{x} - 24 cos 8 z a_{z}) \times e^{- j 6 x} V/m, = \frac{1}{3 π} cos 8 z e^{- j 6 x} a_{y} A/m .

At $z = 0$ : $E_{1} = - 24 e^{- j 6 x} a_{z}$ and $H_{1} = \frac{1}{3 π} e^{- j 6 x} a_{y}$ , so

ρ_{S} J_{S} = - a_{z} \cdot ε_{0} E_{1} = 24 ε_{0} e^{- j 6 x} = 0.212 e^{- j 6 x} nC/ m^{2}, = - a_{z} \times H_{1} = \frac{1}{3 π} e^{- j 6 x} a_{x} = 106 e^{- j 6 x} a_{x} mA/m .

Continuity:

\nabla \cdot J_{S} - j ω ρ_{S} = \frac{\partial J _{S x}}{\partial x} = - j \frac{6}{3 π} e^{- j 6 x} = - j \frac{2}{π} e^{- j 6 x}, = - j \times 3 \times 1 0^{9} \times \frac{24 \times 1 0 ^{- 9}}{36 π} e^{- j 6 x} = - j \frac{2}{π} e^{- j 6 x},

the same.

$E_{x} \propto sin 8 z$ vanishes on the planes $z = - mπ /8 = - 0.393, - 0.785, \dots m$ , and $H_{z} \equiv 0$ everywhere, so a conducting plane there meets only a normal $E$ and a tangential $H$ , as a perfect conductor requires.

Answer

E_{1} H_{1} ρ_{S} J_{S} = (- j 32 sin 8 z a_{x} - 24 cos 8 z a_{z}) \times e^{- j 6 x} V/m, = \frac{1}{3 π} cos 8 z e^{- j 6 x} a_{y} A/m, = 0.212 e^{- j 6 x} nC/ m^{2}, = 106 e^{- j 6 x} a_{x} mA/m, planes z = - mπ /8

Check

With the reference directions of Figure 6.4.10, $E_{∥}^{r} = Γ_{∥} E_{∥}^{i}$ and $Γ_{∥} = - 1$ for $η_{2} = 0$ : the tangential parts of $E^{i}$ and $E^{r}$ cancel and the normal parts add, $- 12 - 12 = - 24$ . Also $∣ J_{S} ∣ = 2∣ H^{i} ∣ = \frac{1}{3 π}$ : the full tangential $H$ doubles at the wall, at any angle.

Remark

Unlike the S wave of Problem 6.4.4, the P wave has an $E$ normal to the wall, so it carries surface charge, and that charge travels along $x$ with the current. Between the wall and a second plate at $z = - π /8$ this field is a TM mode of the parallel-plate waveguide, while the field of Problem 6.4.4 is a TE mode (Section 7.2).