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The formula of Coulomb's law:

If we have two charge q1, q2. We have a formula

1) The magnitude of Coulomb's law: $$F= \frac{k|q_{1}||q_{2}|}{r^{2}}$$

2) The vector form: $$\vec{F_{12}}= \frac{k(q_{1}q_{2})}{r^{2}}\hat{r_{12}}$$

$F_{12}$ is the force on $q_{2}$ due to $q_{1}$.

I don't get idea why we cannot put absolute data in the vector formula (the second formula). Why is it impossible to write:$\vec{F_{12}}= \frac{k(|q_{1}||q_{2}|)}{r^{2}}\hat{r_{12}}$

My suggestion:

$$\vec{F_{12}}= \frac{k(|q_{1}||q_{2}|)}{r^{2}}\hat{r_{12}}$$

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    $\begingroup$ I feel so painful to try to understand your question... could you please try to use Latex? $\endgroup$
    – Shing
    Commented Oct 25, 2016 at 14:39
  • $\begingroup$ Could I ask what latex is in computing !? $\endgroup$ Commented Oct 25, 2016 at 14:42
  • $\begingroup$ I am so sorry for the confusion it's the just that if we have two charges q1 and q2. To measure the magnitude of force, one of the component we use is |q1||q2|. But my question is why in the vector forms we cannot write |q1||q2| but we have to write q1 multiplied by q2 (q1xq2). This is just what I understand from observation ! $\endgroup$ Commented Oct 25, 2016 at 14:44
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    $\begingroup$ @ProtonUpUpDown I have edited your question. Please take a look at the edit and learn Latex asap, cause any question that you ask here must be of that form, both for enhanced presentability and clarity. $\endgroup$ Commented Oct 25, 2016 at 14:45
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    $\begingroup$ @NaveenBalaji: oh thank you very much :). I will learn how to use it. Sorry I have never been interacting on the internet before. $\endgroup$ Commented Oct 25, 2016 at 14:49

2 Answers 2

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If I understand your question correctly, the explanation is the following. The direction of the vector $\vec{F}_{12}$ depends on the sign of $q_1$ and $q_2$. If they have the same sign it will point in one direction, while if they have opposite sign it will point in the opposite direction. Therefore you cannot put the absolute value in the vector expression of Coulomb's law, because you would lose some information about the resulting $\vec{F}_{12}$ vector.

When you only care about the magnitude of this vector, on the other hand, you have to put the absolute value because this has to be a positive number. Note that $|q_1||q_2|$ is the same positive number, independent of the sign of $q_1$ and $q_2$.

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  • $\begingroup$ Sorry I still don't know how to add image in the comment section. This is what I am wondering. I mean if I write the one I suggest. The way is utterly depenedent on the unit vector, which is more convenient sv1.upsieutoc.com/2016/10/25/suggest2.png $\endgroup$ Commented Oct 25, 2016 at 15:32
  • $\begingroup$ I think Steeven's answer addresses this issue. There is a standard definition of the unit vector. Of course you are free to use the convention that you prefer for your calculations, but what is the advantage of having a unit vector that depends on the sign of the charges? $\endgroup$
    – DelCrosB
    Commented Oct 25, 2016 at 16:09
  • $\begingroup$ I have just read again Steeven's answer and your replies again. I understand very well the topic now :). Thank you , DelCrosB $\endgroup$ Commented Oct 26, 2016 at 8:36
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The reason is the definition of $\hat{r}$:

$\hat r$ is a unit vector which is pointing from the other charge to the charge itself.

The vector version of Coulomb's law is:

$$\vec F_1=k\frac{q_1q_2}{r^2}\hat r_{21}$$

Note the difference in notation from your expression: $\vec F_1$ is the force felt by charge 1. $\hat r_{21}$ is the unit vector from charge 2 towards 1.

  • Now, like charges (same sign) repel, so the force will point in the same direction as $\hat r_{21}$ - in other words, away from the other charge.

  • For opposite signs, the $q_1q_2$ term will be negative, so the force will turn around. The force will point in the opposite direction as $\hat r_{21}$, which is towards the other charge. Which it also should, since they attract.

So, you can't add absolute values here. Then the formula would only be correct for like charges. Their signs take care of the correct direction of the force in this formula.

The key point is that the unit vector $\hat r$ doesn't point from the charge itself to the other, but rather from the other charge to the charge itself. Be always very clear about such definitions of each parameter.


When only the magnitude is needed, we do not care about direction. Which means, we do not care about any signs, since the signs only take care of direction. Therefor your expression no. 1) above has absolute values, so that any possible signs are removed. The magnitude is then always positive, and we must just remember that directions cannot be seen here.

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  • $\begingroup$ sv1.upsieutoc.com/2016/10/25/suggest2.png . I mean I want the unit vector to express the direction rather a combination between the unit vector and the pair of q1 and q2. $\endgroup$ Commented Oct 25, 2016 at 15:35
  • $\begingroup$ "I want the unit vector to express the direction " Which direction? The unit vector is expression a direction. For the question in the link: The reason that you cannot add absolute value bars as $$\vec F_1=k\frac{|q_1q_2|}{r^2}\hat r_{21}$$ is that the formula then would not work if the charges where not of opposite signs. Indeed, the formula is working and correct if the signs are opposite, but it is not a general formula if it only works in a certain situation. The general formula must work for any values and signs of the charges. $\endgroup$
    – Steeven
    Commented Oct 25, 2016 at 16:49
  • $\begingroup$ Perhaps there's no standard here, but I take $\vec{r}_{21} = \vec{r}_2 - \vec{r}_1$, the position of 2 relative to 1. It points from 1 to 2. With that we can write the force on 2 due to 1 $\vec{F}_{21} = \ldots\hat{r}_{21}$. The order is always 2,1 in all formulas, and the indices take care of the direction with no need to think. (Provided, of course, that there are no absolute value signs around the charges. We do need to account for the fact that unlike charges repel and like charges attract.) $\endgroup$
    – garyp
    Commented Oct 25, 2016 at 17:28
  • $\begingroup$ @garyp "with no need to think" I would claim here that there certainly is reason to think extra when the same set of indices has different meaning on different parameters. But yes, if there are more than one force affecting the charge, I do though come closer to agreeing. But what is simple and what isn't might be a neverending discussion. Point is to be clear about what exactly the parameters mean, no-matter their notation. $\endgroup$
    – Steeven
    Commented Oct 25, 2016 at 17:50
  • $\begingroup$ I don't see where the same set of indicies have different meanings, so it all makes sense to me. What makes sense to someone else may be entirely different, so I agree that the point is to be clear about the meaning. $\endgroup$
    – garyp
    Commented Oct 25, 2016 at 18:01

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