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| | | ### Statement |
| | | |
| | − | $12.1.13.$ |
| | + | $12.1.13^\ast$. A charged flat capacitor is swung by moving it parallel to the plates. As the oscillation frequency $\nu$ increases, the average intensity $I$ of the electromagnetic waves emitted by the capacitor first increases, then decreases to zero, then increases again, and so on. How can this alternation of radiation intensity be explained? At what frequencies does the capacitor not emit energy? Estimate the frequency at which the 1st and $k$-th emission maxima are observed. |
| | | |
| | − | A charged flat capacitor is swung by moving it parallel to the plates. As the os- |
| | − | cillation frequency ν increases, the average intensity I of the electromagnetic |
| | − | waves emitted by the capacitor first increases, then decreases to zero, then |
| | − | increases again, and so on. At what frequencies does the capacitor not emit |
| | − | energy? Estimate the frequency at which the 1st and $kth$ emission maxima |
| | − | are observed. |
| | + |  |
| | | |
| | | ### 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: |
| | + | We can represent the capacitor plates as two planes with surface charges $+\sigma$ and $-\sigma$. When moving with velocity $v(t)$, they create surface currents $j_1 = \sigma v(t)$ and $j_2 = -\sigma v(t)$. |
| | + | Since the charges are opposite, the currents are directed in opposite directions. This means the electromagnetic waves emitted by them have an intrinsic initial phase difference of $\pi$ (they are emitted in anti-phase). |
| | | |
| | − | $K_+ = \sigma v(t)\,\hat{x}, \qquad K_- = -\sigma v(t)\,\hat{x}, \qquad \sigma = \frac{Q}{S}$ |
| | + | For a distant observer located on a line perpendicular to the plates, the waves from the two plates travel different distances. The path difference is equal to the distance between the plates, $d$. |
| | + | This path difference creates an additional phase shift $\Delta\varphi = \frac{2\pi d}{\lambda} = \frac{2\pi\nu d}{c}$. |
| | | |
| | − | These time-varying currents generate electromagnetic fields that propagate: the capacitor becomes a transmitting antenna. |
| | + | The total intensity of the emitted radiation is determined by the interference of these two waves. Since they start in anti-phase, the total phase difference is $\Phi = \pi + \frac{2\pi\nu d}{c}$. |
| | + | <b>Destructive interference</b> (zero intensity) occurs when the path delay $d$ exactly compensates for the anti-phase nature of the sources. Mathematically, the intensity is proportional to $\sin^2\left(\frac{\pi\nu d}{c}\right)$. The capacitor <b>does not emit</b> energy when this term is zero: |
| | + | $$\frac{\pi\nu d}{c} = k\pi \implies \nu_k = k\frac{c}{d}, \quad \text{where } k = 1, 2, 3, \dots$$ |
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| | − | 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. |
| | + | <b>Constructive interference</b> (intensity maxima) occurs when the path difference corresponds to an odd number of half-wavelengths, reinforcing the anti-phased sources: |
| | + | $$\frac{\pi\nu d}{c} = \left(k - \frac{1}{2}\right)\pi \implies \nu_k \approx \left(k - \frac{1}{2}\right)\frac{c}{d}, \quad \text{where } k = 1, 2, 3, \dots$$ |
| | + | For the 1st maximum ($k=1$): $\nu_1 \approx \frac{c}{2d}$. |
| | | |
| | − | If the wavelength is |
| | − | $\lambda = c/\nu$ |
| | + | <i>Note: The alternation is explained by the interference of waves from the two plates. The fact that the peaks on the graph get progressively higher is because the radiation intensity of an accelerating charge itself grows proportionally to the square of the frequency ($\nu^2$).</i> |
| | | |
| | − | 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 |
| | − | |
| | − | $\boxed{\nu_k' = \frac{c}{d}\,k},\quad k = 1,2,3,\dots$ |
| | − | |
| | − | $\boxed{\nu_k'' = \frac{c}{d}\left(k + \frac{1}{2}\right)},\quad k = 0,1,2,\dots$ |
| | + | Does not emit at $\nu = k\frac{c}{d}$;\ |
| | + | maxima at $\nu_k \approx \left(k - \frac{1}{2}\right)\frac{c}{d}$, where $k = 1, 2, 3, \dots$ |
