| the capacitor E. | | the capacitor E. |
| b. The linear current density on the plate varies sinusoidally with the ampli- | | 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 | | tude i0. Determine in SI and CGS the amplitude of the electric field strength |
| in the wave emitted by this plate. | | in the wave emitted by this plate. |
| c. Determine the reflection coefficient of an electromagnetic wave incident on | | c. Determine the reflection coefficient of an electromagnetic wave incident on |
| a thin conducting film perpendicular to its surface. Film thickness x, number | | a thin conducting film perpendicular to its surface. Film thickness x, number |
| of conduction electrons per unit volume ne, wave frequency v. | | of conduction electrons per unit volume ne, wave frequency v. |
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| ### Solution | | ### Solution |
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| the radiation electric field at a distance x is | | the radiation electric field at a distance x is |
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| $E_{\text{rad}} = -\frac{v}{2c}\,E\!\left(t - \frac{x}{c}\right)$ | | $E_{\text{rad}} = -\frac{v}{2c}\,E\!\left(t - \frac{x}{c}\right)$ |
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| 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. |
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| we start by parts | | we start by parts |
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| a) Parallel‑plate capacitor with constant acceleration a | | a) Parallel‑plate capacitor with constant acceleration a |
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| Data: | | Data: |
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| · 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. |
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| Solution: | | Solution: |
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| Contribution from the upper plate (distance x): | | Contribution from the upper plate (distance x): |
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| $E_1 = -\frac{v}{2c}\,E\!\left(t - \frac{x}{c}\right)$ | | $E_1 = -\frac{v}{2c}\,E\!\left(t - \frac{x}{c}\right)$ |
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| Contribution from the lower plate (distance x + d, opposite charge) | | Contribution from the lower plate (distance x + d, opposite charge) |
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| $ 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)$ |
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| 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 |
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| $ v(t) = a t$ | | $ v(t) = a t$ |
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| The retarded times are | | The retarded times are |
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| $t - x/c and t - (x+d)/c$ | | $t - x/c and t - (x+d)/c$ |
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| $ E_1 = -\frac{a}{2c}\,E\left(t - \frac{x}{c}\right), \qquad$ | | $ E_1 = -\frac{a}{2c}\,E\left(t - \frac{x}{c}\right), \qquad$ |
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| $ 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)$ |
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| $ 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]$ |
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| Simplifying: | | Simplifying: |
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| $ 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}$ |
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| $ \boxed{E_{\text{rad}} = \frac{a d E}{2c^2}}$ | | $ \boxed{E_{\text{rad}} = \frac{a d E}{2c^2}}$ |
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| --- | | --- |
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| b) Plate with sinusoidal current | | b) Plate with sinusoidal current |
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| 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. |
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| Solution: | | Solution: |
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| Relation between surface current and emitted field: | | Relation between surface current and emitted field: |
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| A plate with surface current | | A plate with surface current |
| $ K (A/m) $ | | $ K (A/m) $ |
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| emits a plane wave whose electric field in the perpendicular direction is: | | emits a plane wave whose electric field in the perpendicular direction is: |
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| $ E_{\text{rad}} = \frac{\mu_0 c}{2}\,K$ | | $ E_{\text{rad}} = \frac{\mu_0 c}{2}\,K$ |
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| The amplitude of the surface current is $i_0.$ | | The amplitude of the surface current is $i_0.$ |
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| Expression in SI: | | Expression in SI: |
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| $ 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}$ |
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| it can also be written as: | | it can also be written as: |
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| $\boxed{E_{\text{rad}} = \frac{i_0}{2c\varepsilon_0}} \quad (\text{SI})$ | | $\boxed{E_{\text{rad}} = \frac{i_0}{2c\varepsilon_0}} \quad (\text{SI})$ |
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| Expression in CGS: | | Expression in CGS: |
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| In the Gaussian system, the same relation is: | | In the Gaussian system, the same relation is: |
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| $\boxed{E_{\text{rad}} = \frac{2\pi i_0}{c}} \quad (\text{CGS})$ | | $\boxed{E_{\text{rad}} = \frac{2\pi i_0}{c}} \quad (\text{CGS})$ |
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| --- | | --- |
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| c) Reflection coefficient of a thin conducting film | | c) Reflection coefficient of a thin conducting film |
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| Data: | | Data: |
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| · 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: |
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| $\nu (\omega = 2\pi\nu)$ | | $\nu (\omega = 2\pi\nu)$ |
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| Perpendicular incident wave: | | Perpendicular incident wave: |
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| $ E = E_0 \sin\omega t$ | | $ E = E_0 \sin\omega t$ |
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| Solution: | | Solution: |
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| From the equation of motion | | From the equation of motion |
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| $ 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$ |
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| 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): $K = j x$ |
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| Amplitude of this current | | Amplitude of this current |
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| $ 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 $ |
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| As in part (b): | | As in part (b): |
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| $ 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$. |
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| Substituting | | Substituting |
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| $\mu_0 c = 1/(c\varepsilon_0) and \omega = 2\pi\nu$: | | $\mu_0 c = 1/(c\varepsilon_0) and \omega = 2\pi\nu$: |
| $ 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$ |
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| $ k = \left( \frac{E_{\text{rad}}}{E_0} \right)^2$ | | $ k = \left( \frac{E_{\text{rad}}}{E_0} \right)^2$ |
| $\boxed{k = \left( \frac{n_e e^2 x}{4\pi m_e \nu \varepsilon_0 c} \right)^2}$ | | $\boxed{k = \left( \frac{n_e e^2 x}{4\pi m_e \nu \varepsilon_0 c} \right)^2}$ |
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| #### Answer | | #### Answer |
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| [Insert a concise answer or boxed result] | | [Insert a concise answer or boxed result] |
