Edits to “Statement”, “Solution”

Alexphysics edited Jun 16, 2026, 11:56 PM

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revision #18960 parent #18959 ← older newer →

en/12.1.15.md +47 −18

		@@ -1,17 +1,32 @@
			### Statement

		−	$12.1.15~~.$ [Insert the problem statement]~~
		+	$12.1.15 $

		−	### Solution
		+	Using the formula Eizl= ()E given in the problem 12.1.14. solve the fol-
		+	t−
		+	c
		+	c
		+	lowing problems:
		+	a. Determine the electric field strength in a plane wave emitted by a plane
		+	capacitor as it moves with a constant acceleration a directed parallel to its
		+	plates. The distance between the plates d, the electric field strength inside
		+	the capacitor E.
		+	b. The linear current density on the plate varies sinusoidally with the ampli-
		+	tude i0. Determine in SI and CGS the amplitude of the electric field strength
		+	in the wave emitted by this plate.
		+	c. Determine the reflection coefficient of an electromagnetic wave incident on
		+	a thin conducting film perpendicular to its surface. Film thickness x, number
		+	of conduction electrons per unit volume ne, wave frequency v.
		+
		+	### Solution

		−	~~For a charged plate moving with velocity v,~~ the radiation electric field at a distance x is
		+	the radiation electric field at a distance x is

			$E_{\text{rad}} = -\frac{v}{2c}\,E\!\left(t - \frac{x}{c}\right)$

			where E is the field inside the capacitor and c is the speed of light.
		−	So we start by parts

		−	---
		+	we start by parts

			a) Parallel‑plate capacitor with constant acceleration a

			Data:

			· Plate separation: d
			· Internal field: E
			· Acceleration: a (parallel to the plates)
			· Observation point A at distance x from the capacitor.
		@@ -24,18 +39,24 @@Solution

			Solution:

		−	1. Contribution from the upper plate (distance x):
		+	Contribution from the upper plate (distance x):

		−	E_1 = -\frac{v}{2c}\,E\!\left(t - \frac{x}{c}\right).
		−	2. Contribution from the lower plate (distance x + d, opposite charge)
		+	$E_1 = -\frac{v}{2c}\,E\!\left(t - \frac{x}{c}\right)$
		+
		+	Contribution from the lower plate (distance x + d, opposite charge)
		+
			$ E_2 = \frac{v}{2c}\,E\!\left(t - \frac{x+d}{c}\right)$

			The sign change is because the lower plate has charge -Q, which reverses the direction of the emitted field

		−	~~$ v(t) = a t.~~ The retarded times are ~~t - x/c and t - (x+d)/c$~~
		+	$ v(t) = a t$
		+
		+	The retarded times are

		−	$ E_1 = -\frac{a}{2c}\,E\left(t - \frac{x}{c}\right), \qquad$
		+	$t - x/c and t - (x+d)/c$

		+	$ E_1 = -\frac{a}{2c}\,E\left(t - \frac{x}{c}\right), \qquad$
		+
			$ E_2 = \frac{a}{2c}\,E\left(t - \frac{x+d}{c}\right)$

			$ E_{\text{rad}} = E_1 + E_2 = -\frac{aE}{2c}\left[\left(t - \frac{x}{c}\right) - \left(t - \frac{x+d}{c}\right)\right]$

			Simplifying:

			$ E_{\text{rad}} = -\frac{aE}{2c}\left( \frac{d}{c} \right) = -\frac{a d E}{2c^2}$

			$ \boxed{E_{\text{rad}} = \frac{a d E}{2c^2}}$

			---

			b) Plate with sinusoidal current

			Data:
			Linear current density on the plate: amplitude$ i_0$, sinusoidal variation.
			Find the amplitude of the electric field of the emitted wave, in SI and in CGS.
		@@ -56,23 +77,31 @@Solution

			Solution:

		−	1. Relation between surface current and emitted field:
		+	Relation between surface current and emitted field:

		−	A plate with surface current~~$ K (A/m) $~~
		+	A plate with surface current
		+	$ K (A/m) $
		+
			emits a plane wave whose electric field in the perpendicular direction is:
		+
			$ E_{\text{rad}} = \frac{\mu_0 c}{2}\,K$
		−	The amplitude of the surface current is $i_0.$
		−	2. Expression in SI:

