2 free charges +Q each are placed a distance r apart. A 3rd charge is so placed between them that entire?

Question

system is in equilibirium. What is the location, magnitude and sign of 3rd charge?

Answer 1

Let the two +Q charges be placed left to right with the distance r separating them. Let a third charge of unknown charge q be placed a distance x to the right of the first charge. Now, look at each of the +Q charges and sum the electric force acting on them. For example, the left hand +Q charge experiences a repellent force from the +Q on the right, pushing it towards the left. To balance this out, there must be an attractive force pulling it to the right, so the q charge is negative. However, this is just an observation. In order to maintain the integrity of the algebra, we should continue to assume that q is positive, but recall the negative result llater when we check our answer.. The proportionality factor for electric field strength can be ignored, because it will cancel out, so you can treat Q/r^2 as the value under consideration. In this case, we need Q/r^2 + q/x^2 = 0. However, because the +Q charges are equal in magnitude, I will make the assumption that the distance x = r/2, so Q/r^2 = -4q/r^2 ==> Q = -4q ==> q = -Q/4. Perform the same force balance on the right hand +Q charge to verify.

If the initial charges had been unequal, x would have remained a variable, and the complementary distance (r - x) would have been used for the second force balance in order to form a system of equations.

Answer 2

Too much work. We have a lot of symmetry in this problem. Since the charges are equal, we can infer that the position of the third charge is in the middle.

Now consider one of the original charges. Since the force goes as the square of the distance, the charge that is half as far away needs to be 1/4 the magnitude larger in order to cancel the effect of the more distant one.

Always take advantage of symmetry!