Edit to “Solution”
en/7.2.10.md
+1 −1
| ### Problem | |||
| $7.2.10.$ Where will a thin parallel beam of electrons, accelerated by a potential difference $V_0$, be focused by the electric field created by two concentric spheres of radii $R$ and $R - \Delta$, $\Delta \ll R$? The outer sphere is grounded, the potential of the inner sphere is $V \ll V_0$, and the beam passes through the center of the spheres. | |||
|  | |||
| ### Solution | |||
| @@ -9,7 +9,7 @@Solution | |||
| Since the beam is accelerated by a potential difference $V_0$, the longitudinal velocity of the electrons (with charge $e$ and mass $m$) is: | |||
| $$ U = \sqrt{\frac{2eV_0}{m}} $$ | |||
| − | Since $V \ll V_0$, the change in the longitudinal velocity of the electrons as they pass through the system can be neglected. The velocity $U$ remains practically constant. The outer sphere is grounded, so the electric field exists only in the narrow gap of width $\Delta$ and is directed radially | ||
| + | Since $V \ll V_0$, the change in the longitudinal velocity of the electrons as they pass through the system can be neglected. The velocity $U$ remains practically constant. The outer sphere is grounded, so the electric field exists only in the narrow gap of width $\Delta$ and is directed radially away from the center (assuming $V>0$ so that the force acting on the electron is directed towards the axis). There is no field inside the smaller sphere. | ||
| Let an electron enter the gap at a small distance $x$ from the central axis. The electric field strength in the gap is: | |||
| $$ E = \frac{V}{\Delta} $$ | |||
| Since the electron does not travel exactly along the center, the field has a transverse component perpendicular to the beam axis: | |||
| $$ E_\perp = E \sin\alpha $$ | |||
| For small deviations $\sin\alpha \approx \frac{x}{R}$, then: | |||
| $$ E_\perp = \frac{V}{\Delta} \frac{x}{R} $$ | |||
| The transverse (focusing) force acting on the electron in the first gap is: | |||
| $$ F_\perp = eE_\perp = \frac{Vxe}{R\Delta} $$ | |||
| The time it takes for the electron to pass through the gap is: | |||
| $$ dt = \frac{\Delta}{U} $$ | |||
| The transverse momentum acquired by the electron at the entrance is: | |||
| $$ p_{\perp 1} = F_\perp dt = \left( \frac{Vxe}{R\Delta} \right) \frac{\Delta}{U} = \frac{Vxe}{UR} $$ | |||
| Let's check where the beam would focus if there were no second gap. The transverse velocity of the electron acquired after the first gap is: | |||
| $$ v_{\perp 1} = \frac{p_{\perp 1}}{m} = \frac{Vxe}{mUR} $$ | |||
| The time $t_1$ required for the electron to reach the central axis, covering the distance $x$ with this transverse velocity, is: | |||
| $$ t_1 = \frac{x}{v_{\perp 1}} = \frac{x}{\frac{Vxe}{mUR}} = \frac{mUR}{Ve} $$ | |||
| During this time, the beam, moving with longitudinal velocity $U$, will travel a distance $f_1$, which would be the focal length of the first gap: | |||
| $$ f_1 = U \cdot t_1 = U \left( \frac{mUR}{Ve} \right) = \frac{mU^2 R}{Ve} $$ | |||
| Considering that $mU^2 = 2eV_0$, we obtain: | |||
| $$ f_1 = \frac{2eV_0 R}{Ve} = 2R \frac{V_0}{V} $$ | |||
| Since by the problem condition $V \ll V_0$, then $f_1 \gg 2R$. This means that the beam will not have time to focus inside the cavity and will reach the opposite edge of the inner shell at practically the same distance $x$ from the axis. | |||
| Upon exiting the system, the beam passes through the second gap $\Delta$. The field is again directed towards the center, and the electron receives a second identical transverse momentum: | |||
| $$ p_{\perp 2} = \frac{Vxe}{UR} $$ | |||
| The total transverse momentum after exiting the system is: | |||
| $$ p_\perp = p_{\perp 1} + p_{\perp 2} = \frac{2Vxe}{UR} $$ | |||
| The final acquired transverse velocity of the electron will be: | |||
| $$ v_\perp = \frac{p_\perp}{m} = \frac{2Vxe}{mUR} $$ | |||
| After leaving the spheres, no forces act on the beam. The time $t$ for the electron to finally reach the central axis (covering distance $x$) is: | |||
| $$ t = \frac{x}{v_\perp} = \frac{x}{\frac{2Vxe}{mUR}} = \frac{mUR}{2Ve} $$ | |||
| During this time, the beam, moving with longitudinal velocity $U$, will travel a distance $f$, which is the desired focal length of the entire system: | |||
| $$ f = U \cdot t = U \left( \frac{mUR}{2Ve} \right) = \frac{mU^2 R}{2Ve} $$ | |||
