Solution

We assume that the particles are only acted upon by the force from the electric field. It acts in the direction perpendicular to the plates. Since no forces act in the direction parallel to the plates, the momentum of the particles along this direction is conserved, and therefore, so is the velocity.

We find the deflection in a uniform field from the kinematic formula:
$$y = \frac{1}{2}at^2 \quad (1)$$

We find the acceleration from Newton's second law:
$$a = \frac{F}{m} = \frac{qE}{m} \quad (2)$$

If we denote $L$ as the length of the capacitor along the trajectory, and $v$ as the velocity along the capacitor plates, then the flight time is:
$$t = \frac{L}{v} \quad (3)$$

Substitute (2) and (3) into (1):
$$y = \frac{1}{2} \cdot \frac{qE}{m} \cdot \left(\frac{L}{v}\right)^2 = \frac{qEL^2}{2mv^2}$$

For a proton: $q = e$ and $m = m_p$.
For an $\alpha$-particle (helium nucleus $\text{He}^{2+}$): $q = 2e$ and $m = 4m_p$.

Let's substitute these into the deflection formula:
$$y_p = \frac{eEL^2}{2m_p v^2}$$
$$y_\alpha = \frac{2eEL^2}{2(4m_p)v^2} = \frac{eEL^2}{4m_p v^2}$$

Now let's find the ratio:
$$\frac{y_p}{y_\alpha} = \frac{eEL^2 / (2m_p v^2)}{eEL^2 / (4m_p v^2)} = \frac{4}{2} = 2$$

The deflection of the proton is $2$ times greater than that of the $\alpha$-particle.