Solution

From the grounding conditions (zero potential), the electron acquires velocity $v_0$, and the energy of the electron does not change.

$$e V_0 = \frac{mv_0^2}{2}$$

The charge $q$ on the horizontal plates is derived from the capacitor equation, considering that the slit has the shape of a circle.From the grounding conditions (zero potential), the electron acquires velocity $v_0$, while the electron energy does not change.

$$q = 2 \pi R C V$$

Since the charge distribution has axial symmetry, we consider the interaction of the electron and the total charge of the horizontal plates as the interaction of two point charges.

Since the angle $\alpha$ is small enough under the problem conditions, we assume that the horizontal velocity changes insignificantly.We derive the charge $q$ on the horizontal plates from the capacitor equation, considering that the slit has the shape of a circle.From the grounding conditions (zero potential), the electron acquires velocity $v_0$, and the electron energy does not change.

$$E(\theta) = \frac{kq}{(r/sin(\theta))^2}$$
$$F(\theta)=eE(\theta)=\frac{kq}{(r/sin(\theta))^2}$$
$$dr=vdt$$
$$F_y=Fsin(\theta)$$
$$dP_y=F_ydt=\frac{kqe}{vr}sin(\theta)d\theta$$
$$P_y=\frac{kqe}{vr}\int_{0}^{\pi}sin(\theta)d\theta = \frac{2kqe}{vr}$$
$$P_x = mv$$
$$tan(\alpha)=\frac{P_y}{P_x} = \frac{qe}{2 \pi \epsilon_0 m v_0^2 R}$$
$$\alpha \approx \frac{qe}{2 \pi \epsilon_0 m v_0^2 R}$$

Now substitute the value of $q,e$ and compare with the formula from the problem.

$$\alpha \approx \frac{CV}{2 \epsilon_0 V_0}$$
$$k=\frac{1}{2 \epsilon_0}$$