$7.2.9.$ 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
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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 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.
#### 1. Physical model and superposition principle
−
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 $v$.
+
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).
−
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:
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1. An ideal continuous charged sphere with surface density $\sigma$.
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2. Two small disks with charge density $-\sigma$ located at the entrance and exit holes (they "cut out" the holes in the continuous sphere).
+
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:
+
1. An ideal continuous charged sphere with a surface charge density $+\sigma$.
+
2. Two "virtual" disks with a charge density $-\sigma$, located strictly at the positions of the entrance and exit 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.
+
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 $-\sigma$.
#### 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 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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The cylinder cuts out an area with a charge $\Delta q_1 = \pi x^2 \sigma$ from the charged shell.
+
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).
+
By Gauss's theorem, the flux of the electric field vector through the lateral surface of this cylinder is:
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$$ 2\pi x \int_{1} E_\perp dz = \frac{\pi x^2 \sigma}{\varepsilon_0}$$
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$$ 2\pi x \int E_\perp dz = \frac{\pi x^2 \sigma}{\varepsilon_0}$$
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From this, the integral of the transverse field in the first hole zone is:
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, acquiring 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):
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$$f_1 = vt = v \frac{x}{v_\perp} = x \frac{v}{v_\perp} = x \frac{mv}{p_{\perp 1}}$$
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The time to reach the axis is $t = \frac{x}{v_\perp}$. The focal length after the first hole $f_1$:
+
$$f_1 = vt = v \frac{x}{v_\perp} = x \frac{mv}{p_{\perp 1}} = x \frac{mv}{\frac{q \sigma x}{2\varepsilon_0 v}} = \frac{2mv^2 \varepsilon_0}{q \sigma}$$
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.
+
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.
#### 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, the particle crosses the local field of the second virtual disk ($-\sigma$) and receives a similar transverse momentum directed towards the axis:
$$ f = x \frac{mv}{p_\perp} = x \frac{mv}{\frac{q \sigma x}{\varepsilon_0 v}} = \frac{mv^2 \varepsilon_0}{q \sigma}$$
−
Again substituting $mv^2 = 2qV_0$ and $\sigma = \frac{\varepsilon_0 V}{R}$, we get the final answer:
+
Substituting $mv^2 = 2qV_0$ and $\sigma = \frac{\varepsilon_0 V}{R}$, we obtain the final answer for an empty spherical shell:
$$ f = \frac{(2qV_0) \varepsilon_0}{q \left(\frac{\varepsilon_0 V}{R}\right)} = 2R \frac{V_0}{V}$$
#### Answer
$$f=2R\frac{V_0}{V}$$
*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.*
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### Statement
### Statement
$7.2.9.$ 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 $V \ll V_0$?
@@ -4,61 +4,60 @@Statement
### Solution
### Solution
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 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.
#### 1. Physical model and superposition principle
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 $v$.
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).
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 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:
1. An ideal continuous charged sphere with surface density $\sigma$.
1. An ideal continuous charged sphere with a surface charge 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 "virtual" disks with a charge density $-\sigma$, located strictly at the positions of the entrance and exit 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.
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 $-\sigma$.
#### 2. First transverse momentum at the entrance
#### 2. First transverse momentum at the entrance
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 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.
The cylinder cuts out an area with a charge $\Delta q_1 = \pi x^2 \sigma$ from the charged shell.
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).
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:
$$ 2\pi x \int_{1} E_\perp dz = \frac{\pi x^2 \sigma}{\varepsilon_0}$$
$$ 2\pi x \int E_\perp dz = \frac{\pi x^2 \sigma}{\varepsilon_0}$$
From this, the integral of the transverse field in the first hole zone is:
From this, the integral of the transverse field in the local hole zone is:
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, acquiring a transverse velocity $v_\perp = \frac{p_{\perp 1}}{m}$ directed towards the beam axis.
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 to reach the axis is $t = \frac{x}{v_\perp}$. The focal length after the first hole $f_1$:
$$f_1 = vt = v \frac{x}{v_\perp} = x \frac{v}{v_\perp} = x \frac{mv}{p_{\perp 1}}$$
$$f_1 = vt = v \frac{x}{v_\perp} = x \frac{mv}{p_{\perp 1}} = x \frac{mv}{\frac{q \sigma x}{2\varepsilon_0 v}} = \frac{2mv^2 \varepsilon_0}{q \sigma}$$
Substituting the found momentum:
The kinetic energy is given by the accelerating voltage: $mv^2 = 2qV_0$.
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.
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.
#### 4. Second momentum and total focal length
#### 4. Second momentum and total focal length
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, the particle crosses the local field of the second virtual disk ($-\sigma$) and receives a similar transverse momentum directed towards the axis:
$$ f = x \frac{mv}{p_\perp} = x \frac{mv}{\frac{q \sigma x}{\varepsilon_0 v}} = \frac{mv^2 \varepsilon_0}{q \sigma}$$
$$ f = x \frac{mv}{p_\perp} = x \frac{mv}{\frac{q \sigma x}{\varepsilon_0 v}} = \frac{mv^2 \varepsilon_0}{q \sigma}$$
Again substituting $mv^2 = 2qV_0$ and $\sigma = \frac{\varepsilon_0 V}{R}$, we get the final answer:
Substituting $mv^2 = 2qV_0$ and $\sigma = \frac{\varepsilon_0 V}{R}$, we obtain the final answer for an empty spherical shell:
$$ f = \frac{(2qV_0) \varepsilon_0}{q \left(\frac{\varepsilon_0 V}{R}\right)} = 2R \frac{V_0}{V}$$
$$ f = \frac{(2qV_0) \varepsilon_0}{q \left(\frac{\varepsilon_0 V}{R}\right)} = 2R \frac{V_0}{V}$$
#### Answer
#### Answer
$$f=2R\frac{V_0}{V}$$
$$f=2R\frac{V_0}{V}$$
*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.*
*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.*
