Solution

The solution presented in some sources solely considers the device as a "velocity selector" where electric and magnetic forces perfectly cancel each other out ($evB = eE$), resulting in straight-line motion. However, this is only a special case. To find the *full* range of conditions, we must analyze the electron's dynamics, assuming an infinitely long capacitor where the electron must not hit the plates.

Let's establish a coordinate system: the $x$-axis is directed along the initial velocity $v$ (to the right), the $y$-axis is directed vertically upwards, and the magnetic field $\vec{B}$ is along the $z$-axis (pointing towards us).
The electric field $E = V/h$. For an electron (charge $-e$) to have a chance to pass, the electric force must oppose the upward magnetic Lorentz force, meaning the vector $\vec{E}$ must be directed upwards.

Let's write the equations of motion (Newton's second law projections) for an electron of mass $m_e$.
Along the $x$-axis, the only force is the projection of the magnetic force arising from the vertical velocity component:
$$m_e \frac{dv_x}{dt} = -eB v_y = -eB \frac{dy}{dt}$$

Multiplying by $dt$ and integrating with initial conditions (at $y = 0$, $v_x = v$):
$$m_e(v_x - v) = -eBy \implies v_x = v - \frac{eB}{m_e}y \quad (1)$$

Along the $y$-axis, the electron experiences both the downward electric force and the magnetic force from the horizontal velocity:
$$m_e \frac{dv_y}{dt} = eB v_x - eE$$

Substitute $v_x$ from equation $(1)$:
$$m_e \frac{dv_y}{dt} = eB \left( v - \frac{eB}{m_e}y \right) - eE = e(Bv - E) - \frac{e^2B^2}{m_e}y$$

This equation describes a harmonic oscillator. Let's find the equilibrium coordinate $y_0$ where the vertical acceleration is zero:
$$e(Bv - E) - \frac{e^2B^2}{m_e}y_0 = 0 \implies y_0 = \frac{m_e(Bv - E)}{eB^2}$$

Since the electron enters the capacitor at $y = 0$ with zero vertical velocity, this point is an extreme point of its oscillation. The maximum deflection from the central axis will be:
$$y_{max} = 2y_0 = \frac{2m_e(Bv - E)}{eB^2}$$

For the electron to pass through the capacitor without hitting the plates, its maximum absolute deflection must not exceed half the distance between the plates:
$$|y_{max}| \le \frac{h}{2}$$
$$\left| \frac{2m_e(Bv - E)}{eB^2} \right| \le \frac{h}{2}$$

Substituting $E = V/h$ and dividing the term inside the modulus by $B$, we obtain the final condition for the initial velocity $v$:
$$\left| v - \frac{V}{Bh} \right| \le \frac{eBh}{4m_e}$$

*Note: The trivial straight-line case $v = V/(Bh)$ mentioned in simplified solutions corresponds to the center of this valid velocity range, where the amplitude of oscillations is zero.*