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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:

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

Spoilt Milk
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PandoraU.U.D
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    I feel so painful to try to understand your question... could you please try to use Latex? – Shing Oct 25 '16 at 14:39
  • Could I ask what latex is in computing !? – PandoraU.U.D Oct 25 '16 at 14:42
  • 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 ! – PandoraU.U.D Oct 25 '16 at 14:44
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    @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. – Spoilt Milk Oct 25 '16 at 14:45
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    @NaveenBalaji: oh thank you very much :). I will learn how to use it. Sorry I have never been interacting on the internet before. – PandoraU.U.D Oct 25 '16 at 14:49
  • May I ask why the notation $_{12}$ is only on some of the $F$ and $r$ parameters? – Steeven Oct 25 '16 at 14:49
  • @Steeven I've reformatted. Well if u mean why in the literal sense it's because they are vectors. They have both direction and magnitude. Hence $\vec{F_{12}}$ is the force vector having some magnitude acting in the direction $1→2$ – Spoilt Milk Oct 25 '16 at 14:55
  • @NaveenBalaji and ProtonUpUpDown, in that case the written equation 2) is incorrect. The force is not necessarily from one charge towards the other, so the notation $\vec F_{12}$ makes no sense. If $\vec r_{12}$ is the unit vector from charge 1 to 2, then the force should be named $\vec F_2$ only, since it is the force, which charge 2 feels. This is the usual notation. See https://en.wikipedia.org/wiki/Coulomb%27s_law. – Steeven Oct 25 '16 at 15:05
  • All in all, @ProtonUpUpDown, you can solve your own question by simply making it clear for yourself what exactly $\hat r_{12}$ is and then what the force is (and which direction it has). – Steeven Oct 25 '16 at 15:06
  • P/s: F2 is force on q2 due to q1: http://sv1.upsieutoc.com/2016/10/25/suggest2.png – PandoraU.U.D Oct 25 '16 at 15:30
  • I understand the unit vector matter. But I mean. I mean can we not put the information q1 and q2 away and put it into |q1||q2| and then using the unit vector to express the direction !? – PandoraU.U.D Oct 25 '16 at 15:44
  • @ProtonUpUpDown Yes, you can, but then you might have to choose a different unit vector depending on the situation: You can use the unit vector you have when the charges are opposite, but then you must suddenly turn it around and use an opposite unit vector, when the charges have same signs. How is it more convenient that you have to remove signs and then choose another vector instead of just plugging in the actual charge values with their signs? – Steeven Oct 25 '16 at 17:00
  • Oooh ! I get it ! Thank you very much. It's like I have just witnessed magic :). Thanks a lot, Mr Steeven. – PandoraU.U.D Oct 26 '16 at 08:35

2 Answers2

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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 $q_2$.

DelCrosB
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  • 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 http://sv1.upsieutoc.com/2016/10/25/suggest2.png – PandoraU.U.D Oct 25 '16 at 15:32
  • 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? – DelCrosB Oct 25 '16 at 16:09
  • I have just read again Steeven's answer and your replies again. I understand very well the topic now :). Thank you , DelCrosB – PandoraU.U.D Oct 26 '16 at 08:36
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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.

Steeven
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  • http://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. – PandoraU.U.D Oct 25 '16 at 15:35
  • "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. – Steeven Oct 25 '16 at 16:49
  • 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.) – garyp Oct 25 '16 at 17:28
  • @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. – Steeven Oct 25 '16 at 17:50
  • 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. – garyp Oct 25 '16 at 18:01
  • @garyp Well, the notation ${21}$ indicates direction for $\vec r{21}$ but not for $\vec F_{21}$. – Steeven Oct 25 '16 at 18:05
  • Depends on how you look at it. Force on 2 due to 1. Position of 2 relative to 1. To me that seems very ... satisfying. – garyp Oct 25 '16 at 18:10
  • @garyp In any case this will never be a fruitful discussion ☺️ Let's not discuss this further – Steeven Oct 25 '16 at 18:25