3
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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:

diagram of charges and displacement vector

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. $$\lVert\vec{F}\rVert = \left\lVert\frac{1}{4\pi\epsilon_0}\frac{q_1 q_2}{r^2}\hat{r}\right\rVert = \left\lvert\frac{1}{4\pi\epsilon_0}\frac{q_1 q_2}{r^2}\right\rvert \underbrace{\left\lVert\hat{r}\right\rVert}_{1} = \frac{1}{4\pi\epsilon_0}\frac{\lvert q_1\rvert\,\lvert q_2\rvert}{r^2}$$

$\endgroup$
3
$\begingroup$

Your equation (2) does not allow opposite charges to attract, it only allows for charges to repel each other (or only to attract each other, depending on your interpretation). So that cannot work.
Let me take your example. Take two charges $q_1=-q_2$ on a one dimensional line at a distance of $l$, $q_1$ at $x=0$, $q_2$ at $x=l$. Now according to your equation (2), the force between them is: $$ \vec{F} = \frac{1}{4\pi\epsilon_{0}}\cdot\frac{|q_1|^2}{l^2}\cdot \vec{e}_x $$ If this is the force on $q_1$, this is correct - but then we obtain the same force for $q_1=q_2$ in which case we would need repulsion. If this is the force on $q_2$, it is wrong because they should attract.

I think (and it took me a bit of thinking and editing) that the main problem here is that we need to specify very carefully about the force on which charge we are talking.

(1) is the standard version for the force one of the charges feels from the other if the other one is sitting at the origin and is correct in classical electrodynamics. It does however depend on the coordinate system and on putting one charge (the one we do not care about) in the origin if I am allowed to interpret $r$ as the radial component of spherical coordinates and $\hat{r}$ as the corresponding unit vector. For better generality I do however prefer $$ \vec{F}= \frac{1}{4\pi\epsilon_{0}}\cdot\frac{q_1\cdot q_2}{|\vec{r}_1-\vec{r}_2|^3}\cdot (\vec{r}_1-\vec{r}_2) $$ This is the force that charge $q_2$ exerts on $q_1$, while the one that charge $q_1$ exerts on $q_2$ is given by: $$ \vec{F}'= \frac{1}{4\pi\epsilon_{0}}\cdot\frac{q_1\cdot q_2}{|\vec{r}_1-\vec{r}_2|^3}\cdot (\vec{r}_2-\vec{r}_1) $$ Observe that the direction of the force depends on which charge we are talking about. This becomes especially important when we choose e.g. $q_1$ to be sitting at the origin and use spherical coordinates to express the force onto $q_2$. Then the force needs to be outward for same charges and inward for opposite charges. This is exactly what your formula (1) achieves. The change in direction comes from the prefactor which becomes negative or positive.

$\endgroup$
1
$\begingroup$

Which is the correct vector equation for electrostatic force (1) or (2)?

Equation 1 is correct. In explicit notation (with subscripts) the equation is $$ \vec{F_{12}} = \frac{1}{4\pi\epsilon_0}\frac{q_1 q_2}{|\vec{r_{12}}|^2}\hat{r_{12}} $$ where

  • $\vec{F_{12}}$ is the force particle 1 exerts on particle 2
  • $\vec{r}_{12} = \vec{r}_2 - \vec{r}_1$ is the displacement needed to move from particle 1 to particle 2
  • $\hat{r}_{12} = \frac{\vec{r}_{12}}{|\vec{r}_{12}|}$ is the direction of of $\vec{r}_{12}$

I've found that (1) leads to contradictions, if either $q_1$ or $q_2$ is negative, as the unit vector $\hat{r}_{12}$ needs to have one of its entries negated (corresponding to a geometrical 'change in direction' of the vector) in order for $\vec{F}_{12}$ to be logical.

