Why is there no electric potential energy when it is directly above an electric dipole? If you have two opposite charges of equal magnitude, why is the potential energy of another random particle that is directly above or below the middle distance between the two charges have 0 electric potential energy?
Summing up the electric potential energy from both charges yields $\frac{kQq}{r} - \frac{kQq}{r} = 0$. Why is this the case? If the particle is positive and I released it at rest, it would instantly move with the electric field lines thus gaining a kinetic energy.
 A: Your intuition is fine - the particle will move just as you said. There are two problems with your reasoning, which lead to the conflict with your intuition.
First problem: It seems that you're imagining that "zero potential energy" means that an object won't feel any force, and won't move. That simply isn't the case - it's the spatial derivative of the potential energy (for example, $\frac{dU}{dr}$), which determines the force on the particle. To provide an everyday example, if we define gravitational potential energy on the floor to be zero, an object sitting on the floor will fall if you open a trap door under it. That's because even though it has zero potential energy, the derivative of its potential energy with respect to height ($\frac{dU}{dh}$ is not zero, and therefore it feels a force.
Second problem: potential energy isn't defined unless you specify a reference point. You can pick any spot anywhere to have zero potential energy. A common choice for this sort of problem is to define the infinite distance away from the system to have a potential energy of zero. In that case, if you imagine bringing a test charge from infinity to the position you describe, along its path, the charge is being attracted by one charge, and repelled by the other, in exactly equal amounts along the path it takes along the midline of your situation. Therefore, it requires no energy to move a particle along this path, and its potential energy remains the same as it was at an infinite distance, which we defined to be zero.
Third probelm: your equation $U(r) = \frac{kQq}{r} - \frac{kQq}{r}$, is a little deceptive. Your variable $r$ isn't a real coordinate, since it simultaneously is representing the distance from the negative and positive charge at the same time; that equation is only coincidentally accurate on the midline, and nowhere else. A globally accurate equation would be $U(x, y, z) = \frac{kQq}{\sqrt{x^2 + y^2 + (z-d)^2})} - \frac{kQq}{\sqrt{x^2 + y^2 + (z+d)^2}}$, where $x$, $y$, and $z$ are cartesian coordinates, and the charges are each a distance $d$ from the midpoint. You can see that this potential energy is only zero at infinity and wherever $z = 0$, and furthermore even where $z = 0$, the spatial derivatives are not zero, which is why there is a force there.
You can visualize both the electric field lines and equipotential lines with this diagram:   (image credit)
You can see that along the midline, there is an equipotential line that extends straight out to infinity, and yet there is a nonzero electric field everywhere along that line.
