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+### Statement
+
+$7.2.13.$ Determine the potential difference on the capacitor plates if a ribbon beam of protons perpendicular to the plates and passing through two narrow parallel slits is focused at a distance $l$ from the second plate. The protons were accelerated by the potential difference $V_0$. Distance between capacitor plates $d$. The first lining is grounded, $l \gg d$.
+
+### Solution
+
+Consider a proton moving at a small transverse distance $x$ from the beam axis.
+
+Let the first plate be grounded and let the potential of the second plate be $V$. The electric field between the plates is therefore approximately
+
+$$
+E=\frac{V}{d}.
+$$
+
+Before entering the capacitor, the protons were accelerated through a potential difference $V_0$, so their initial kinetic energy is
+
+$$
+K_0=eV_0.
+$$
+
+Hence,
+
+$$
+\frac{mv_0^2}{2}=eV_0.
+$$
+
+We assume $V>0$, so a positive proton moving from the first plate toward the second one is decelerated.
+
+According to the condition,
+
+$$
+l\gg d.
+$$
+
+Therefore, the focusing action is weak, and we will eventually be able to use the approximation
+
+$$
+\frac{V}{V_0}\ll1.
+$$
+
+For convenience, introduce
+
+$$
+\varepsilon=\frac{V}{V_0}.
+$$
+
+---
+
+#### 1. Deflection at the first slit
+
+Far from the slit, the electric field is perpendicular to the capacitor plates. Near a narrow slit, however, the field lines bend, producing a small transverse component $E_{\perp}$.
+
+Consider the fringe-field region near the first slit. For a proton at a transverse distance $x$ from the axis, the integral of the transverse field component along the longitudinal direction is of order
+
+$$
+\int E_{\perp}\,dz=\frac{V}{d}x.
+$$
+
+The change in transverse momentum is
+
+$$
+\Delta p_{\perp}=\int F_{\perp}\,dt.
+$$
+
+Since
+
+$$
+F_{\perp}=eE_{\perp},
+$$
+
+we have
+
+$$
+\Delta p_{\perp}=e\int E_{\perp}\,dt.
+$$
+
+The fringe-field region is narrow, so the longitudinal speed may be treated as approximately constant while the proton crosses it:
+
+$$
+dt=\frac{dz}{v}.
+$$
+
+Therefore,
+
+$$
+\Delta p_{\perp}=\frac{e}{v}\int E_{\perp}\,dz.
+$$
+
+Using the field integral obtained above,
+
+$$
+|\Delta p_{\perp}|=\frac{eVx}{dv}.
+$$
+
+Let the initial transverse distance from the beam axis be $x_0$. At the first slit the proton speed is $v_0$, so
+
+$$
+p_{\perp1}=\frac{eVx_0}{dv_0}.
+$$
+
+Since the deflection angle is small,
+
+$$
+\theta_1\simeq\frac{p_{\perp1}}{mv_0}.
+$$
+
+Hence,
+
+$$
+\theta_1=\frac{eVx_0}{dmv_0^2}.
+$$
+
+Using
+
+$$
+mv_0^2=2eV_0,
+$$
+
+we obtain
+
+$$
+\theta_1=\frac{Vx_0}{2V_0d}.
+$$
+
+Thus,
+
+$$
+\boxed{\theta_1=\frac{\varepsilon x_0}{2d}}.
+$$
+
+The first slit slightly deflects the proton away from the axis.
+
+---
+
+#### 2. Motion between the plates
+
+Between the plates, the electric field is approximately uniform and directed along the longitudinal direction. Therefore, there is no transverse force, and the transverse momentum remains constant:
+
+$$
+p_{\perp}=\text{const}.
+$$
+
+However, the longitudinal speed decreases because the positive proton moves toward a higher electric potential.
+
+At a distance $z$ from the first plate, the potential is
+
+$$
+\varphi(z)=\frac{V}{d}z.
+$$
+
+Energy conservation gives
+
+$$
+\frac{mv^2(z)}{2}+e\varphi(z)=eV_0.
+$$
+
+Therefore,
+
+$$
+\frac{mv^2(z)}{2}=eV_0-e\frac{V}{d}z.
+$$
+
+Factoring out $eV_0$,
+
+$$
+\frac{mv^2(z)}{2}=eV_0\left(1-\varepsilon\frac{z}{d}\right).
+$$
+
+Hence,
+
+$$
+v(z)=v_0\sqrt{1-\varepsilon\frac{z}{d}}.
