| @@ -1,153 +1,36 @@ |
| | | ### Statement |
| | | |
| | − | $12.1.15 $ |
| | + | $12.1.15^\ast$. Using the formula $E_{\text{rad}} = (v_{t-x/c}/c)E$ given in problem 12.1.14, solve the following problems:\ |
| | + | <b>a.</b> 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 is $d$, the electric field strength inside the capacitor is $E$.\ |
| | + | <b>b.</b> The linear current density on the plate varies sinusoidally with amplitude $i_0$. Determine in SI and CGS the amplitude of the electric field strength in the wave emitted by this plate.\ |
| | + | <b>c.</b> 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 $n_e$, wave frequency $\nu$. |
| | | |
| | − | Using the formula given in the problem 12.1.14. solve the following |
| | + | ### Solution |
| | | |
| | − | 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 |
| | + | <b>a)</b> The field inside the capacitor is $E$. Each of the two plates individually creates a field $E_0 = E/2$. The plates are oppositely charged, so their emitted fields have opposite signs. |
| | + | Let the plate closer to the observer be at distance $x$, and the farther one at $x+d$. According to the formula, the total radiated field is: |
| | + | $$E_{\text{rad}} = \frac{1}{c} \left( v_{t - \frac{x+d}{c}} \frac{E}{2} - v_{t - \frac{x}{c}} \frac{E}{2} \right) = \frac{E}{2c} \left( v\left(t - \frac{x+d}{c}\right) - v\left(t - \frac{x}{c}\right) \right)$$ |
| | + | Since the capacitor moves with constant acceleration $a$, then $v(t) = at$. The difference in velocities for times differing by $\Delta t = d/c$ is $\Delta v = a \frac{d}{c}$. |
| | + | Then the magnitude of the radiated field is: |
| | + | $$E_{\text{rad}} = \frac{E}{2c} \left( a \frac{d}{c} \right) = \frac{a d E}{2 c^2}$$ |
| | | |
| | − | the radiation electric field at a distance x is |
| | + | <b>b)</b> The surface current $i = i_0 \sin(\omega t)$ creates a magnetic field $B$ near the plate. |
| | + | In <b>SI</b>: By Ampere's law, $B = \frac{\mu_0 i}{2}$. The emitted electric field is related to the magnetic field as $E_{\text{rad}} = c B = \frac{c \mu_0 i_0}{2}$. Since $c^2 = \frac{1}{\mu_0 \varepsilon_0}$, the amplitude is: |
| | + | $$E_{\text{rad}} = \frac{i_0}{2 \varepsilon_0 c}$$ |
| | + | In <b>CGS</b>: The magnetic field of the current is $B = \frac{2\pi i}{c}$. The emitted electric field is $E_{\text{rad}} = B$, hence the amplitude is: |
| | + | $$E_{\text{rad}} = \frac{2\pi i_0}{c}$$ |
| | | |
| | − | $E_{\text{rad}} = -\frac{v}{2c}\,E\!\left(t - \frac{x}{c}\right)$ |
| | + | <b>c)</b> The incident wave $E_{\text{inc}} = E_0 \cos(2\pi\nu t)$ forces the film's electrons to move. The equation of motion for an electron is $m_e \dot{v} = e E_0 \cos(2\pi\nu t)$. |
| | + | Electron velocity: $v(t) = \frac{e E_0}{m_e \cdot 2\pi\nu} \sin(2\pi\nu t)$. |
| | + | The current density in the film is $j = n_e e v$, and the surface current (current per unit width) is $K = j x$: |
| | + | $$K(t) = \frac{n_e e^2 x E_0}{2\pi m_e \nu} \sin(2\pi\nu t)$$ |
| | + | Amplitude of the surface current $K_0 = \frac{n_e e^2 x E_0}{2\pi m_e \nu}$. |
| | + | The film acts as an emitting plate from part (b). The amplitude of the reflected wave (in SI) is: |
| | + | $$E_{\text{ref}} = \frac{K_0}{2 \varepsilon_0 c} = \frac{n_e e^2 x E_0}{4\pi \varepsilon_0 c m_e \nu}$$ |
| | + | The reflection coefficient $R$ is the ratio of the intensities (squares of the field amplitudes) of the reflected and incident waves: |
| | + | $$R = \left( \frac{E_{\text{ref}}}{E_0} \right)^2 = \left( \frac{n_e e^2 x}{4 \pi \varepsilon_0 m_e c \nu} \right)^2$$ |
| | | |
| | − | where E is the field inside the capacitor and c is the speed of light. |
| | − | |
| | − | 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. |
| | − | |
| | − | Solution: |
| | − | |
| | − | Contribution from the upper plate (distance x): |
| | − | |
| | − | $E_1 = -\frac{v}{2c}\,E\!\left(t - \frac{x}{c}\right)$ |
| | − | |
| | − | 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 |
| | − | |
| | − | $t - x/c and t - (x+d)/c$ |
| | − | |
| | − | $ 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. |
| | − | |
| | − | Solution: |
| | − | |
| | − | Relation between surface current and emitted field: |
| | − | |
| | − | 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 $i_0.$ |
| | − | |
| | − | 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: |
| | − | |
| | − | $\boxed{E_{\text{rad}} = \frac{i_0}{2c\varepsilon_0}} \quad (\text{SI})$ |
| | − | |
| | − | 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] |
| | + | a. $E_{\text{rad}} = \frac{a d E}{2 c^2}$\ |
| | + | b. $E_{\text{rad}} = \frac{i_0}{2 \varepsilon_0 c}$ (SI); $E_{\text{rad}} = \frac{2\pi i_0}{c}$ (CGS)\ |
| | + | c. $R = \left( \frac{n_e e^2 x}{4 \pi \varepsilon_0 m_e c \nu} \right)^2$ |
