Whenever we talk about a force, there are always two objects pushing or pulling against each other. According to Newton's Third Law, every object which is pushed is also pushing, and when someone pulls on a rope, the rope automatically pulls back. However, in most circumstances our focus is on a single object and the forces acting on it, and we ignore the forces that all the other objects in the problem may feel. When we do this, we've divided the objects in the problem into a target which feels forces, and various sources which provide the forces. This distinction is artificially imposed by the problem solver—in practice every object is a source and a target—but is a very useful distinction to make.
In the figure to the right we have two positive point charges repelling one another. We've designated the charge on the right as the target, the other charge as the source, and we will be calculating the force of the source on the target. To facilitate this, we define the vector \(\vec d\) which starts at the source and goes to the target. Its magnitude \(d\) is the distance between the two charges.
This vector \(\vec d\) is closely related to the direction of the force on the target, as seen in the figure to the right. If the charges have the same sign (so that q_{s}q_{t} > 0) then the force points in the same direction as \(\vec d\); if the charges have opposite signs (q_{s}q_{t}<0), then the force points in the direction of \(-\vec d\). In short, the force always points in the direction \(q_sq_t\vec d\).
This insight leads us to the vector version of Coulomb's Law:
$$\vec F=k{q_sq_t\over d^2}{\vec d\over d}$$
The left half of the equation is just what we saw before, except I've removed the absolute value signs from the numerator. I then multiplied by \(\vec d\) to give the direction of the force, but I had to divide this by \(d\) to keep the magnitude of the force the same:
$$\begin{eqnarray}
\abs{\vec F}&=&\abs{k{q_sq_t\over d^2}{\vec d\over d}}\\
&=&k{\abs{q_sq_t}\over d^2}{\abs{\vec d}\over d}\\
&=&k{\abs{q_sq_t}\over d^2}{d\over d}\\
\implies \abs{\vec F}&=&k{\abs{q_sq_t}\over d^2}\\
\end{eqnarray}$$
Consider a positive 3 nC charge and a negative –2 µC charge sitting on the \(x\) axis as shown in the figure. Let's find the force on both charges, by allowing each one to be the target in turn.
Find the force on the negative charge, on the right.
Suppose the charge on the right is the target, and the charge on the left is the source. Then the vector \(\vec d\), shown in green, is 5 meters long and points in the positive x direction, so \(\vec d=5\hat x\u{m}\), and
$$\vec F=\left(9\ten9\u{Nm^2\over C^2}\right)
\frac{({\color{blue}{3\ten{-9}\u{C}}})({\color{red}{-2\ten{-6}\u{C}}})}
{\color{green}{(5\u{m})^3}}
(\color{green}{5\hat x\u{m}})$$
$$=\boxed{2.2\ten{-6}\u{N}(-\hat x)}$$
The minus sign properly belongs to the \(\hat x\) in this expression, as it tells us that the force on the target charge points in the \(-\hat x\) direction, that is to the left, which agrees with our knowledge that the negative target should be attracted to the positive source.
Find the force on the positive charge, on the left.
If we make the positive charge the target, then the vector \(\vec d\) points in the opposite direction: \(\vec d=-5\hat x\u{m}\), and
$$\vec F=\left(9\ten9\u{Nm^2\over C^2}\right)
\frac{(\color{red}{-2\ten{-6}\u{C}})(\color{blue}{3\ten{-9}\u{C}})}
{\color{green}{(5\u{m})^3}}
(\color{green}{-5\hat x\u{m}})$$
$$=\boxed{2.2\ten{-6}\u{N}\,(\hat x)}$$
The only difference with part (a) is the negative sign in \(\vec d\), which makes the force point to the right (\(+\hat x\)), which again agrees with our understanding that the positive target should be attracted to the negative source. The magnitudes of the two forces are the same because of Newton's Third Law.
Now let's consider an example where the two charges don't lie on a single axis, as shown.
Find the force on the charge in the upper-left corner.
We need to write the vector \(\vec d\) in component form: that is, we need to find a way to get from the source to the target by only moving along grid lines (i.e. horizontally or vertically). In this case, we can start at the source and move 5 meters to the left and then 5 meters up to get to the target. Therefore
\(\vec d=(5(-\hat x)+5(-\hat y))\u{m}\).
Once we have this vector we can get the distance d between the charges using the Pythagorean Theorem:
$$d=\abs{\vec d}=\sqrt{(-5\u{m})^2+(-5\u{m})^2}=7.07\u{m}$$
and so
$$
\begin{eqnarray*}
\vec F &=& {\left(9\ten9\u{Nm^2\over C^2}\right)}\frac{(-5\ten{-9}\u{C})(-2\ten{-6}\u{C})}{(7.07\u{m})^2}\frac{(-5\hat x-5\hat y)\u{m}}{7.07\u{m}}\\
&=& {\left(9\ten9\u{Nm^2\over C^2}\right)}\frac{1\ten{-14}\u{C^2}}{(7.07\u{m})^3}(-5\hat x-5\hat y)\u{m}\\
&=&(2.55\ten{-7}\u{N})(-5\hat x-5\hat y)\\
&=&\boxed{\left[1.27\ten{-6}(-\hat x)+1.27\ten{-6}(-\hat y)\right]\u{N}}\\
\end{eqnarray*}$$
This force points upward and to the left, which means the target feels a force away from the source, as it should.
In practice, it is usually convenient to rewrite Coulomb's Law slightly by combining the two denominators, giving us