		+	The amplitude of the surface current is $i_0.$
		+
		+	Expression in SI:
		+
			$ E_{\text{rad}} = \frac{\mu_0 c}{2}\,i_0
			$ Using $\mu_0 c = \frac{1}${c\varepsilon_0}$

			it can also be written as:

		−	$ \boxed{E_{\text{rad}} = \frac{i_0}{2c\varepsilon_0}} \quad (\text{SI})$
		−	3. Expression in CGS:
		−	In the Gaussian system, the same relation is:
		−	$ \boxed{E_{\text{rad}} = \frac{2\pi i_0}{c}} \quad (\text{CGS})$
		+	$\boxed{E_{\text{rad}} = \frac{i_0}{2c\varepsilon_0}} \quad (\text{SI})$
		+
		+	Expression in CGS:
		+
		+	In the Gaussian system, the same relation is:
		+
		+	$\boxed{E_{\text{rad}} = \frac{2\pi i_0}{c}} \quad (\text{CGS})$

			---

			c) Reflection coefficient of a thin conducting film

			Data:

			· Film thickness: x
			· Density of conduction electrons: $n_e$
			· Electron charge: e, mass:$ m_e$
			· Frequency of the incident wave:

			$\nu (\omega = 2\pi\nu)$

			Perpendicular incident wave:

			$ E = E_0 \sin\omega t$

			Solution:

			From the equation of motion

			$ m_e \dot{v} = e E_0 \sin\omega t:
			v(t) = -\frac{e E_0}{m_e \omega} \cos\omega t$

			Current density: $ j = n_e e v$.
			Thickness x → surface current (per unit width): $K = j x$

			Amplitude of this current

			$ K_0 = n_e e \cdot \frac{e E_0}{m_e \omega} \cdot x = \frac{n_e e^2 x}{m_e \omega} E_0 $

			As in part (b):

			$ E_{\text{rad}} = \frac{\mu_0 c}{2} K_0 = \frac{\mu_0 c}{2} \cdot \frac{n_e e^2 x}{m_e \omega} E_0$.

			Substituting

			$\mu_0 c = 1/(c\varepsilon_0) and \omega = 2\pi\nu$:
			$ E_{\text{rad}} = \frac{n_e e^2 x}{4\pi m_e \nu \varepsilon_0 c}\,E_0$

			$ k = \left( \frac{E_{\text{rad}}}{E_0} \right)^2$
			$\boxed{k = \left( \frac{n_e e^2 x}{4\pi m_e \nu \varepsilon_0 c} \right)^2}$

			#### Answer

			[Insert a concise answer or boxed result]
		unchanged lines 44

	@@ -1,17 +1,32 @@
	### Statement		### Statement

	$12.1.15~~.$ [Insert the problem statement]~~		$12.1.15 $

	### Solution		Using the formula Eizl= ()E given in the problem 12.1.14. solve the fol-
			t−
			c
			c
			lowing problems:
			a. Determine the electric field strength in a plane wave emitted by a plane
			capacitor as it moves with a constant acceleration a directed parallel to its
			plates. The distance between the plates d, the electric field strength inside
			the capacitor E.
			b. The linear current density on the plate varies sinusoidally with the ampli-
			tude i0. Determine in SI and CGS the amplitude of the electric field strength
			in the wave emitted by this plate.
			c. Determine the reflection coefficient of an electromagnetic wave incident on
			a thin conducting film perpendicular to its surface. Film thickness x, number
			of conduction electrons per unit volume ne, wave frequency v.

