Edits to “Statement”, “Solution”, “Answer”
en/7.1.2.md
+4 −12
| @@ -1,13 +1,9 @@ | |||
| ### Statement | |||
| − | $7.1.2.$ [Insert the problem statement] | ||
| − | |||
| − | ### Solution | ||
| − | |||
| − | ### Statement | ||
| − | |||
| $7.1.2.$ An electron enters a region of a uniform electric field of strength $200\text{ V/m}$ with a velocity of $10^7\text{ m/s}$. The velocity is directed along the electric field. For how long will the electron remain in the region of this field? Determine at what distance from the entry point into the field the electron will exit it, if it enters at an angle of $45^\circ$ to the field direction. | |||
| + |  | ||
| + | |||
| ### Solution | |||
| <b>First part:</b> | |||
| The negatively charged electron moves in the direction of the electric field lines, which means the field exerts a retarding (decelerating) force on it. According to Newton's second law: | |||
| $$a = \frac{F}{m} = \frac{-eE}{m}$$ | |||
| where $m$ is the mass of the electron, and $e$ is the elementary charge. | |||
| The velocity of the electron changes from $v$ to $0$ and back to $-v$. Then, according to kinematics, the total time the electron spends under the influence of the field is twice the stopping time: | |||
| @@ -19,7 +15,7 @@Solution | |||
| $$t = 2 \cdot \left(\frac{-v}{a}\right) = \frac{2mv}{eE}$$ | |||
| Substituting the numerical values of the quantities ($m_e \approx 9.1 \cdot 10^{-31}\text{ kg}$, $e \approx 1.6 \cdot 10^{-19}\text{ C}$), we get: | |||
| − | $$t \approx 5.69 \cdot 10^{-7}\text{ s} \approx 0.5 | ||
| + | $$t \approx 5.69 \cdot 10^{-7}\text{ s} \approx 0.56\text{ }\mu\text{s}$$ | ||
| <b>Second part:</b> | |||
| When the electron enters at an angle $\alpha$ to the field lines, the projection of its velocity onto the direction perpendicular to the field is conserved during the motion, since no forces act on the electron in this direction. Let's call this component $v_{\perp} = v \sin\alpha$ (relative to the plane perpendicular to the field) or simply use the fact that the transverse component is constant. | |||
| <i>Note: The calculation below defines the angle relative to the plane bounding the field, but mathematically $\sin 45^\circ = \cos 45^\circ$, so the exact trigonometric function assignment yields the same result. Let's express the parallel component to the field as $v_{\parallel} = v \sin\alpha$ and perpendicular as $v_{\perp} = v \cos\alpha$ as per the image logic.</i> | |||
| Then the distance the electron will travel along the boundary plane before exiting is: | |||
| $$x = v_{\perp} \cdot t_1 = v \cos\alpha \cdot t_1$$ | |||
| The time spent in the field depends only on the component of velocity parallel to the field ($v_{\parallel} = v \sin\alpha$): | |||
| $$t_1 = \frac{2mv_{\parallel}}{eE} = \frac{2mv \sin\alpha}{eE}$$ | |||
| Substitute $t_1$ into the equation for $x$: | |||
| $$x = v \cos\alpha \cdot \frac{2mv \sin\alpha}{eE} = \frac{mv^2 \sin 2\alpha}{eE}$$ | |||
| Substituting the numerical values ($\alpha = 45^\circ$, $\sin 90^\circ = 1$), we obtain: | |||
| @@ -38,9 +34,5 @@Solution | |||
| $$x \approx 2.84\text{ m} \approx 2.8\text{ m}$$ | |||
| #### Answer | |||
| − | $t \approx 0.5 | ||
| + | $t \approx 0.56\text{ }\mu\text{s}$ | ||
| $x \approx 2.8\text{ m}$ | |||
| − | |||
| − | #### Answer | ||
| − | |||
| − | [Insert a concise answer or boxed result] | ||
| @@ -1,13 +1,9 @@ | |||
| ### Statement | ### Statement | ||
| $7.1.2.$ [Insert the problem statement] | |||
| ### Solution | |||
| ### Statement | |||
| $7.1.2.$ An electron enters a region of a uniform electric field of strength $200\text{ V/m}$ with a velocity of $10^7\text{ m/s}$. The velocity is directed along the electric field. For how long will the electron remain in the region of this field? Determine at what distance from the entry point into the field the electron will exit it, if it enters at an angle of $45^\circ$ to the field direction. | $7.1.2.$ An electron enters a region of a uniform electric field of strength $200\text{ V/m}$ with a velocity of $10^7\text{ m/s}$. The velocity is directed along the electric field. For how long will the electron remain in the region of this field? Determine at what distance from the entry point into the field the electron will exit it, if it enters at an angle of $45^\circ$ to the field direction. | ||
|  | |||
| ### Solution | ### Solution | ||
| <b>First part:</b> | <b>First part:</b> | ||
