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| ### Statement |
| ### Problem |
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| $7.2.9.$ A thin parallel beam of charged particles accelerated by a potential difference $V_0$ passes through the center of a uniformly charged spherical cavity. At what distance will this beam focus if the potential at the center of the sphere is $V \ll V_0$? |
| $7.2.9.$ A thin parallel beam of charged particles, accelerated by a potential difference $V_0$, passes through the center of a uniformly charged spherical cavity. At what distance will this beam focus if the potential at the center of the sphere is $V \ll V_0$? |
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| ### Solution |
| ### Solution |
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| In this solution, the problem is considered in strict accordance with the text of the condition: the model is a uniformly charged empty spherical shell, all of whose charge is concentrated on the surface, with two small holes for the beam to pass through. |
| In this solution, the problem is considered in strict accordance with the text of the condition: the model is a uniformly charged empty spherical shell, the entire charge of which is concentrated on the surface, with two small holes for the beam to pass through. |
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| #### 1. Physical model and superposition principle |
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| Since $V \ll V_0$, the change in the longitudinal kinetic energy of the beam is negligible ($qV \ll qV_0$). The particle velocity $v$ along the axis can be considered constant. Furthermore, given the enormous speed of the beam and the traditional assumptions of such problems, we neglect the influence of the external scattering field at large distances from the sphere, focusing only on the sharp local field "kicks" directly at the holes (the aperture lens effect). |
| <b>1. Physical model and the principle of superposition</b>\ |
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| Since $V \ll V_0$, the change in the longitudinal kinetic energy of the beam is negligible ($qV \ll qV_0$). The particle velocity $v$ along the axis can be considered constant. Furthermore, given the enormous speed of the beam and the traditional assumptions of such problems, we neglect the influence of the external diverging field at large distances from the sphere, focusing our attention only on the sharp local "kicks" of the field directly in the holes (the aperture lens effect). |
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| For a rigorous calculation of the local fields, we apply the superposition principle. We represent the real shell with two holes as a superposition of two ideal systems: |
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| 1. An ideal continuous charged sphere with surface density $+\sigma$. |
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| 2. Two "virtual" disks with charge density $-\sigma$, located strictly at the entrance and exit holes. |
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| Inside the ideal continuous sphere, the electric field is strictly zero. Consequently, all the focusing transverse field inside the cavity near the holes is created exclusively by these two virtual disks with charge $-\sigma$. |
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| To strictly calculate the local fields, we apply the superposition principle. Let us represent the real shell with two holes as a superposition of two ideal systems: |
| <b>2. First transverse momentum at the entrance</b>\ |
| 1. An ideal continuous charged sphere with a surface charge density $+\sigma$. |
| Let the particle fly at a distance $x$ from the symmetry axis of the beam ($x \ll R$). We find the transverse momentum received from the local field of the first hole. |
| 2. Two "virtual" disks with a charge density $-\sigma$, located strictly at the positions of the entrance and exit holes. |
| Let us select a small Gaussian cylinder of radius $x$ and thickness $dz$, covering only the area of the entrance hole itself. Within our mathematical superposition model, the charge of the virtual disk inside this cylinder is equal in absolute value to $|\Delta q| = \pi x^2 \sigma$. The negative sign of this effective charge means that the field is directed towards the axis (creating a focusing effect). |
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| Inside an ideal continuous sphere, the electric field is strictly zero. Therefore, all the focusing transverse field inside the cavity near the holes is created exclusively by these two virtual disks with a charge of $-\sigma$. |
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| #### 2. First transverse momentum at the entrance |
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| Let a particle fly at a distance $x$ from the beam's axis of symmetry ($x \ll R$). Let's find the transverse momentum acquired from the local field of the first hole. |
