Правка разделов «Statement», «Answer»
en/14.4.10.md
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| @@ -1,7 +1,12 @@ | |||
| ### Statement | |||
| − | $14.4.10.$ | ||
| + | $14.4.10.$ | ||
| + | What is the maximum velocity that a particle with a rest mass m and charge q | ||
| + | can acquire, born with zero velocity in an alternating sinusoidal electric field | ||
| + | with an amplitude of intensity E and a frequency ω ? | ||
| + | |||
| + | |||
| ### 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: | |||
| $\boxed{v_{\max} = \frac{c}{\sqrt{1 + \left(\dfrac{m c \omega}{2 q E}\right)^2}}}$ | |||
| @@ -31,4 +36,6 @@Solution | |||
| #### Answer | |||
| − | |||
| + | the maximum velocity is | ||
| + | |||
| + | $\boxed{v_{\max} = \frac{c}{\sqrt{1 + \left(\dfrac{m c \omega}{2 q E}\right)^2}}}$ | ||
| @@ -1,7 +1,12 @@ | |||
| ### Statement | ### Statement | ||
| $14.4.10.$ |
$14.4.10.$ | ||
| What is the maximum velocity that a particle with a rest mass m and charge q | |||
| can acquire, born with zero velocity in an alternating sinusoidal electric field | |||
| with an amplitude of intensity E and a frequency ω ? | |||
| ### Solution | ### Solution | ||
| The particle is born at rest inside a sinusoidal electric field $\mathcal{E}(t) = E \sin(\omega t)$ | 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 | The electric force F = qE \sin(\omega t) modifies its relativistic momentum | ||
| $ p = \dfrac{mv}{\sqrt{1 - v^2/c^2}} $ | $ p = \dfrac{mv}{\sqrt{1 - v^2/c^2}} $ | ||
| according to Newton's second law | according to Newton's second law | ||
| $\frac{dp}{dt} = qE \sin(\omega t)$ | $\frac{dp}{dt} = qE \sin(\omega t)$ | ||
| Integrating with the initial condition $p(0) = 0$ we obtain | Integrating with the initial condition $p(0) = 0$ we obtain | ||
| $p(t) = \frac{qE}{\omega}\bigl[1 - \cos(\omega t)\bigr]$ | $p(t) = \frac{qE}{\omega}\bigl[1 - \cos(\omega t)\bigr]$ | ||
| whose maximum value is reached when $\cos(\omega t) = -1$ | whose maximum value is reached when $\cos(\omega t) = -1$ | ||
| $p_{\max} = \frac{2qE}{\omega}$ | $p_{\max} = \frac{2qE}{\omega}$ | ||
| From the definition of p, the velocity is solved for in terms of momentum: | From the definition of p, the velocity is solved for in terms of momentum: | ||
| $v = \frac{p/m}{\sqrt{1 + \dfrac{p^2}{m^2 c^2}}}$ | $v = \frac{p/m}{\sqrt{1 + \dfrac{p^2}{m^2 c^2}}}$ | ||
| Substituting$ p_{\max}$ and simplifying algebraically, we arrive at the maximum velocity, which is: | Substituting$ p_{\max}$ and simplifying algebraically, we arrive at the maximum velocity, which is: | ||
| $\boxed{v_{\max} = \frac{c}{\sqrt{1 + \left(\dfrac{m c \omega}{2 q E}\right)^2}}}$ | $\boxed{v_{\max} = \frac{c}{\sqrt{1 + \left(\dfrac{m c \omega}{2 q E}\right)^2}}}$ | ||
| @@ -31,4 +36,6 @@Solution | |||
| #### Answer | #### Answer | ||
| the maximum velocity is | |||
| $\boxed{v_{\max} = \frac{c}{\sqrt{1 + \left(\dfrac{m c \omega}{2 q E}\right)^2}}}$ | |||