Why electric potential at equatorial point of a dipole is 0? Well ok I know that electric potential at equatorial point of a dipole is 0. It is a derived conclusion.
Q1. But how this can be explained "theoretically"?
Another question that electric potential at infinity is 0.
So if I bring a positive charge from infinity to the equatorial point of the dipole then the work done by me to bring it is 0. How could this be possible if I am true?
Q2. If I am not then what is the work done in this case?
Please help 
 A: If you look at the electric filed lines you will note that a positive charge moving in the direction of the blue arrow does have a force on it but that force is always at right angles to its motion.  Hence no work is done moving the charge.

A: 
Q1. But how this can be explained "theoretically"?

I assume your question is about the concept of 'electric potential ' due to a distribution of charges and in the present case 'a dipole'.
The best way is to imagine an unit positive charge being carried/moved from infinity to a point on the equatorial line of the dipole.
Naturally your probe charge will be moving in the field of dipole charges , which are equal and opposite and 
at any segment of the equatorial line say dy ,the effective field of forces or the net electric field is sum of the two forces acting due to the two charges of the dipole, which are opposite in character but equal in magnitude -
so if  the effect of one is attractive then the effect of other one of the pair will be repulsive and 
the segment dy is symmetrically placed with respect to the dipole charges ,therefore   the two forces will be equal in magnitude but acting in 
such directions that
 their resultant will be normal to dy , and the displacement of unit positive charge being normal to net electric field can lead to a zero work done. 
If work done during translation along all such segments be added together
then the sum will be zero. so net work done will vanish and the potential will be zero.
The above displacement  was done on equatorial line but in conservative force field of electric charges the path is not important -the end points are important -the end points are infinity and point P at the equatorial line and potential at both points are zero.
the electric  potential difference  between two points are independent of actual path traversed by the charge
therefore any curved path between infinity and the point P should give us the same potential difference-which in the presence case comes to zero.
One can ask what is happening ?
Actually the potential due to one charge of the dipole is just equal and opposite to that of due to other charge  on any point on the equatorial line,therefore the potential of a dipole vanishes on any point on the equatorial line.
the above is due to symmetry of the charges of dipole and their opposite character.
A: Electric Potential is a scalar quantity. On equatorial position of the dipole, the distance from the positive charge and the negative charge is equal and they have equal, but opposite magnitude. Therefore, on equatorial position net charge is zero, q=0. This implies that no work will be done in moving a charge along the equitorial line of the dipole. This implies that electric potential will be zero. Here is a mathematical proof of the answer:

Also, work done from bringing a charge from infinity to equatorial position of a dipole is zero. Because potential is constant. And if potential is constant electric field is zero. This implies that work done is zero.