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| Let's isolate a small Gaussian cylinder of radius $x$ and thickness $dz$, encompassing only the region of the entrance hole itself. Within our mathematical superposition model, this cylinder contains the charge of the virtual disk, equal in magnitude to $|\Delta q| = \pi x^2 \sigma$. The negative sign of this effective charge means the field is directed towards the axis (creating a focusing effect). |
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| 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: |
| $$ 2\pi x \int E_\perp dz = \frac{\pi x^2 \sigma}{\varepsilon_0} $$ |
| $$ 2\pi x \int E_\perp dz = \frac{\pi x^2 \sigma}{\varepsilon_0} $$ |
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| Hence, the integral of the transverse field in the zone of the local hole is: |
| From this, the integral of the transverse field in the local hole zone is: |
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| $$ \int E_\perp dz = \frac{\sigma x}{2\varepsilon_0} $$ |
| $$ \int E_\perp dz = \frac{\sigma x}{2\varepsilon_0} $$ |
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| The transverse momentum that the particle acquires by penetrating this local field is: |
| The transverse momentum the particle acquires by breaking through this local field is: |
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| $$ p_{\perp 1} = \int q E_\perp \frac{dz}{v} = \frac{q}{v} \int E_\perp dz = \frac{q \sigma x}{2\varepsilon_0 v} $$ |
| $$ p_{\perp 1} = \int q E_\perp \frac{dz}{v} = \frac{q}{v} \int E_\perp dz = \frac{q \sigma x}{2\varepsilon_0 v} $$ |
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| #### 3. Focus check inside the cavity |
| <b>3. Checking focusing inside the cavity</b>\ |
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| Having received the first transverse momentum, the particle flies deep into the cavity, acquiring 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, acquiring a transverse velocity $v_\perp = \frac{p_{\perp 1}}{m}$ directed towards the beam axis. |
| The time of flight to the axis is $t = \frac{x}{v_\perp}$. The focal length after the first hole is $f_1$: |
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| $$ f_1 = v t = v \frac{x}{v_\perp} = x \frac{m v}{p_{\perp 1}} = x \frac{m v}{\frac{q \sigma x}{2\varepsilon_0 v}} = \frac{2 m v^2 \varepsilon_0}{q \sigma} $$ |
| The time to reach the axis is $t = \frac{x}{v_\perp}$. The focal length after the first hole $f_1$: |
| The kinetic energy is given by the accelerating voltage: $mv^2 = 2qV_0$. |
| $$f_1 = vt = v \frac{x}{v_\perp} = x \frac{mv}{p_{\perp 1}} = x \frac{mv}{\frac{q \sigma x}{2\varepsilon_0 v}} = \frac{2mv^2 \varepsilon_0}{q \sigma}$$ |
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| The kinetic energy is given by the accelerating voltage: $mv^2 = 2qV_0$. |
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| The potential of the sphere is $V = \frac{\sigma R}{\varepsilon_0}$, which implies $\sigma = \frac{\varepsilon_0 V}{R}$. |
| The potential of the sphere is $V = \frac{\sigma R}{\varepsilon_0}$, which implies $\sigma = \frac{\varepsilon_0 V}{R}$. |
| Substituting into $f_1$: |
| Substituting into $f_1$: |
| $$f_1 = \frac{2(2qV_0) \varepsilon_0}{q \left(\frac{\varepsilon_0 V}{R}\right)} = 4R \frac{V_0}{V}$$ |
| $$ f_1 = \frac{2(2qV_0)\varepsilon_0}{q\left(\frac{\varepsilon_0 V}{R}\right)} = 4R \frac{V_0}{V} $$ |
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| Since $V \ll V_0$, then $f_1 \gg R$. The beam flies too fast and does not have time to focus inside the cavity, reaching the exit hole at practically the same distance $x$ from the axis. |
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| Since $V \ll V_0$, then $f_1 \gg R$. The beam travels too fast and does not have time to focus inside the cavity, reaching the exit hole at practically the same distance $x$ from the axis. |
| <b>4. Second momentum and final focal length</b>\ |
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| #### 4. Second momentum and total focal length |
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| Flying through the exit hole, the particle crosses the local field of the second virtual disk ($-\sigma$) and receives a similar transverse momentum directed towards the axis: |
| Flying through the exit hole, the particle crosses the local field of the second virtual disk ($-\sigma$) and receives a similar transverse momentum directed towards the axis: |
| $$ p_{\perp 2} = \frac{q \sigma x}{2\varepsilon_0 v} $$ |
| $$ p_{\perp 2} = \frac{q \sigma x}{2\varepsilon_0 v} $$ |
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| The total transverse momentum after exiting is: |
| The total transverse momentum after exiting: |
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| $$ p_\perp = p_{\perp 1} + p_{\perp 2} = \frac{q \sigma x}{\varepsilon_0 v} $$ |
| $$ p_\perp = p_{\perp 1} + p_{\perp 2} = \frac{q \sigma x}{\varepsilon_0 v} $$ |
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| The final focal length $f$ of the entire system is: |
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| $$ f = x \frac{m v}{p_\perp} = x \frac{m v}{\frac{q \sigma x}{\varepsilon_0 v}} = \frac{m v^2 \varepsilon_0}{q \sigma} $$ |
