A proton and an $\alpha$ -particle both initially at rest are accelerated in a region of electric potential difference V and gravity free space. The proton is projected with gained kinetic energy against a uniform constant electric field and comes to momentary rest after travelling a distance $S_0$ . If the $\alpha$ -particle is also projected with gained kinetic energy against same electric field, it will come to momentary rest after travelling through:

Not really. The electric dipole moment vector directs or points from the negative charge to the positive charge. But the electric field lines that a dipole creates will point away from the positive and move to the negative charge.

Yes, the cube, which is a closed surface containing only one electric dipole will make electric flux zero. This follows Gauss's Law when the total charge inside it is zero. The field lines entering the surface will exit, and that would result in zero net flux. 

The magnitude of each charge and the distance that separates them. 

Gauss Law is only concerned with the total enclosed charge that finally tells us the total flux. The charges outside may change field patterns. They not affect the total flux. It's actually incorrect to assume the field due to the external charges should also affect the flux through the Gaussian sur

Gauss Law does not directly give the electric field in all cases. It can only be used in calculations for symmetrical surfaces: spherical, cylindrical, or planar.