			### Solution

	~~For a charged plate moving with velocity v,~~ the radiation electric field at a distance x is		the radiation electric field at a distance x is

	$E_{\text{rad}} = -\frac{v}{2c}\,E\!\left(t - \frac{x}{c}\right)$		$E_{\text{rad}} = -\frac{v}{2c}\,E\!\left(t - \frac{x}{c}\right)$

	where E is the field inside the capacitor and c is the speed of light.		where E is the field inside the capacitor and c is the speed of light.
	So we start by parts

	---		we start by parts

	a) Parallel‑plate capacitor with constant acceleration a		a) Parallel‑plate capacitor with constant acceleration a

	Data:		Data:

	· Plate separation: d		· Plate separation: d
	· Internal field: E		· Internal field: E
	· Acceleration: a (parallel to the plates)		· Acceleration: a (parallel to the plates)
	· Observation point A at distance x from the capacitor.		· Observation point A at distance x from the capacitor.
	@@ -24,18 +39,24 @@Solution

	Solution:		Solution:

	1. Contribution from the upper plate (distance x):		Contribution from the upper plate (distance x):

	E_1 = -\frac{v}{2c}\,E\!\left(t - \frac{x}{c}\right).		$E_1 = -\frac{v}{2c}\,E\!\left(t - \frac{x}{c}\right)$
	2. Contribution from the lower plate (distance x + d, opposite charge)
			Contribution from the lower plate (distance x + d, opposite charge)

	$ E_2 = \frac{v}{2c}\,E\!\left(t - \frac{x+d}{c}\right)$		$ E_2 = \frac{v}{2c}\,E\!\left(t - \frac{x+d}{c}\right)$

	The sign change is because the lower plate has charge -Q, which reverses the direction of the emitted field		The sign change is because the lower plate has charge -Q, which reverses the direction of the emitted field

	~~$ v(t) = a t.~~ The retarded times are ~~t - x/c and t - (x+d)/c$~~		$ v(t) = a t$

			The retarded times are

	$ E_1 = -\frac{a}{2c}\,E\left(t - \frac{x}{c}\right), \qquad$		$t - x/c and t - (x+d)/c$

			$ E_1 = -\frac{a}{2c}\,E\left(t - \frac{x}{c}\right), \qquad$

	$ E_2 = \frac{a}{2c}\,E\left(t - \frac{x+d}{c}\right)$		$ E_2 = \frac{a}{2c}\,E\left(t - \frac{x+d}{c}\right)$

	$ E_{\text{rad}} = E_1 + E_2 = -\frac{aE}{2c}\left[\left(t - \frac{x}{c}\right) - \left(t - \frac{x+d}{c}\right)\right]$		$ E_{\text{rad}} = E_1 + E_2 = -\frac{aE}{2c}\left[\left(t - \frac{x}{c}\right) - \left(t - \frac{x+d}{c}\right)\right]$

	Simplifying:		Simplifying:

	$ E_{\text{rad}} = -\frac{aE}{2c}\left( \frac{d}{c} \right) = -\frac{a d E}{2c^2}$		$ E_{\text{rad}} = -\frac{aE}{2c}\left( \frac{d}{c} \right) = -\frac{a d E}{2c^2}$

	$ \boxed{E_{\text{rad}} = \frac{a d E}{2c^2}}$		$ \boxed{E_{\text{rad}} = \frac{a d E}{2c^2}}$

	---		---

	b) Plate with sinusoidal current		b) Plate with sinusoidal current

	Data:		Data:
	Linear current density on the plate: amplitude$ i_0$, sinusoidal variation.		Linear current density on the plate: amplitude$ i_0$, sinusoidal variation.
	Find the amplitude of the electric field of the emitted wave, in SI and in CGS.		Find the amplitude of the electric field of the emitted wave, in SI and in CGS.
	@@ -56,23 +77,31 @@Solution

	Solution:		Solution:

	1. Relation between surface current and emitted field:		Relation between surface current and emitted field:

	A plate with surface current~~$ K (A/m) $~~		A plate with surface current
			$ K (A/m) $

	emits a plane wave whose electric field in the perpendicular direction is:		emits a plane wave whose electric field in the perpendicular direction is:

	$ E_{\text{rad}} = \frac{\mu_0 c}{2}\,K$		$ E_{\text{rad}} = \frac{\mu_0 c}{2}\,K$
	The amplitude of the surface current is $i_0.$
	2. Expression in SI:

			The amplitude of the surface current is $i_0.$

			Expression in SI:

