Figure1 Q. Three charges of equal magnitude are anchored at the vertices of an equilateral triangle of side length a, as shown in the diagram.

Find the electric field at point P. P is located midway between the negative charges and at the origin of the coordinate system. Express your answers in terms of q, a, 0, , and .

Wowsers, what an exciting question! Don't worry, there's a lot of little things going on, but if you take it slowly & keep things organized, there'll be no problem at all.

To find the electric field at point P we'll be using the equation

where the direction of the electric field is determined by what the sign of the charge is.

In the topic of electricity and magnetism, frequently you'll find that the problem simplifies greatly because of the symmetry of the question.
Writers of physics question can be mean, but very rarely are they purely evil.
Frequently, a difficult problem is actually simple because of symmetry. Look for that in the other problems you'll find in this section.

First, look at the two negative charges on either side of point P. They have the same magnitude of charge, are the same distance from that point and are directly opposite each other. Without lifiting a finger, you should recognize that the electric field due to those two charge will, at that point, cancel out. WHEW! All we need to worry about is that positive charge on top.

First, we'll deal with the magnitude of the E-field due to the positive charge. We'll just utilize that equation from above. The only difficulty will be in determining the distance from the charge to P. not a big problem, though... Just look at the diagram.

Figure 2

The distance r is nothing more than one of the legs of a right triangle. The hypotenuse of the triangle is a and it's base is a/2. Applying a quick Pythagorean to it


And then, with this distance found, find the magnitude of the electric field,
The electric field due to positive charges always points AWAY from the charge. So, in terms of the unit vectors, away from that charge in the direction of P would be in the negative direction.

So our final answer to this part of the question would be

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