Why is the Electric Field Zero at this point? In a Phet Lab simulation, which is all I have at the moment to do my learning on electric field forces, the point represented as colored on the diagram below is shown to have an electric field magnitude of 0. I can see why from the arrowhead diagrams, but how can I explain this? Is it simply that the repulsion of the equidistant point charges creates a dead zone?

 A: 
Is it simply that the repulsion of the equidistant point charges creates a dead zone? 

It has nothing to do with repulsion or attraction. To find the electric field at some location due to a set of point charges, you have to add the electric field contribution due to each of the point charges. You also have to remember that the electric field is a vector, so you have to vector addition, not arithmetic addition.
The electric field due to a positive point charge points radially away from the charge, and the electric field due to a negative point charge points radially toward the charge. 
In your case, you have like charges, both positive and of equal amount. That means that along the line connecting the charges, the left charge contributes a rightward electric field, and the right charge contributes a leftward electric field. They are equidistant so they both have magnitude $$E = \frac{kq}{r^2}.$$
Doing the vector addition gives you $$ \vec{E}=\frac{kq}{r^2}(right)+\frac{kq}{r^2}(left)=0$$.
To demonstrate that it has nothing to do with attraction or repulsion of the charges themselves, consider a positive charge of $q$ and a negative charge of $-2q$, separated by distance $R$. There is a spot along the line connecting the charges, just to the "far" side of the positive charge (on the side away from the negative charge) where the electric field is zero.  
In general, the zero field point for opposite sign charges will be on the "outside" of the smaller magnitude charge.  The zero field point for like sign charges will be between the charges, closer to the smaller charge (and in the middle for equal charges). There is no zero-field point for a pair of equal-magnitude-but-opposite-sign charges.
A: Electric field is zero in that point because the sum of electric field vectors have same intensity and direction, but are opposite. That point is halfway between two like charges.
A: One particularly easy way to see that the electric field must vanish at that point is by the use of symmetry.
Let's say, in your picture, that the $z$ axis points up, the $y$ axis points right, and the $x$ axis points out of the screen, and set the origin at the point in question. The physical setup is invariant under reflections through all three of these axes about the origin (just imagine putting a mirror at the origin perpendicular to any of these axes). Thus, the electric field must be invariant under these transformations as well. Let's say it has a nonzero component in the $z$ direction, $E_z$. Then upon reflecting through the $z$-axis, this will transform to $-E_z$. Thus, by symmetry, we must have $E_z=-E_z=0$. The same logic applies to all three components, and we have $E_x=E_y=E_z=0$.
Symmetry arguments like these are both incredibly intuitive and incredibly powerful. Knowing how to apply them and always having the word "symmetry" at the front of your mind will serve you well.
Disclaimer: This would not work if the "vector" in question were instead a magnetic field, an angular momentum, or any vector constructed from a curl or a cross product. This is because these quantities are psuedovectors, and don't change sign under a reflection of the coordinate system.
A: With a little hesitance, I guess  electric field of two equal but opposite charges  is zero. Electric field due to proton is $E_p= +\frac{kq}{r^2}$, and due to electron that will be $E_e=-\frac{kq}{r^2}$. Their vector sum will be $E_{\rm net}= \frac{kq}{r^2} - \frac{kq}{r^2} = 0$. If  my reasoning is wrong, please correct me.
A: 
Electric field at a distance midway between two equal and opposite charges or between two equal and same charges
