Новое решение
en/14.4.10.md
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| + | ### Statement | ||
| + | |||
| + | $14.4.10.$ [Insert the problem statement] | ||
| + | |||
| + | ### Solution | ||
| + | |||
| + | The particle is born at rest inside a sinusoidal electric field $\mathcal{E}(t) = E \sin(\omega t)$ | ||
| + | The electric force F = qE \sin(\omega t) modifies its relativistic momentum | ||
| + | |||
| + | $ p = \dfrac{mv}{\sqrt{1 - v^2/c^2}} $ | ||
| + | |||
| + | according to Newton's second law | ||
| + | |||
| + | $\frac{dp}{dt} = qE \sin(\omega t)$ | ||
| + | |||
| + | Integrating with the initial condition $p(0) = 0$ we obtain | ||
| + | |||
| + | $p(t) = \frac{qE}{\omega}\bigl[1 - \cos(\omega t)\bigr]$ | ||
| + | |||
| + | whose maximum value is reached when $\cos(\omega t) = -1$ | ||
| + | |||
| + | $p_{\max} = \frac{2qE}{\omega}$ | ||
| + | |||
| + | From the definition of p, the velocity is solved for in terms of momentum: | ||
| + | |||
| + | $v = \frac{p/m}{\sqrt{1 + \dfrac{p^2}{m^2 c^2}}}$ | ||
| + | |||
| + | Substituting$ p_{\max}$ and simplifying algebraically, we arrive at the maximum velocity, which is: | ||
| + | |||
| + | $\boxed{v_{\max} = \frac{c}{\sqrt{1 + \left(\dfrac{m c \omega}{2 q E}\right)^2}}}$ | ||
| + | |||
| + | #### Answer | ||
| + | |||
| + | [Insert a concise answer or boxed result] | ||
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| ### Statement | |||
| $14.4.10.$ [Insert the problem statement] | |||
| ### Solution | |||
| The particle is born at rest inside a sinusoidal electric field $\mathcal{E}(t) = E \sin(\omega t)$ | |||
| The electric force F = qE \sin(\omega t) modifies its relativistic momentum | |||
| $ p = \dfrac{mv}{\sqrt{1 - v^2/c^2}} $ | |||
| according to Newton's second law | |||
| $\frac{dp}{dt} = qE \sin(\omega t)$ | |||
| Integrating with the initial condition $p(0) = 0$ we obtain | |||
| $p(t) = \frac{qE}{\omega}\bigl[1 - \cos(\omega t)\bigr]$ | |||
| whose maximum value is reached when $\cos(\omega t) = -1$ | |||
| $p_{\max} = \frac{2qE}{\omega}$ | |||
| From the definition of p, the velocity is solved for in terms of momentum: | |||
| $v = \frac{p/m}{\sqrt{1 + \dfrac{p^2}{m^2 c^2}}}$ | |||
| Substituting$ p_{\max}$ and simplifying algebraically, we arrive at the maximum velocity, which is: | |||
| $\boxed{v_{\max} = \frac{c}{\sqrt{1 + \left(\dfrac{m c \omega}{2 q E}\right)^2}}}$ | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||