| Substituting the initial kinetic energy $mU^2 = 2eV_0$, we finally obtain: | |||
| $$ f = \frac{2eV_0 R}{2Ve} = R \frac{V_0}{V} $$ | |||
| #### Answer | |||
| $$ f = R\frac{V_0}{V} $$ | |||
| --- | |||
| #### Note: Why is the answer in the textbook $2R(V_0/V)^2$? | |||
| Our result $f = R(V_0/V)$ (linear dependence on the potential ratio) is absolutely correct for the model of <b>ideal transparent grids</b>. In such a model, the field in the gap is strictly radial and drops off abruptly. | |||
| However, the authors of the textbook implicitly assume a more realistic system: solid metal spheres with <b>open apertures</b> for the beam to pass through (aperture lenses). In such a system, fringe effects arise: | |||
| 1. <b>First-order effect zeroes out:</b> Near the apertures, the electric field "bulges" out. Diverging sections of the field appear. According to the integral theorem of electron optics, for any gridless system (where the potential on the axis changes smoothly from one constant to another), the converging effect of the central part of the field is perfectly and completely compensated by the diverging effect of the fringe fields. That is, focusing of the first order of smallness ($V/V_0$) becomes strictly equal to zero. | |||
| 2. <b>Second-order focusing:</b> Since the first-order effect vanishes, microscopic changes in the longitudinal velocity come into play. Inside the gap, the kinetic energy of the electron changes by $\sim eV$. Because of this, the electron spends different amounts of time in the converging and diverging parts of the aperture field. This time imbalance creates a weak net focusing proportional to the square of the velocity change, i.e., $(V/V_0)^2$. Strict integration of the paraxial ray equation for such a field yields the exact coefficient, leading to the textbook authors' answer $f = 2R(V_0/V)^2$. | |||
| unchanged lines 60 | |||
| ### Problem | ### Problem | ||
| $7.2.10.$ Where will a thin parallel beam of electrons, accelerated by a potential difference $V_0$, be focused by the electric field created by two concentric spheres of radii $R$ and $R - \Delta$, $\Delta \ll R$? The outer sphere is grounded, the potential of the inner sphere is $V \ll V_0$, and the beam passes through the center of the spheres. | $7.2.10.$ Where will a thin parallel beam of electrons, accelerated by a potential difference $V_0$, be focused by the electric field created by two concentric spheres of radii $R$ and $R - \Delta$, $\Delta \ll R$? The outer sphere is grounded, the potential of the inner sphere is $V \ll V_0$, and the beam passes through the center of the spheres. | ||
|  |  | ||
| ### Solution | ### Solution | ||
| @@ -9,7 +9,7 @@Solution | |||
| Since the beam is accelerated by a potential difference $V_0$, the longitudinal velocity of the electrons (with charge $e$ and mass $m$) is: | Since the beam is accelerated by a potential difference $V_0$, the longitudinal velocity of the electrons (with charge $e$ and mass $m$) is: | ||
| $$ U = \sqrt{\frac{2eV_0}{m}} $$ | $$ U = \sqrt{\frac{2eV_0}{m}} $$ | ||
| Since $V \ll V_0$, the change in the longitudinal velocity of the electrons as they pass through the system can be neglected. The velocity $U$ remains practically constant. The outer sphere is grounded, so the electric field exists only in the narrow gap of width $\Delta$ and is directed radially |
Since $V \ll V_0$, the change in the longitudinal velocity of the electrons as they pass through the system can be neglected. The velocity $U$ remains practically constant. The outer sphere is grounded, so the electric field exists only in the narrow gap of width $\Delta$ and is directed radially away from the center (assuming $V>0$ so that the force acting on the electron is directed towards the axis). There is no field inside the smaller sphere. | ||
| Let an electron enter the gap at a small distance $x$ from the central axis. The electric field strength in the gap is: | Let an electron enter the gap at a small distance $x$ from the central axis. The electric field strength in the gap is: | ||
| $$ E = \frac{V}{\Delta} $$ | $$ E = \frac{V}{\Delta} $$ | ||
| Since the electron does not travel exactly along the center, the field has a transverse component perpendicular to the beam axis: | Since the electron does not travel exactly along the center, the field has a transverse component perpendicular to the beam axis: | ||
