Section 5.5 · Problems · 2 of 6

Problem 5.5.2: Relative permittivity from the wave equation

Problem 5.5.2

Relative permittivity from the wave equation

If the field

E = 45 sin (1 0^{9} t) cos (5 z) a_{y} V/m

is present in a nonmagnetic, lossless and source-free medium, find the relative permittivity using the wave equation. Also write $E$ as a sum of two travelling waves, and find the magnetic field $H$ from Faraday's law.

Try it yourself first.

Strategy

$E_{y}$ is a Cartesian component, so it obeys the scalar wave equation Eq. (5.5.14). Each derivative of a sine or a cosine gives back the function times a constant, so the wave equation becomes one algebraic condition on $μ ε$ . Faraday's law then gives $H$ by one integration in time (the recipe of Section 5.5.3).

1. Divergence

$E_{y}$ depends on $z$ and $t$ only, so $\nabla \cdot E = \partial E_{y} / \partial y = 0$ , as a source-free field must.

2. Wave equation

With $μ = μ_{0}$ and $ε = ε_{r} ε_{0}$ ,

\frac{\partial ^{2} E _{y}}{\partial z ^{2}} - μ_{0} ε_{0} ε_{r} \frac{\partial ^{2} E _{y}}{\partial t ^{2}} = 0, \frac{\partial ^{2} E _{y}}{\partial z ^{2}} = - 25 E_{y}, \frac{\partial ^{2} E _{y}}{\partial t ^{2}} = - 1 0^{18} E_{y},

so that

- 25 + μ_{0} ε_{0} ε_{r} \times 1 0^{18} = 0 ⟹ ε_{r} = \frac{25 c ^{2}}{1 0 ^{18}} = \frac{25 \times 9 \times 1 0 ^{16}}{1 0 ^{18}} .

Answer

ε_{r} = 2.25, v = \frac{ω}{k} = \frac{1 0 ^{9}}{5} = 2 \times 1 0^{8} m/s

3. Two travelling waves

With $sin A cos B = \frac{1}{2} [sin (A - B) + sin (A + B)]$ ,

E = 22.5 [sin (1 0^{9} t - 5 z) + sin (1 0^{9} t + 5 z)] a_{y} V/m,

two waves of amplitude $22.5 V/m$ , one towards $+ z$ and one towards $- z$ , which together form a standing wave. The frequency is $f = 1 0^{9} / (2 π) = 159.2 MHz$ and the wavelength $λ = 2 π /5 = 1.257 m$ .

4. Magnetic field

For a field with only a $y$ component depending on $z$ , $\nabla \times E = - a_{x} \partial E_{y} / \partial z = 225 sin (1 0^{9} t) sin (5 z) a_{x}$ , and Faraday's law $\nabla \times E = - μ_{0} \partial H / \partial t$ gives, integrating in time without a static part,

H = \frac{225}{μ _{0} \times 1 0 ^{9}} cos (1 0^{9} t) sin (5 z) a_{x} .

Answer

H = 0.179 cos (1 0^{9} t) sin (5 z) a_{x} A/m

Check

Each travelling part has $H = E / η$ with $η = 120 π / 2.25 = 251 Ω$ (Exercise 5.1.1), so $22.5/251 = 0.0895 A/m$ , and the two parts add to $2 \times 0.0895 = 0.179 A/m$ at the maxima of $sin 5 z$ . The generalised Ampère law holds too: $\nabla \times H = a_{y} \partial H_{x} / \partial z = 0.895 cos (1 0^{9} t) cos (5 z) a_{y}$ and $ε \partial E_{y} / \partial t = 2.25 ε_{0} \times 45 \times 1 0^{9} cos (1 0^{9} t) cos (5 z) = 0.895 cos (1 0^{9} t) cos (5 z)$ .

(a) E_y at three instants. — (a) $E_{y}$ at three instants.

(b) H_x at three instants. — (a) $E_{y}$ at three instants.

Remark

The wave equation fixes only the product $μ ε$ , which is why the problem has to say that the medium is nonmagnetic. $H$ is a quarter period and a quarter wavelength away from $E$ (Figure 5.5.14): where $E_{y}$ has its maxima ( $5 z = nπ$ ), $H_{x}$ is zero. The energy sloshes between $E$ and $H$ inside each quarter wavelength and none is carried on average, as for any standing wave (Section 6.3).

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