	$ E_{\text{rad}} = \frac{\mu_0 c}{2}\,i_0		$ E_{\text{rad}} = \frac{\mu_0 c}{2}\,i_0
	$ Using $\mu_0 c = \frac{1}${c\varepsilon_0}$		$ Using $\mu_0 c = \frac{1}${c\varepsilon_0}$

	it can also be written as:		it can also be written as:

	$ \boxed{E_{\text{rad}} = \frac{i_0}{2c\varepsilon_0}} \quad (\text{SI})$		$\boxed{E_{\text{rad}} = \frac{i_0}{2c\varepsilon_0}} \quad (\text{SI})$
	3. Expression in CGS:
	In the Gaussian system, the same relation is:		Expression in CGS:
	$ \boxed{E_{\text{rad}} = \frac{2\pi i_0}{c}} \quad (\text{CGS})$
			In the Gaussian system, the same relation is:

			$\boxed{E_{\text{rad}} = \frac{2\pi i_0}{c}} \quad (\text{CGS})$

	---		---

	c) Reflection coefficient of a thin conducting film		c) Reflection coefficient of a thin conducting film

	Data:		Data:

	· Film thickness: x		· Film thickness: x
	· Density of conduction electrons: $n_e$		· Density of conduction electrons: $n_e$
	· Electron charge: e, mass:$ m_e$		· Electron charge: e, mass:$ m_e$
	· Frequency of the incident wave:		· Frequency of the incident wave:

	$\nu (\omega = 2\pi\nu)$		$\nu (\omega = 2\pi\nu)$

	Perpendicular incident wave:		Perpendicular incident wave:

	$ E = E_0 \sin\omega t$		$ E = E_0 \sin\omega t$

	Solution:		Solution:

	From the equation of motion		From the equation of motion

	$ m_e \dot{v} = e E_0 \sin\omega t:		$ m_e \dot{v} = e E_0 \sin\omega t:
	v(t) = -\frac{e E_0}{m_e \omega} \cos\omega t$		v(t) = -\frac{e E_0}{m_e \omega} \cos\omega t$

	Current density: $ j = n_e e v$.		Current density: $ j = n_e e v$.
	Thickness x → surface current (per unit width): $K = j x$		Thickness x → surface current (per unit width): $K = j x$

	Amplitude of this current		Amplitude of this current

	$ K_0 = n_e e \cdot \frac{e E_0}{m_e \omega} \cdot x = \frac{n_e e^2 x}{m_e \omega} E_0 $		$ K_0 = n_e e \cdot \frac{e E_0}{m_e \omega} \cdot x = \frac{n_e e^2 x}{m_e \omega} E_0 $

	As in part (b):		As in part (b):

	$ E_{\text{rad}} = \frac{\mu_0 c}{2} K_0 = \frac{\mu_0 c}{2} \cdot \frac{n_e e^2 x}{m_e \omega} E_0$.		$ E_{\text{rad}} = \frac{\mu_0 c}{2} K_0 = \frac{\mu_0 c}{2} \cdot \frac{n_e e^2 x}{m_e \omega} E_0$.

	Substituting		Substituting

	$\mu_0 c = 1/(c\varepsilon_0) and \omega = 2\pi\nu$:		$\mu_0 c = 1/(c\varepsilon_0) and \omega = 2\pi\nu$:
	$ E_{\text{rad}} = \frac{n_e e^2 x}{4\pi m_e \nu \varepsilon_0 c}\,E_0$		$ E_{\text{rad}} = \frac{n_e e^2 x}{4\pi m_e \nu \varepsilon_0 c}\,E_0$

	$ k = \left( \frac{E_{\text{rad}}}{E_0} \right)^2$		$ k = \left( \frac{E_{\text{rad}}}{E_0} \right)^2$
	$\boxed{k = \left( \frac{n_e e^2 x}{4\pi m_e \nu \varepsilon_0 c} \right)^2}$		$\boxed{k = \left( \frac{n_e e^2 x}{4\pi m_e \nu \varepsilon_0 c} \right)^2}$

	#### Answer		#### Answer

	[Insert a concise answer or boxed result]		[Insert a concise answer or boxed result]
	unchanged lines 44