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| In this solution, the problem is considered strictly in accordance with the text of the condition: the model is a uniformly charged spherical cavity (an empty shell inside), all of whose charge is concentrated on the surface, and which has 2 small holes for the beam of charged particles to pass through. | | In this solution, the problem is considered strictly in accordance with the text of the condition: the model is a uniformly charged spherical cavity (an empty shell inside), all of whose charge is concentrated on the surface, and which has 2 small holes for the beam of charged particles to pass through. |
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| #### 1. Problem model | | #### 1. Problem model |
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| First, let's justify why the longitudinal velocity of the particles $v$ can be considered constant. The initial kinetic energy of the particles is $qV_0$. When passing through the field of the charged sphere, the particle's energy changes by an amount of the order of $qV$. Since by condition $V \ll V_0$, the change in energy $qV \ll qV_0$ is negligible. We are justified in assuming that the particle flies through the entire system with a constant longitudinal velocity $v$. | | First, let's justify why the longitudinal velocity of the particles $v$ can be considered constant. The initial kinetic energy of the particles is $qV_0$. When passing through the field of the charged sphere, the particle's energy changes by an amount of the order of $qV$. Since by condition $V \ll V_0$, the change in energy $qV \ll qV_0$ is negligible. We are justified in assuming that the particle flies through the entire system with a constant longitudinal velocity $v$. |
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| To calculate the electric fields, we apply the "superposition principle". Let us represent the real shell with two holes as a superposition of two systems: | | To calculate the electric fields, we apply the "superposition principle". Let us represent the real shell with two holes as a superposition of two systems: |
| 1. An ideal continuous charged sphere with surface density $\sigma$. | | 1. An ideal continuous charged sphere with surface density $\sigma$. |
| 2. Two small disks with charge density $-\sigma$ located at the entrance and exit holes (they "cut out" the holes in the continuous sphere). | | 2. Two small disks with charge density $-\sigma$ located at the entrance and exit holes (they "cut out" the holes in the continuous sphere). |
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| The field inside an ideal continuous sphere is strictly zero. Therefore, all the electric field inside the cavity is created exclusively by these two local disks. The field of each disk is maximal near the hole itself and rapidly decays with distance from it. Therefore, the change in the particle's momentum occurs not smoothly, but in the form of two sharp "kicks" at the moments of passing through the holes. | | The field inside an ideal continuous sphere is strictly zero. Therefore, all the electric field inside the cavity is created exclusively by these two local disks. The field of each disk is maximal near the hole itself and rapidly decays with distance from it. Therefore, the change in the particle's momentum occurs not smoothly, but in the form of two sharp "kicks" at the moments of passing through the holes. |
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| #### 2. First transverse momentum at the entrance | | #### 2. First transverse momentum at the entrance |
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| Let the particle fly at a distance $x$ from the beam's axis of symmetry ($x \ll R$). To find the total transverse momentum acquired while passing through the first hole, we apply Gauss's theorem. Let's isolate an imaginary cylinder of radius $x$, whose axis coincides with the particle's trajectory, covering only the region of the first hole (from the space far outside to a region deep enough inside the cavity where the field of the first disk has already decayed). | | Let the particle fly at a distance $x$ from the beam's axis of symmetry ($x \ll R$). To find the total transverse momentum acquired while passing through the first hole, we apply Gauss's theorem. Let's isolate an imaginary cylinder of radius $x$, whose axis coincides with the particle's trajectory, covering only the region of the first hole (from the space far outside to a region deep enough inside the cavity where the field of the first disk has already decayed). |
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| The cylinder cuts out an area with a charge $\Delta q_1 = \pi x^2 \sigma$ from the charged shell. | | The cylinder cuts out an area with a charge $\Delta q_1 = \pi x^2 \sigma$ from the charged shell. |
| By Gauss's theorem, the flux of the electric field vector through the lateral surface of this cylinder is: | | By Gauss's theorem, the flux of the electric field vector through the lateral surface of this cylinder is: |
| $$ 2\pi x \int_{1} E_\perp dz = \frac{\pi x^2 \sigma}{\varepsilon_0} $$ | | $$ 2\pi x \int_{1} E_\perp dz = \frac{\pi x^2 \sigma}{\varepsilon_0} $$ |
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| From this, the integral of the transverse field in the first hole zone is: | | From this, the integral of the transverse field in the first hole zone is: |
