Новое решение
en/14.4.13.md
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| + | ### Statement | ||
| + | |||
| + | $14.4.13.$ [Insert the problem statement] | ||
| + | |||
| + | ### Solution | ||
| + | |||
| + | We start from Newton's second law in its relativistic form. Since the electric force is constant and equal to $eE$, the linear momentum p of the electron grows linearly with time: | ||
| + | |||
| + | $\frac{dp}{dt} = eE \quad\Rightarrow\quad p = eEt$ | ||
| + | |||
| + | because the electron is born at rest $(p(0)=0)$. | ||
| + | |||
| + | Now we need to relate momentum to velocity. In special relativity, this relation is | ||
| + | |||
| + | $p = \frac{mv}{\sqrt{1 - v^2/c^2}} \quad\Rightarrow\quad v = \frac{p/m}{\sqrt{1 + p^2/(m^2c^2)}}$ | ||
| + | |||
| + | The distance traveled x is obtained by integrating velocity over time. Using $p = eEt$ and the change of variable $dp = eE\$,dt, we write the accumulated distance up to time$ \tau $as | ||
| + | |||
| + | $l = \int_0^\tau v\,dt = \int_0^{p_0} \frac{p/m}{\sqrt{1 + p^2/(m^2c^2)}}\,\frac{dp}{eE}, | ||
| + | \qquad p_0 = eE\tau$ | ||
| + | |||
| + | This integral is standard. Making the substitution $ u = 1 + p^2/(m^2c^2)$, it is solved directly and yields: | ||
| + | |||
| + | $l = \frac{mc^2}{eE}\left( \sqrt{1 + \frac{p_0^2}{m^2c^2}} - 1 \right)$ | ||
| + | |||
| + | In this expression, the distance l already appears as a function of time$ \tau (through p_0)$ Now we isolate $\tau$ We move the term -1 to the other side and square: | ||
| + | |||
| + | $\left(\frac{eEl}{mc^2} + 1\right)^2 = 1 + \frac{p_0^2}{m^2c^2}$ | ||
| + | |||
| + | We subtract 1 and expand the square of the parenthesis | ||
| + | |||
| + | $\frac{p_0^2}{m^2c^2} = \left(\frac{eEl}{mc^2}\right)^2 + 2\frac{eEl}{mc^2}$ | ||
| + | |||
| + | We substitute $p_0 = eE\tau$ and cancel the common factors | ||
| + | |||
| + | $\frac{e^2E^2\tau^2}{m^2c^2} = \frac{eEl}{mc^2}\left( \frac{eEl}{mc^2} + 2 \right)$ | ||
| + | |||
| + | We multiply both sides by $m^2c^2/(e^2E^2) $ and simplify | ||
| + | |||
| + | $\tau^2 = \frac{ml}{eE}\left( 2 + \frac{eEl}{mc^2} \right)$ | ||
| + | |||
| + | Finally, we take the positive square root, since time is positive, and obtain: | ||
| + | |||
| + | $\boxed{\tau = \sqrt{\frac{ml}{eE}\left(2 + \frac{eEl}{mc^2}\right)}}$ | ||
| + | |||
| + | If we apply the numerical data whut the formula we have : | ||
| + | |||
| + | $\boxed{\tau \approx 4.7 \times 10^{-9}\ \text{s} \;(4.7\ \text{ns})}$ | ||
| + | |||
| + | The electron travels 1 meter in approximately 4.7 nanoseconds, reaching speeds close to that of light due to the intense electric field. | ||
| + | |||
| + | |||
| + | #### Answer | ||
| + | |||
| + | [Insert a concise answer or boxed result] | ||
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| ### Statement | |||
| $14.4.13.$ [Insert the problem statement] | |||
| ### Solution | |||
| We start from Newton's second law in its relativistic form. Since the electric force is constant and equal to $eE$, the linear momentum p of the electron grows linearly with time: | |||
| $\frac{dp}{dt} = eE \quad\Rightarrow\quad p = eEt$ | |||
| because the electron is born at rest $(p(0)=0)$. | |||
| Now we need to relate momentum to velocity. In special relativity, this relation is | |||
| $p = \frac{mv}{\sqrt{1 - v^2/c^2}} \quad\Rightarrow\quad v = \frac{p/m}{\sqrt{1 + p^2/(m^2c^2)}}$ | |||
| The distance traveled x is obtained by integrating velocity over time. Using $p = eEt$ and the change of variable $dp = eE\$,dt, we write the accumulated distance up to time$ \tau $as | |||
| $l = \int_0^\tau v\,dt = \int_0^{p_0} \frac{p/m}{\sqrt{1 + p^2/(m^2c^2)}}\,\frac{dp}{eE}, | |||
| \qquad p_0 = eE\tau$ | |||
| This integral is standard. Making the substitution $ u = 1 + p^2/(m^2c^2)$, it is solved directly and yields: | |||
| $l = \frac{mc^2}{eE}\left( \sqrt{1 + \frac{p_0^2}{m^2c^2}} - 1 \right)$ | |||
| In this expression, the distance l already appears as a function of time$ \tau (through p_0)$ Now we isolate $\tau$ We move the term -1 to the other side and square: | |||
| $\left(\frac{eEl}{mc^2} + 1\right)^2 = 1 + \frac{p_0^2}{m^2c^2}$ | |||
| We subtract 1 and expand the square of the parenthesis | |||
| $\frac{p_0^2}{m^2c^2} = \left(\frac{eEl}{mc^2}\right)^2 + 2\frac{eEl}{mc^2}$ | |||
| We substitute $p_0 = eE\tau$ and cancel the common factors | |||
| $\frac{e^2E^2\tau^2}{m^2c^2} = \frac{eEl}{mc^2}\left( \frac{eEl}{mc^2} + 2 \right)$ | |||
| We multiply both sides by $m^2c^2/(e^2E^2) $ and simplify | |||
| $\tau^2 = \frac{ml}{eE}\left( 2 + \frac{eEl}{mc^2} \right)$ | |||
| Finally, we take the positive square root, since time is positive, and obtain: | |||
| $\boxed{\tau = \sqrt{\frac{ml}{eE}\left(2 + \frac{eEl}{mc^2}\right)}}$ | |||
| If we apply the numerical data whut the formula we have : | |||
| $\boxed{\tau \approx 4.7 \times 10^{-9}\ \text{s} \;(4.7\ \text{ns})}$ | |||
| The electron travels 1 meter in approximately 4.7 nanoseconds, reaching speeds close to that of light due to the intense electric field. | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||