| The negatively charged electron moves in the direction of the electric field lines, which means the field exerts a retarding (decelerating) force on it. According to Newton's second law: | The negatively charged electron moves in the direction of the electric field lines, which means the field exerts a retarding (decelerating) force on it. According to Newton's second law: | ||
| $$a = \frac{F}{m} = \frac{-eE}{m}$$ | $$a = \frac{F}{m} = \frac{-eE}{m}$$ | ||
| where $m$ is the mass of the electron, and $e$ is the elementary charge. | where $m$ is the mass of the electron, and $e$ is the elementary charge. | ||
| The velocity of the electron changes from $v$ to $0$ and back to $-v$. Then, according to kinematics, the total time the electron spends under the influence of the field is twice the stopping time: | The velocity of the electron changes from $v$ to $0$ and back to $-v$. Then, according to kinematics, the total time the electron spends under the influence of the field is twice the stopping time: | ||
| @@ -19,7 +15,7 @@Solution | |||
| $$t = 2 \cdot \left(\frac{-v}{a}\right) = \frac{2mv}{eE}$$ | $$t = 2 \cdot \left(\frac{-v}{a}\right) = \frac{2mv}{eE}$$ | ||
| Substituting the numerical values of the quantities ($m_e \approx 9.1 \cdot 10^{-31}\text{ kg}$, $e \approx 1.6 \cdot 10^{-19}\text{ C}$), we get: | Substituting the numerical values of the quantities ($m_e \approx 9.1 \cdot 10^{-31}\text{ kg}$, $e \approx 1.6 \cdot 10^{-19}\text{ C}$), we get: | ||
| $$t \approx 5.69 \cdot 10^{-7}\text{ s} \approx 0.5 |
$$t \approx 5.69 \cdot 10^{-7}\text{ s} \approx 0.56\text{ }\mu\text{s}$$ | ||
| <b>Second part:</b> | <b>Second part:</b> | ||
| When the electron enters at an angle $\alpha$ to the field lines, the projection of its velocity onto the direction perpendicular to the field is conserved during the motion, since no forces act on the electron in this direction. Let's call this component $v_{\perp} = v \sin\alpha$ (relative to the plane perpendicular to the field) or simply use the fact that the transverse component is constant. | When the electron enters at an angle $\alpha$ to the field lines, the projection of its velocity onto the direction perpendicular to the field is conserved during the motion, since no forces act on the electron in this direction. Let's call this component $v_{\perp} = v \sin\alpha$ (relative to the plane perpendicular to the field) or simply use the fact that the transverse component is constant. | ||
| <i>Note: The calculation below defines the angle relative to the plane bounding the field, but mathematically $\sin 45^\circ = \cos 45^\circ$, so the exact trigonometric function assignment yields the same result. Let's express the parallel component to the field as $v_{\parallel} = v \sin\alpha$ and perpendicular as $v_{\perp} = v \cos\alpha$ as per the image logic.</i> | <i>Note: The calculation below defines the angle relative to the plane bounding the field, but mathematically $\sin 45^\circ = \cos 45^\circ$, so the exact trigonometric function assignment yields the same result. Let's express the parallel component to the field as $v_{\parallel} = v \sin\alpha$ and perpendicular as $v_{\perp} = v \cos\alpha$ as per the image logic.</i> | ||
| Then the distance the electron will travel along the boundary plane before exiting is: | Then the distance the electron will travel along the boundary plane before exiting is: | ||
| $$x = v_{\perp} \cdot t_1 = v \cos\alpha \cdot t_1$$ | $$x = v_{\perp} \cdot t_1 = v \cos\alpha \cdot t_1$$ | ||
| The time spent in the field depends only on the component of velocity parallel to the field ($v_{\parallel} = v \sin\alpha$): | The time spent in the field depends only on the component of velocity parallel to the field ($v_{\parallel} = v \sin\alpha$): | ||
| $$t_1 = \frac{2mv_{\parallel}}{eE} = \frac{2mv \sin\alpha}{eE}$$ | $$t_1 = \frac{2mv_{\parallel}}{eE} = \frac{2mv \sin\alpha}{eE}$$ | ||
| Substitute $t_1$ into the equation for $x$: | Substitute $t_1$ into the equation for $x$: | ||
| $$x = v \cos\alpha \cdot \frac{2mv \sin\alpha}{eE} = \frac{mv^2 \sin 2\alpha}{eE}$$ | $$x = v \cos\alpha \cdot \frac{2mv \sin\alpha}{eE} = \frac{mv^2 \sin 2\alpha}{eE}$$ | ||
| Substituting the numerical values ($\alpha = 45^\circ$, $\sin 90^\circ = 1$), we obtain: | Substituting the numerical values ($\alpha = 45^\circ$, $\sin 90^\circ = 1$), we obtain: | ||
| @@ -38,9 +34,5 @@Solution | |||
| $$x \approx 2.84\text{ m} \approx 2.8\text{ m}$$ | $$x \approx 2.84\text{ m} \approx 2.8\text{ m}$$ | ||
| #### Answer | #### Answer | ||
| $t \approx 0.5 |
$t \approx 0.56\text{ }\mu\text{s}$ | ||
| $x \approx 2.8\text{ m}$ | $x \approx 2.8\text{ m}$ | ||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||