| $$ \int_{1} E_\perp dz = \frac{\sigma x}{2\varepsilon_0} $$ | | $$ \int_{1} E_\perp dz = \frac{\sigma x}{2\varepsilon_0} $$ |
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| The transverse momentum acquired by the particle when breaking through the local field of the entrance hole is: | | The transverse momentum acquired by the particle when breaking through the local field of the entrance hole is: |
| $$ p_{\perp 1} = \int q E_\perp \frac{dz}{v} = \frac{q}{v} \int_{1} E_\perp dz = \frac{q \sigma x}{2\varepsilon_0 v} $$ | | $$ p_{\perp 1} = \int q E_\perp \frac{dz}{v} = \frac{q}{v} \int_{1} E_\perp dz = \frac{q \sigma x}{2\varepsilon_0 v} $$ |
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| #### 3. Focus check inside the cavity | | #### 3. Focus check inside the cavity |
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| Having received the first transverse momentum, the particle flies deeper into the cavity. It acquires a transverse velocity $v_\perp = \frac{p_{\perp 1}}{m}$ directed towards the beam axis. | | Having received the first transverse momentum, the particle flies deeper into the cavity. It acquires a transverse velocity $v_\perp = \frac{p_{\perp 1}}{m}$ directed towards the beam axis. |
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| The time it would take for the particle to reach the axis is $t = \frac{x}{v_\perp}$. During this same time, it will fly horizontally a distance $f_1$ (the focal length after the first hole): | | The time it would take for the particle to reach the axis is $t = \frac{x}{v_\perp}$. During this same time, it will fly horizontally a distance $f_1$ (the focal length after the first hole): |
| $$f_1 = vt = v \frac{x}{v_\perp} = x \frac{v}{v_\perp} = x \frac{mv}{p_{\perp 1}}$$ | | $$f_1 = vt = v \frac{x}{v_\perp} = x \frac{v}{v_\perp} = x \frac{mv}{p_{\perp 1}}$$ |
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| Substituting the found momentum: | | Substituting the found momentum: |
| $$f_1 = x \frac{mv}{\frac{q \sigma x}{2\varepsilon_0 v}} = \frac{2mv^2 \varepsilon_0}{q \sigma}$$ | | $$f_1 = x \frac{mv}{\frac{q \sigma x}{2\varepsilon_0 v}} = \frac{2mv^2 \varepsilon_0}{q \sigma}$$ |
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| The kinetic energy of the particles is given by the accelerating voltage: $mv^2 = 2qV_0$. | | The kinetic energy of the particles is given by the accelerating voltage: $mv^2 = 2qV_0$. |
| The potential at the center of the charged sphere is $V = \frac{\sigma R}{\varepsilon_0}$, which implies $\sigma = \frac{\varepsilon_0 V}{R}$. | | The potential at the center of the charged sphere is $V = \frac{\sigma R}{\varepsilon_0}$, which implies $\sigma = \frac{\varepsilon_0 V}{R}$. |
| Substituting this into the formula for $f_1$: | | Substituting this into the formula for $f_1$: |
| $$f_1 = \frac{2(2qV_0) \varepsilon_0}{q \left(\frac{\varepsilon_0 V}{R}\right)} = 4R \frac{V_0}{V}$$ | | $$f_1 = \frac{2(2qV_0) \varepsilon_0}{q \left(\frac{\varepsilon_0 V}{R}\right)} = 4R \frac{V_0}{V}$$ |
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| By condition $V \ll V_0$, which means $f_1 \gg R$. The beam travels too fast and does not have time to focus inside the cavity. The particles reach the exit hole at practically the same distance $x$ from the axis. | | By condition $V \ll V_0$, which means $f_1 \gg R$. The beam travels too fast and does not have time to focus inside the cavity. The particles reach the exit hole at practically the same distance $x$ from the axis. |
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| #### 4. Second momentum and total focal length | | #### 4. Second momentum and total focal length |
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| Flying through the exit hole at a distance $x$, the particle crosses the local field of the second "cut-out" disk and receives a second, identical transverse momentum: | | Flying through the exit hole at a distance $x$, the particle crosses the local field of the second "cut-out" disk and receives a second, identical transverse momentum: |
| $$ p_{\perp 2} = \frac{q \sigma x}{2\varepsilon_0 v} $$ | | $$ p_{\perp 2} = \frac{q \sigma x}{2\varepsilon_0 v} $$ |
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| The total transverse momentum after leaving the target: | | The total transverse momentum after leaving the target: |
| $$ p_\perp = p_{\perp 1} + p_{\perp 2} = \frac{q \sigma x}{\varepsilon_0 v} $$ | | $$ p_\perp = p_{\perp 1} + p_{\perp 2} = \frac{q \sigma x}{\varepsilon_0 v} $$ |
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| The final focal length $f$ of the entire system: | | The final focal length $f$ of the entire system: |
| $$ f = x \frac{mv}{p_\perp} = x \frac{mv}{\frac{q \sigma x}{\varepsilon_0 v}} = \frac{mv^2 \varepsilon_0}{q \sigma} $$ | | $$ f = x \frac{mv}{p_\perp} = x \frac{mv}{\frac{q \sigma x}{\varepsilon_0 v}} = \frac{mv^2 \varepsilon_0}{q \sigma} $$ |
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| Again substituting $mv^2 = 2qV_0$ and $\sigma = \frac{\varepsilon_0 V}{R}$, we get the final answer: | | Again substituting $mv^2 = 2qV_0$ and $\sigma = \frac{\varepsilon_0 V}{R}$, we get the final answer: |
| $$ f = \frac{(2qV_0) \varepsilon_0}{q \left(\frac{\varepsilon_0 V}{R}\right)} = 2R \frac{V_0}{V} $$ | | $$ f = \frac{(2qV_0) \varepsilon_0}{q \left(\frac{\varepsilon_0 V}{R}\right)} = 2R \frac{V_0}{V} $$ |
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