| $$ E_\perp = E \sin\alpha $$ | $$ E_\perp = E \sin\alpha $$ | ||
| For small deviations $\sin\alpha \approx \frac{x}{R}$, then: | For small deviations $\sin\alpha \approx \frac{x}{R}$, then: | ||
| $$ E_\perp = \frac{V}{\Delta} \frac{x}{R} $$ | $$ E_\perp = \frac{V}{\Delta} \frac{x}{R} $$ | ||
| The transverse (focusing) force acting on the electron in the first gap is: | The transverse (focusing) force acting on the electron in the first gap is: | ||
| $$ F_\perp = eE_\perp = \frac{Vxe}{R\Delta} $$ | $$ F_\perp = eE_\perp = \frac{Vxe}{R\Delta} $$ | ||
| The time it takes for the electron to pass through the gap is: | The time it takes for the electron to pass through the gap is: | ||
| $$ dt = \frac{\Delta}{U} $$ | $$ dt = \frac{\Delta}{U} $$ | ||
| The transverse momentum acquired by the electron at the entrance is: | The transverse momentum acquired by the electron at the entrance is: | ||
| $$ p_{\perp 1} = F_\perp dt = \left( \frac{Vxe}{R\Delta} \right) \frac{\Delta}{U} = \frac{Vxe}{UR} $$ | $$ p_{\perp 1} = F_\perp dt = \left( \frac{Vxe}{R\Delta} \right) \frac{\Delta}{U} = \frac{Vxe}{UR} $$ | ||
| Let's check where the beam would focus if there were no second gap. The transverse velocity of the electron acquired after the first gap is: | Let's check where the beam would focus if there were no second gap. The transverse velocity of the electron acquired after the first gap is: | ||
| $$ v_{\perp 1} = \frac{p_{\perp 1}}{m} = \frac{Vxe}{mUR} $$ | $$ v_{\perp 1} = \frac{p_{\perp 1}}{m} = \frac{Vxe}{mUR} $$ | ||
| The time $t_1$ required for the electron to reach the central axis, covering the distance $x$ with this transverse velocity, is: | The time $t_1$ required for the electron to reach the central axis, covering the distance $x$ with this transverse velocity, is: | ||
| $$ t_1 = \frac{x}{v_{\perp 1}} = \frac{x}{\frac{Vxe}{mUR}} = \frac{mUR}{Ve} $$ | $$ t_1 = \frac{x}{v_{\perp 1}} = \frac{x}{\frac{Vxe}{mUR}} = \frac{mUR}{Ve} $$ | ||
| During this time, the beam, moving with longitudinal velocity $U$, will travel a distance $f_1$, which would be the focal length of the first gap: | During this time, the beam, moving with longitudinal velocity $U$, will travel a distance $f_1$, which would be the focal length of the first gap: | ||
| $$ f_1 = U \cdot t_1 = U \left( \frac{mUR}{Ve} \right) = \frac{mU^2 R}{Ve} $$ | $$ f_1 = U \cdot t_1 = U \left( \frac{mUR}{Ve} \right) = \frac{mU^2 R}{Ve} $$ | ||
| Considering that $mU^2 = 2eV_0$, we obtain: | Considering that $mU^2 = 2eV_0$, we obtain: | ||
| $$ f_1 = \frac{2eV_0 R}{Ve} = 2R \frac{V_0}{V} $$ | $$ f_1 = \frac{2eV_0 R}{Ve} = 2R \frac{V_0}{V} $$ | ||
| Since by the problem condition $V \ll V_0$, then $f_1 \gg 2R$. This means that the beam will not have time to focus inside the cavity and will reach the opposite edge of the inner shell at practically the same distance $x$ from the axis. | Since by the problem condition $V \ll V_0$, then $f_1 \gg 2R$. This means that the beam will not have time to focus inside the cavity and will reach the opposite edge of the inner shell at practically the same distance $x$ from the axis. | ||
| Upon exiting the system, the beam passes through the second gap $\Delta$. The field is again directed towards the center, and the electron receives a second identical transverse momentum: | Upon exiting the system, the beam passes through the second gap $\Delta$. The field is again directed towards the center, and the electron receives a second identical transverse momentum: | ||
| $$ p_{\perp 2} = \frac{Vxe}{UR} $$ | $$ p_{\perp 2} = \frac{Vxe}{UR} $$ | ||
| The total transverse momentum after exiting the system is: | The total transverse momentum after exiting the system is: | ||
| $$ p_\perp = p_{\perp 1} + p_{\perp 2} = \frac{2Vxe}{UR} $$ | $$ p_\perp = p_{\perp 1} + p_{\perp 2} = \frac{2Vxe}{UR} $$ | ||
| The final acquired transverse velocity of the electron will be: | The final acquired transverse velocity of the electron will be: | ||
| $$ v_\perp = \frac{p_\perp}{m} = \frac{2Vxe}{mUR} $$ | $$ v_\perp = \frac{p_\perp}{m} = \frac{2Vxe}{mUR} $$ | ||
| After leaving the spheres, no forces act on the beam. The time $t$ for the electron to finally reach the central axis (covering distance $x$) is: | After leaving the spheres, no forces act on the beam. The time $t$ for the electron to finally reach the central axis (covering distance $x$) is: | ||