+$$
+
+The transverse momentum remains unchanged after the first slit, so the transverse velocity is
+
+$$
+v_{\perp}=\theta_1v_0.
+$$
+
+Therefore, the slope of the trajectory inside the capacitor is
+
+$$
+\frac{dx}{dz}=\frac{v_{\perp}}{v(z)}.
+$$
+
+Substituting the expressions for $v_{\perp}$ and $v(z)$,
+
+$$
+\frac{dx}{dz}=\frac{\theta_1}{\sqrt{1-\varepsilon z/d}}.
+$$
+
+The transverse coordinate just before the second slit is therefore determined by
+
+$$
+x_2-x_0=\int_0^d\frac{\theta_1\,dz}{\sqrt{1-\varepsilon z/d}}.
+$$
+
+Taking $\theta_1$ outside the integral,
+
+$$
+x_2-x_0=\theta_1\int_0^d\frac{dz}{\sqrt{1-\varepsilon z/d}}.
+$$
+
+The integral is
+
+$$
+\int_0^d\frac{dz}{\sqrt{1-\varepsilon z/d}}=\frac{2d}{\varepsilon}\left(1-\sqrt{1-\varepsilon}\right).
+$$
+
+Thus,
+
+$$
+x_2-x_0=\theta_1\frac{2d}{\varepsilon}\left(1-\sqrt{1-\varepsilon}\right).
+$$
+
+Using
+
+$$
+\theta_1=\frac{\varepsilon x_0}{2d},
+$$
+
+we get
+
+$$
+x_2-x_0=x_0\left(1-\sqrt{1-\varepsilon}\right).
+$$
+
+Therefore,
+
+$$
+\boxed{x_2=x_0\left(2-\sqrt{1-\varepsilon}\right)}.
+$$
+
+---
+
+#### 3. Trajectory angle before the second slit
+
+At the second plate, the proton kinetic energy is
+
+$$
+K_2=e(V_0-V).
+$$
+
+Therefore,
+
+$$
+\frac{mv_2^2}{2}=e(V_0-V).
+$$
+
+Using $\varepsilon=V/V_0$,
+
+$$
+v_2=v_0\sqrt{1-\varepsilon}.
+$$
+
+Since the transverse momentum remains constant between the plates,
+
+$$
+\theta_{2-}=\frac{p_{\perp1}}{mv_2}.
+$$
+
+But
+
+$$
+\frac{p_{\perp1}}{mv_0}=\theta_1.
+$$
+
+Hence,
+
+$$
+\theta_{2-}=\theta_1\frac{v_0}{v_2}.
+$$
+
+Therefore,
+
+$$
+\boxed{\theta_{2-}=\frac{\varepsilon x_0}{2d\sqrt{1-\varepsilon}}}.
+$$
+
+---
+
+#### 4. Deflection at the second slit
+
+The second slit deflects the proton in the opposite direction, toward the beam axis.
+
+Using the same expression for the transverse momentum change,
+
+$$
+|\Delta p_{\perp2}|=\frac{eVx_2}{dv_2}.
+$$
+
+The corresponding angular change is
+
+$$
+|\Delta\theta_2|=\frac{|\Delta p_{\perp2}|}{mv_2}.
+$$
+
+Thus,
+
+$$
+|\Delta\theta_2|=\frac{eVx_2}{dmv_2^2}.
+$$
+
+Since
+
+$$
+mv_2^2=2e(V_0-V),
+$$
+
+we obtain
+
+$$
+|\Delta\theta_2|=\frac{Vx_2}{2d(V_0-V)}.
+$$
+
+In terms of $\varepsilon$,
+
+$$
+\boxed{|\Delta\theta_2|=\frac{\varepsilon x_2}{2d(1-\varepsilon)}}.
+$$
+
+Because the second slit deflects the proton toward the axis,
+
+$$
+\theta_{\text{out}}=\theta_{2-}-|\Delta\theta_2|.
+$$
+
+Substituting the expressions obtained above,
+
+$$
+\theta_{\text{out}}=\frac{\varepsilon x_0}{2d\sqrt{1-\varepsilon}}-\frac{\varepsilon x_2}{2d(1-\varepsilon)}.
+$$
+
+Using
+
+$$
+x_2=x_0\left(2-\sqrt{1-\varepsilon}\right),
+$$
+
+we find
+
+$$
+\theta_{\text{out}}=\frac{\varepsilon x_0}{2d}\left[\frac{1}{\sqrt{1-\varepsilon}}-\frac{2-\sqrt{1-\varepsilon}}{1-\varepsilon}\right].
