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en/12.1.13.md
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| + | ### Statement | ||
| + | |||
| + | $12.1.13.$ [Insert the problem statement] | ||
| + | |||
| + | ### Solution | ||
| + | |||
| + | The capacitor has two parallel plates with opposite charges $\pm Q$ and area S. When oscillating along the x-axis (parallel to the plates) with velocity $v(t)$, each plate is equivalent to a surface current: | ||
| + | |||
| + | $K_+ = \sigma v(t)\,\hat{x}, \qquad K_- = -\sigma v(t)\,\hat{x}, \qquad \sigma = \frac{Q}{S}$. | ||
| + | |||
| + | These time-varying currents generate electromagnetic fields that propagate: the capacitor becomes a transmitting antenna. | ||
| + | |||
| + | Each plate emits electromagnetic waves. For a distant observer in the direction perpendicular to the plates (z-axis), the waves emitted by the front and back plates travel different paths. The path difference is the separation between the plates, d. | ||
| + | |||
| + | If the wavelength is $\lambda = c/\nu$ | ||
| + | the phase difference between the two waves at the observation point is: | ||
| + | |||
| + | $\Delta\phi = \frac{2\pi}{\lambda}\,d = \frac{2\pi\nu}{c}\,d$. | ||
| + | |||
| + | Constructive interference (intensity maximum): | ||
| + | It occurs when the waves arrive in phase: | ||
| + | $ \Delta\phi = 2\pi k \quad\Rightarrow\quad \frac{2\pi\nu d}{c} = 2\pi k \quad\Rightarrow\quad$ | ||
| + | |||
| + | $ \boxed{\nu_k' = \frac{c}{d}\,k},\quad k = 1,2,3,\dots$ | ||
| + | |||
| + | These are the frequencies of the emission maxima. | ||
| + | The first maximum $(k = 1) is \nu_1' = c/d$ | ||
| + | |||
| + | Destructive interference (zero intensity): | ||
| + | It occurs when the waves arrive in opposite phase: | ||
| + | $ \Delta\phi = (2k+1)\pi \quad\Rightarrow\quad \frac{2\pi\nu d}{c} = (2k+1)\pi \quad\Rightarrow\quad$ | ||
| + | $ \boxed{\nu_k'' = \frac{c}{d}\left(k + \frac{1}{2}\right)},\quad k = 0,1,2,\dots$ | ||
| + | |||
| + | At these frequencies, the contributions from the two plates cancel out and the capacitor emits no energy (in that main direction). That is why the average intensity drops to zero. | ||
| + | |||
| + | As the frequency \nu increases from zero, the phase difference \Delta\phi grows. For very low frequencies, | ||
| + | |||
| + | $\Delta\phi \approx 0$ | ||
| + | and there is reinforcement. When reaching | ||
| + | $\Delta\phi = \pi $(first cancellation,$ k = 0)$ | ||
| + | the intensity goes to zero. Then, upon reaching $ \Delta\phi = 2\pi (k = 1)$, | ||
| + | it becomes maximum again, and so on periodically. The intensity oscillates between high values and zeros, giving rise to the observed alternation. | ||
| + | |||
| + | #### Answer | ||
| + | |||
| + | [Insert a concise answer or boxed result] | ||
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| ### Statement | |||
| $12.1.13.$ [Insert the problem statement] | |||
| ### Solution | |||
| The capacitor has two parallel plates with opposite charges $\pm Q$ and area S. When oscillating along the x-axis (parallel to the plates) with velocity $v(t)$, each plate is equivalent to a surface current: | |||
| $K_+ = \sigma v(t)\,\hat{x}, \qquad K_- = -\sigma v(t)\,\hat{x}, \qquad \sigma = \frac{Q}{S}$. | |||
| These time-varying currents generate electromagnetic fields that propagate: the capacitor becomes a transmitting antenna. | |||
| Each plate emits electromagnetic waves. For a distant observer in the direction perpendicular to the plates (z-axis), the waves emitted by the front and back plates travel different paths. The path difference is the separation between the plates, d. | |||
| If the wavelength is $\lambda = c/\nu$ | |||
| the phase difference between the two waves at the observation point is: | |||
| $\Delta\phi = \frac{2\pi}{\lambda}\,d = \frac{2\pi\nu}{c}\,d$. | |||
| Constructive interference (intensity maximum): | |||
| It occurs when the waves arrive in phase: | |||
| $ \Delta\phi = 2\pi k \quad\Rightarrow\quad \frac{2\pi\nu d}{c} = 2\pi k \quad\Rightarrow\quad$ | |||
| $ \boxed{\nu_k' = \frac{c}{d}\,k},\quad k = 1,2,3,\dots$ | |||
| These are the frequencies of the emission maxima. | |||
| The first maximum $(k = 1) is \nu_1' = c/d$ | |||
| Destructive interference (zero intensity): | |||
| It occurs when the waves arrive in opposite phase: | |||
| $ \Delta\phi = (2k+1)\pi \quad\Rightarrow\quad \frac{2\pi\nu d}{c} = (2k+1)\pi \quad\Rightarrow\quad$ | |||
| $ \boxed{\nu_k'' = \frac{c}{d}\left(k + \frac{1}{2}\right)},\quad k = 0,1,2,\dots$ | |||
| At these frequencies, the contributions from the two plates cancel out and the capacitor emits no energy (in that main direction). That is why the average intensity drops to zero. | |||
| As the frequency \nu increases from zero, the phase difference \Delta\phi grows. For very low frequencies, | |||
| $\Delta\phi \approx 0$ | |||
| and there is reinforcement. When reaching | |||
| $\Delta\phi = \pi $(first cancellation,$ k = 0)$ | |||
| the intensity goes to zero. Then, upon reaching $ \Delta\phi = 2\pi (k = 1)$, | |||
| it becomes maximum again, and so on periodically. The intensity oscillates between high values and zeros, giving rise to the observed alternation. | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||