The figure shows a region of length 'l' with a uniform magnetic field of 0.3 T in it and a proton entering the region with velocity 4 * 10⁵ ms⁻¹ making an angle 60° with the field. If the proton completes 10 revolution by the time it cross the region shown, 'l' is close to (mass of proton = 1.67 * 10⁻²⁷ kg, charge of the proton = 1.6 * 10⁻¹⁹C)

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.