In the vector equation for the electrostatic force:
$$\vec{F} = \frac{1}{4\pi\epsilon_{0}}\cdot\frac{q_1\cdot q_2}{r^2}\cdot \hat{r} \tag{1}$$
Is the absolute value of $q_1$ and $q_2$ taken? As $(1)$ would then become :
$$\vec{F} = \frac{1}{4\pi\epsilon_{0}}\cdot\frac{|q_1|\cdot |q_2|}{r^2}\cdot \hat{r} \tag{2}$$
I've found that $(1)$ leads to contradictions, if either $q_1$ or $q_2$ is negative, as the unit vector $\hat{r}$ needs to have one of its entries negated (corresponding to a geometrical 'change in direction' of the vector) in order for $\vec{F}$ to be logical.
If $(2)$ is used however, no contradiction arises, and the unit vector does not need to have any of it's entries negated, thus we don't need to 'change the direction' of the vector as our result for the force $\vec{F}$ 'makes sense'
An easy way to test what I've said, is to assume $\vec{F}, \hat{r} \in \mathbb{R^1}$ (a one-dimensional space: a line) and $q_1 = – \lambda, q_2 = \lambda \ni \lambda \in \mathbb{R}$
Which is the correct vector equation for electrostatic force $(1)$ or $(2)$?
My university makes use of $(1)$, in formula sheets for lectures/tests/exams, but not $(2)$ and our lecturer has not explicitly stated that the absolute value of $q_1$ and $q_2$ must be taken, which is why I've asked this question.
Best Answer
It's important to be clear on what these variables mean: specifically, this equation is for calculating the force exerted on one charge by the other. The unit vector $\hat{r}$ points from the charge exerting the force to the charge experiencing the force. E.g. if you're calculating the force experienced by $q_1$, you should have this image in mind:
Now, the force on $q_1$ might be toward or away from $q_2$, depending on the signs of the charges. Clearly, to represent a force away from $q_2$, $\hat{r}$ should be multiplied by something positive, and to represent a force toward $q_2$, $\hat{r}$ should be multiplied by something negative. You know that like charges repel and opposite charges attract, so if $q_1$ and $q_2$ have the same sign, $\hat{r}$ needs to have a positive coefficient, and if they have opposite signs, $\hat{r}$ should have a negative coefficient.
Your formula (1) does exactly this. The coefficient $\frac{1}{4\pi\epsilon_0}\frac{q_1 q_2}{r^2}$ is positive when $q_1$ and $q_2$ have the same sign, and negative when they don't. If you take the absolute values, as in formula (2), the coefficient is positive in all cases, so that will give the wrong result for opposite-signed charges.
I'm not quite sure how you found that equation (1) leads to contradictions; there's no need to alter the unit vector $\hat{r}$.
You could take the opposite convention for the direction of $\hat{r}$, i.e. take it to point from the charge experiencing the force to the charge exerting the force. This is not conventional, and it will confuse a lot of people, but it is physically valid. In that case, you would need to put an overall negative sign in the formula, like this: $$\vec{F} = -\frac{1}{4\pi\epsilon_0}\frac{q_1 q_2}{r^2}\hat{r}\tag{opposite convention}$$ But still, the same argument as above will show you that the results you get are incorrect if you take the absolute values of the charges.
What is often done is to use Coulomb's law to calculate the magnitude of the force only, and then figure out the direction by physical reasoning. If you calculate the magnitude, $\lVert \vec{F}\rVert$, then you do take the absolute value of everything. $$\lVert\vec{F}\rVert = \left\lVert\frac{1}{4\pi\epsilon_0}\frac{q_1 q_2}{r^2}\hat{r}\right\rVert = \left\lvert\frac{1}{4\pi\epsilon_0}\frac{q_1 q_2}{r^2}\right\rvert \underbrace{\left\lVert\hat{r}\right\rVert}_{1} = \frac{1}{4\pi\epsilon_0}\frac{\lvert q_1\rvert\,\lvert q_2\rvert}{r^2}$$