In electrostatics, $\nabla\times\mathbf E=0$. Why is this? I can understand why this is mathematically but I do not understand the actual why, like in words why this is.
 A: If the curl of a static electric field were nonzero, you could move a point charge around a closed circuit and it would have more energy when it got back to its starting point than it did when it was there last time.  This would be a violation of conservation of energy.
A: Well, in short, there is no deeper understanding as to why the curl of the electric field is zero. The Maxwell equations are the most basic laws of electrodynamics and when you constrain yourself to the static case, that the curl of the electric field vanishes is one of the statements of the Maxwell equations.
However, you can try to better understand this law by studying it in different equivalent formulations. For example, within electrostatics, Coloumb's law and the Maxwell equations are equivalent. Thus, you can see the vanishing curl of the electric field as an aspect of the Coloumb's law if you wish. This is easy to intuit. Any configuration of electric fields can be imagined to be conjured up via superposing multiple point electric charges (either of finite or infinite charge density). Since the electric field of a point charge is radial, its curl is trivially zero. Since the curl is a linear operator, the curl of the superposition of various electric fields whose curls are zero is also zero. 
But, in the end, I would say that this description is really just a way to intuit. In the larger context, the Maxwell equations are the most basic principles of electrodynamics and thus, it is in vain to "explain" them in terms of some other results of electrodynamics. 