| $$ t = \frac{x}{v_\perp} = \frac{x}{\frac{2Vxe}{mUR}} = \frac{mUR}{2Ve} $$ | $$ t = \frac{x}{v_\perp} = \frac{x}{\frac{2Vxe}{mUR}} = \frac{mUR}{2Ve} $$ | ||
| During this time, the beam, moving with longitudinal velocity $U$, will travel a distance $f$, which is the desired focal length of the entire system: | During this time, the beam, moving with longitudinal velocity $U$, will travel a distance $f$, which is the desired focal length of the entire system: | ||
| $$ f = U \cdot t = U \left( \frac{mUR}{2Ve} \right) = \frac{mU^2 R}{2Ve} $$ | $$ f = U \cdot t = U \left( \frac{mUR}{2Ve} \right) = \frac{mU^2 R}{2Ve} $$ | ||
| Substituting the initial kinetic energy $mU^2 = 2eV_0$, we finally obtain: | Substituting the initial kinetic energy $mU^2 = 2eV_0$, we finally obtain: | ||
| $$ f = \frac{2eV_0 R}{2Ve} = R \frac{V_0}{V} $$ | $$ f = \frac{2eV_0 R}{2Ve} = R \frac{V_0}{V} $$ | ||
| #### Answer | #### Answer | ||
| $$ f = R\frac{V_0}{V} $$ | $$ f = R\frac{V_0}{V} $$ | ||
| --- | --- | ||
| #### Note: Why is the answer in the textbook $2R(V_0/V)^2$? | #### Note: Why is the answer in the textbook $2R(V_0/V)^2$? | ||
| Our result $f = R(V_0/V)$ (linear dependence on the potential ratio) is absolutely correct for the model of <b>ideal transparent grids</b>. In such a model, the field in the gap is strictly radial and drops off abruptly. | Our result $f = R(V_0/V)$ (linear dependence on the potential ratio) is absolutely correct for the model of <b>ideal transparent grids</b>. In such a model, the field in the gap is strictly radial and drops off abruptly. | ||
| However, the authors of the textbook implicitly assume a more realistic system: solid metal spheres with <b>open apertures</b> for the beam to pass through (aperture lenses). In such a system, fringe effects arise: | However, the authors of the textbook implicitly assume a more realistic system: solid metal spheres with <b>open apertures</b> for the beam to pass through (aperture lenses). In such a system, fringe effects arise: | ||
| 1. <b>First-order effect zeroes out:</b> Near the apertures, the electric field "bulges" out. Diverging sections of the field appear. According to the integral theorem of electron optics, for any gridless system (where the potential on the axis changes smoothly from one constant to another), the converging effect of the central part of the field is perfectly and completely compensated by the diverging effect of the fringe fields. That is, focusing of the first order of smallness ($V/V_0$) becomes strictly equal to zero. | 1. <b>First-order effect zeroes out:</b> Near the apertures, the electric field "bulges" out. Diverging sections of the field appear. According to the integral theorem of electron optics, for any gridless system (where the potential on the axis changes smoothly from one constant to another), the converging effect of the central part of the field is perfectly and completely compensated by the diverging effect of the fringe fields. That is, focusing of the first order of smallness ($V/V_0$) becomes strictly equal to zero. | ||
| 2. <b>Second-order focusing:</b> Since the first-order effect vanishes, microscopic changes in the longitudinal velocity come into play. Inside the gap, the kinetic energy of the electron changes by $\sim eV$. Because of this, the electron spends different amounts of time in the converging and diverging parts of the aperture field. This time imbalance creates a weak net focusing proportional to the square of the velocity change, i.e., $(V/V_0)^2$. Strict integration of the paraxial ray equation for such a field yields the exact coefficient, leading to the textbook authors' answer $f = 2R(V_0/V)^2$. | 2. <b>Second-order focusing:</b> Since the first-order effect vanishes, microscopic changes in the longitudinal velocity come into play. Inside the gap, the kinetic energy of the electron changes by $\sim eV$. Because of this, the electron spends different amounts of time in the converging and diverging parts of the aperture field. This time imbalance creates a weak net focusing proportional to the square of the velocity change, i.e., $(V/V_0)^2$. Strict integration of the paraxial ray equation for such a field yields the exact coefficient, leading to the textbook authors' answer $f = 2R(V_0/V)^2$. | ||
| unchanged lines 60 | |||