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| Substituting $mv^2 = 2qV_0$ and $\sigma = \frac{\varepsilon_0 V}{R}$, we obtain the final answer for an empty spherical shell: |
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| $$ f = \frac{(2qV_0)\varepsilon_0}{q\left(\frac{\varepsilon_0 V}{R}\right)} = 2R \frac{V_0}{V} $$ |
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| The final focal length $f$ of the entire system: |
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| $$ f = x \frac{mv}{p_\perp} = x \frac{mv}{\frac{q \sigma x}{\varepsilon_0 v}} = \frac{mv^2 \varepsilon_0}{q \sigma} $$ |
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| Substituting $mv^2 = 2qV_0$ and $\sigma = \frac{\varepsilon_0 V}{R}$, we obtain the final answer for an empty spherical shell: |
| <b>5. Rigorous calculation for the alternative model (solid sphere)</b><br> |
| $$ f = \frac{(2qV_0) \varepsilon_0}{q \left(\frac{\varepsilon_0 V}{R}\right)} = 2R \frac{V_0}{V} $$ |
| The official textbook provides the answer $f = 0.5 R \frac{V_0}{V}$. This answer not only refers to a different physical model (a solid charged sphere pierced completely through, instead of a hollow shell) but also contains a gross error. Let us derive the exact answer for a solid sphere. |
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| If the beam passes from $-\infty$ to $+\infty$ through a <b>solid</b> charged sphere with volume density $\rho$, we must consider the action of the field along the entire path (both inside and outside the sphere). We select an infinite Gaussian cylindrical flow tube of small radius $x$, coaxial with the beam. Inside this tube, the charge is located only within the bounds of the sphere itself (on the segment $[-R, R]$). |
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| The charge inside the tube is: $\Delta q = \pi x^2 \cdot 2R \cdot \rho$. |
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| Applying Gauss's theorem to this infinite tube, we obtain the exact integral of the transverse field: |
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| $$ 2\pi x \int_{-\infty}^{+\infty} E_\perp dz = \frac{\pi x^2 \cdot 2R \cdot \rho}{\varepsilon_0} \implies \int_{-\infty}^{+\infty} E_\perp dz = \frac{\rho R x}{\varepsilon_0} $$ |
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| The total transverse momentum acquired by the particle is: |
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| $$ p_\perp = \int_{-\infty}^{+\infty} q E_\perp \frac{dz}{v} = \frac{q \rho R x}{\varepsilon_0 v} $$ |
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| The focal length for a solid sphere (in terms of density $\rho$): |
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| $$ f_{solid} = \frac{m v^2 x}{p_\perp} = \frac{m v^2 \varepsilon_0 v}{q \rho R v} = \frac{2 \varepsilon_0 V_0}{\rho R} \tag{1} $$ |
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| Now it is necessary to rigorously express the potential at the center of the solid sphere $V$ in terms of $\rho$. To do this, we integrate the radial field $E(r)$ from infinity to the center: |
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| Outside the sphere ($r > R$): $E_{out} = \frac{\rho R^3}{3\varepsilon_0 r^2}$ |
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| Inside the sphere ($r < R$): $E_{in} = \frac{\rho r}{3\varepsilon_0}$ |
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| $$ V = \int_{0}^{\infty} E(r) dr = \int_{0}^{R} \frac{\rho r}{3\varepsilon_0} dr + \int_{R}^{\infty} \frac{\rho R^3}{3\varepsilon_0 r^2} dr $$ |
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| $$ V = \frac{\rho R^2}{6\varepsilon_0} + \frac{\rho R^3}{3\varepsilon_0} \left( -\frac{1}{r} \right)\Bigg|_R^\infty = \frac{\rho R^2}{6\varepsilon_0} + \frac{\rho R^2}{3\varepsilon_0} = \frac{\rho R^2}{2\varepsilon_0} $$ |
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| From this we express the density: $\rho = \frac{2\varepsilon_0 V}{R^2}$. |
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| Substituting $\rho$ into formula (1): |
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| $$ f_{solid} = \frac{2 \varepsilon_0 V_0}{\left( \frac{2\varepsilon_0 V}{R^2} \right) R} = R \frac{V_0}{V} $$ |
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| <i>(The textbook authors obtained the incorrect coefficient of 0.5 exclusively because at the final step, instead of the sphere's potential, they mistakenly used the formula for the potential difference of an infinite solid cylinder $V = \frac{\rho R^2}{4\varepsilon_0}$).</i> |
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| #### Answer |
| #### Answer |
| $$f=2R\frac{V_0}{V}$$ |
| For a spherical cavity: $$f = 2R\frac{V_0}{V}$$ |
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| <i>Note: This result is a strict consequence for an empty spherical shell. The answer in the official solutions manual is likely obtained for an alternative model, possibly - a solid charged sphere with a through cylindrical channel. |
| For a solid sphere: $$f = R \frac{V_0}{V}$$ |
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| <i>Note: The textbook answer $0.5 R\frac{V_0}{V}$ is obtained due to the erroneous application of the infinite cylinder potential formula to a solid sphere.</i> |