Nope, $\hat{r}_{12}$ doesn't change - it only depends on the position of the two particles ($\vec{r}_2 - \vec{r}_1$), i.e. it does not depend on the charge of the particle. On the other hand, $\vec{F}$ does depend on the charge of the particles (as evidenced by equation 1).

If one of $q_1$ or $q_2$ is negative, then $\vec{F}_{12} \propto -\hat{r}_{12}$. Using your test as an example, with $q_1 = -\lambda$ and $q_2 = +\lambda$, we would expect opposite charges to attract. $$ \vec{F}_{12} = -\frac{1}{4\pi\epsilon_0}\frac{\lambda^2}{|\vec{r}_{12}|^2}\hat{r}_{12} $$ This means that $\vec{F}$ is in the opposite direction to $\hat{r}$, which is correct for an attractive force (particle 2 experiences a force in the opposite direction of $\hat{r_{12}}$, pulling it towards particle 1). If in another test scenario we have $q_1 = -\lambda$ and $q_2 = -\lambda$, the negatives cancel, leaving you with $$ \vec{F}_{12} = +\frac{1}{4\pi\epsilon_0}\frac{\lambda^2}{|\vec{r}_{12}|^2}\hat{r}_{12} $$ which is a repulsive force (particle 2 experiences a force in the direction of $\hat{r_{12}}$, pushing it away from particle 1).

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'

The unit vector's components don't change, and now you have the contradiction of not being able to have attractive forces ($\vec{F}_{12} \propto -\hat{r}_{12}$).

$\endgroup$
  • 1
    $\begingroup$ If you replaced $|\vec{r}|$ with $|\vec{r_{12}}|$ and if needed explicitly define $\vec{r_{12}} = \vec{r_1} - \vec{r_2}$, then I think all ambiguity vanishes. I know that this is the core of your argument, and I agree with you. I'm suggesting a notation that is even more explicit might help the OP. $\endgroup$ – garyp Aug 7 '16 at 16:37
  • $\begingroup$ Excellent point, thank you @garyp! I've followed your suggestions and edited my answer so that $\vec{r}_{12}$ is properly defined and consistent subscript notation is used throughout. It's much better now thank you :) N.B. I defined $\vec{r_{12}} = \vec{r_2} -\vec{ r_1}$, because I think it represents the displacement required to move from point 1 to point 2. $\endgroup$ – Judge Aug 7 '16 at 18:23
  • $\begingroup$ Your choice of 12 is ok, but it's more common to see 21: the vector at 2 from 1. Similar to $F_{ab}$ being the force on $a$ due to $b$. But it doesn't matter if the reader is paying attention. $\endgroup$ – garyp Aug 7 '16 at 21:18
1
$\begingroup$

enter image description here

$\mathbf{f}_{\imath\jmath}=$ force exerted on charged particle $\:\imath\:$ from charged particle $\:\jmath\:$
$\mathbf{n}_{\imath\jmath}=$ unit norm vector from charged particle $\:\imath\:$ to charged particle $\:\jmath\:$ $=\dfrac{\left(\mathbf{r}_\jmath-\mathbf{r}_\imath\right)}{\Vert \mathbf{r}_\jmath-\mathbf{r}_\imath\Vert}$

\begin{equation} \mathbf{f}_{21}=-\mathbf{f}_{12} =\dfrac{1}{4\pi\epsilon_{o}}\dfrac{q_{1}\;q_{2}}{\Vert \mathbf{r}_2-\mathbf{r}_1\Vert^{3}}\left(\mathbf{r}_2-\mathbf{r}_1\right)=\dfrac{1}{4\pi\epsilon_{o}}\dfrac{q_{1}\;q_{2}}{\Vert \mathbf{r}_2-\mathbf{r}_1\Vert^{2}}\mathbf{n}_{12} \tag{001} \end{equation}

Note : In Figure the force vectors $\:\mathbf{f}_{21},\mathbf{f}_{12} \:$ are drawn as if $\:q_{1}q_{2} <0\:$.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.