+$$
+
+Let
+
+$$
+s=\sqrt{1-\varepsilon}.
+$$
+
+Then
+
+$$
+1-\varepsilon=s^2.
+$$
+
+The expression in brackets becomes
+
+$$
+\frac{1}{s}-\frac{2-s}{s^2}.
+$$
+
+Combining the terms,
+
+$$
+\frac{1}{s}-\frac{2-s}{s^2}=\frac{s-(2-s)}{s^2}=\frac{2s-2}{s^2}.
+$$
+
+Therefore,
+
+$$
+\theta_{\text{out}}=\frac{\varepsilon x_0}{d}\frac{s-1}{s^2}.
+$$
+
+Returning to $\varepsilon$,
+
+$$
+\boxed{\theta_{\text{out}}=-\frac{\varepsilon x_0}{d}\frac{1-\sqrt{1-\varepsilon}}{1-\varepsilon}}.
+$$
+
+The minus sign shows that after the second slit the proton travels toward the beam axis.
+
+---
+
+#### 5. Using the condition $l\gg d$
+
+Since
+
+$$
+l\gg d,
+$$
+
+the system acts as a weak electrostatic lens, so
+
+$$
+\varepsilon=\frac{V}{V_0}\ll1.
+$$
+
+Expand the square root:
+
+$$
+\sqrt{1-\varepsilon}=1-\frac{\varepsilon}{2}-\frac{\varepsilon^2}{8}+\ldots
+$$
+
+Hence,
+
+$$
+1-\sqrt{1-\varepsilon}=\frac{\varepsilon}{2}+\frac{\varepsilon^2}{8}+\ldots
+$$
+
+Since the expression for $\theta_{\text{out}}$ already contains one factor of $\varepsilon$, it is sufficient to keep only
+
+$$
+1-\sqrt{1-\varepsilon}\simeq\frac{\varepsilon}{2},
+$$
+
+and also
+
+$$
+1-\varepsilon\simeq1.
+$$
+
+Therefore,
+
+$$
+\theta_{\text{out}}\simeq-\frac{\varepsilon x_0}{d}\frac{\varepsilon}{2}.
+$$
+
+Thus,
+
+$$
+\boxed{\theta_{\text{out}}\simeq-\frac{\varepsilon^2x_0}{2d}}.
+$$
+
+In magnitude,
+
+$$
+\boxed{|\theta_{\text{out}}|\simeq\frac{x_0}{2d}\left(\frac{V}{V_0}\right)^2}.
+$$
+
+Thus, the first-order effects of the two slits nearly cancel, and the resulting focusing effect is of second order in $V/V_0$.
+
+---
+
+#### 6. Focusing condition
+
+Beyond the second plate, the electric field is negligible, so the proton travels in a straight line.
+
+If it reaches the beam axis at a distance $l$ from the second plate, then for a small angle,
+
+$$
+|\theta_{\text{out}}|\simeq\frac{x_2}{l}.
+$$
+
+Since
+
+$$
+x_2=x_0\left[1+O(\varepsilon)\right],
+$$
+
+while the angle itself is already of order $\varepsilon^2$, we may write, to the required accuracy,
+
+$$
+x_2\simeq x_0.
+$$
+
+Therefore,
+
+$$
+|\theta_{\text{out}}|\simeq\frac{x_0}{l}.
+$$
+
+On the other hand,
+
+$$
+|\theta_{\text{out}}|\simeq\frac{x_0}{2d}\left(\frac{V}{V_0}\right)^2.
+$$
+
+Equating these expressions,
+
+$$
+\frac{x_0}{l}=\frac{x_0}{2d}\left(\frac{V}{V_0}\right)^2.
+$$
+
+Canceling $x_0$,
+
+$$
+\frac{1}{l}=\frac{1}{2d}\left(\frac{V}{V_0}\right)^2.
+$$
+
+Hence,
+
+$$
+\left(\frac{V}{V_0}\right)^2=\frac{2d}{l}.
+$$
+
+Therefore,
+
+$$
+\boxed{V=V_0\sqrt{\frac{2d}{l}}}.
+$$
+
+#### Answer
+
+$$
+\boxed{V=V_0\sqrt{\frac{2d}{l}}}
+$$
+
+The first slit slightly defocuses the beam, while the second slit focuses it slightly more strongly. The first-order effects in $V/V_0$ nearly cancel, so the net focusing effect is of order $(V/V_0)^2$.