Put an electron halfway between the screens. It is subject to a force F directed toward the positive screen, so it accelerates in that direction. Once it's through the screen, the force changes sign, but quickly drops to zero (for infinitely large screens, it is exactly zero everywhere outside the space bounded by the two screens). The work done by the electric field is
W = F * d / q
where
d = distance from the electron's start position to the positive screen
q = the electron's charge
All this work is now kinetic energy carried by the electron, so we also have
W = (m_e/2) * v_e^2
where
m_e = mass of the electron
v_e = final speed of the electron
Therefore,
v_e = sqrt( 2 * F * d / (m_e * q) )
If we repeat the experiment with a proton, it accelerates in the opposite direction, toward the negative screen, and its mass m_p is larger, but other than that it all goes as with the electron; the absolute values of q and F (which only depends on electric field strength and q) are the same. So,
v_p = sqrt( 2 * F * d / (m_p * q) )
Divide the two equations to get
v_e / v_p = sqrt( m_p / m_e )
Momentum is conserved, so we also have
M * v + m_e * v_e - m_p * v_p = 0
where
M = mass of the screens (and rest of the spacecraft)
v = speed gained by the screens (and rest of the spacecraft)
and the minus sign accounts for the opposite direction of travel of electron and proton. Use the expression for v_e / v_p to eliminate v_p, solve for v, and find
v = (sqrt(m_p / m_e) - 1) * (m_e / M) * v_e
In numbers, m_p ~ 1837 * m_e, so
v ~ 41.86 * (m_e / M) * v_e
and Zubrin's argument is essentially correct.