The problem I am working on is, "In the figure below, determine the point (other than infinity) at which the electric field is zero. (Let $q_1 = -2.45\ \mu C$ and $q_2 = 6.5\ \mu C$)
Here is a little commentary my author gives on this problem:
Each charged particle produces a field that gets weaker farther away, so the
net field due to both charges approaches zero as the distance goes to infinity in any direction. We are asked for the point at which the nonzero fields of the two particles add to zero as oppositely directed vectors of equal magnitude.
The electric field lines are represented by the curved lines in the diagram. The field of positive charge $q_2$ points radially away from its location. Negative charge $q_1$ creates a field pointing radially toward its location. These two fields are directed along different lines at any point in the plane except for points along the line joining the particles; the two fields cannot add to zero, except at some location along this line. To the right of the positive charge on this line, the fields are in opposite directions but the field from the larger magnitude of the positive charge dominates. In between the two particles, the fields are in the same direction and add together. To the left of the negative charge, the fields are in opposite directions and at some point they will add to zero such that $E = E_1 + E_2 = 0$.
For the first selection, are they saying that the particles together create one field? How is that so? As for the second selection, I honestly do not know what is it is saying.
Also, I know that the electric force that two particles exert on each other are equal in magnitude and opposite, but the electric fields aren't equal and opposite. So, when I find that the distance between the two particles where the two electrics fields are equal and opposite and they cancel, what is happening physically? What does it mean for electric fields to cancel each other out?
Best Answer
You are looking for the point where a test charge doesn't feel any force, because it's attraction to, say $q_1$, cancels out with repulsion from $q_2$. I don't see your figure, so let's assume that $q_1$ and $q_2$ are on the $x$-axis, with $x$-coordinates $x_1$ and $x_2$. "By symmetry" (as explained in the commentary) you will guess that the point your are looking for is also on the $x$-axis. Then you have to solve the equation
$\frac{q_1{\rm sign}(x-x_1)}{(x-x_1)^2} + \frac{q_2{\rm sign}(x-x_2)}{(x-x_2)^2}=0$.
For test charges between the two particles, the forces coming from the two particles point in the same direction, so they can never cancel out. So we have to be either to the right or to the left of both particles, where the signs in the equation at the same. We can cancel them from the equation and get a quadratic equation that is easy to solve. Identify the right solution! (the quadratic equation give a fake solution that corresponds to the point between the two charges, where the forces have the same size, but also the same direction, so they don't cancel out).