A sample of an ideal gas is taken through the cyclic process ABCA as shown in figure. It absorbs, 40 J of heat during the part AB, no heat during BC and rejects 60 J of heat during CA. A work of 50 J is done on the during the part BC. The internal energy of the gas at A is 1560 J. The work done by the gas during the part CA is:

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.