Question

A positive charge 1.1X10^-11 C is located 10^-2 m away from a negative charge of the same magnitude. Point P is exactly half way between them --what is the E field at point P?

Answer 1

Being "a" the positive charge, and "b" the negative charge, and the distance between them is r=10^-2m

Suppose that both charges are located along the x axis, and the point P is the origin (x=0). The negative charge is located on the negative side of the x axis, and the positive charge is located on the positive side of the x axis. This is just a supposition.

The electric field generated by each point charge, at point P is

the negative sign is because the field generated by the positive charge points "away" from it. So for an observer at point P, the field points towards the origin.

the negative sign is because the field generated by the negative charge points "towards the charge" (remember that a negative charge acts like a sink. So for an observer at point P, the field points towards the origin (the negative charge, b, is closer to the origin than the point P, which is in the middle between the charges).

Then, at point P, the net electric field will be the sum

Is like the positive charge pushes towards the origin, and the negative charge pulls towards itself. For an observer at point P, the sum of both effects is a total effect towards the negative side of the x axis. If we change the suppositions about the locations of the charges, being charges of the same magnitude, and different sign, the field at point P will always point towards the negative charge. I would gladly explain this in more detail using the electric forces if you post a new question about it.