Solution

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

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 toward the center (assuming $V>0$). 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:
$$f = \frac{2eV_0 R}{2Ve} = R \frac{V_0}{V}$$