Why do like charges repel and opposite charges attract?
$\begingroup$
$\endgroup$
2
asked May 22, 2015 at 3:45
-
$\begingroup$ How we call them is not important. The important thing is whether two charges are like charges or alike charges. Experiments tell us like charges repel and alike charges attract. $\endgroup$– M. ZengCommented May 22, 2015 at 3:59
-
4$\begingroup$ Duplicate, this ones got some amazing answers: physics.stackexchange.com/q/80807 $\endgroup$– Hritik NarayanCommented May 22, 2015 at 5:39
Add a comment
|
4 Answers
$\begingroup$
$\endgroup$
3
There are many different levels of explanation for this question. Strangely enough most of them will dive into quantum electrodynamics, Feynman diagrams and exchange of virtual photons...
I will try a simpler path that still carries some explanation.
When you put two charges at a distance, they deform the -- otherwise flat -- electromagnetic (EM) potential field. Depending on whether the two charges have the same sign or not, the EM field will be deformed differently.
Quantitatively, the deformation is measured by a local change in the EM field, and considering the static setup we consider, this change is solely measured by the electric field $\mathbf{E} \equiv -\mathbf{\nabla} \phi$ generated by this system of charges.
Deforming the EM field costs some energy that is stored as a curvature term of a electrostatic potential sheet if you will.
As you may know it formally reads:
\begin{equation} \mathcal{E}_{elec} = \frac{\varepsilon_0}{2}\int d^3r \: \mathbf{E}^2 \end{equation}
In our case we have that:
\begin{equation} \mathbf{E}(\mathbf{r}) = \frac{q_1 (\mathbf{r}-\mathbf{r}_1)}{4\pi \varepsilon_0 |\mathbf{r}-\mathbf{r}_1|^2} + \frac{q_2 (\mathbf{r}-\mathbf{r}_2)}{4\pi \varepsilon_0 |\mathbf{r}-\mathbf{r}_2|^2} \end{equation}
so that
\begin{equation} \mathbf{E}^2 = \frac{q_1^2}{(4\pi \varepsilon_0)^2} + \frac{q_2^2}{(4\pi \varepsilon_0)^2} + \frac{2 q_1 q_2 (\mathbf{r}-\mathbf{r}_1) \cdot (\mathbf{r}-\mathbf{r}_2)}{(4\pi \varepsilon_0)^2 |\mathbf{r}-\mathbf{r}_1|^2|\mathbf{r}-\mathbf{r}_2|^2} \end{equation}
Upon integration over the whole available volume (often infinite) one finds that:
\begin{equation} \mathcal{E}_{elec} = \epsilon_1 + \epsilon_2 + \frac{q_1q_2}{4\pi \varepsilon_0 |\mathbf{r}_1-\mathbf{r}_2|} \end{equation}
where $\epsilon_{1,2} \equiv \int d^3r \:q^2_{1,2}/2r^2 \varepsilon_0$ is the deformation energy created by the single charge $q_{1,2}$ had it been alone in the universe.
The total energy is thus expressed as the sum of the individual contributions coming from each particles plus a correction due to the fact that, when the charges are close enough, the EM field deformations generated by one charge will be affected by the deformations created by the other.
As we can see, the sign of this corrective term is that of the product $q_1 q_2$ and is negative whenever the charges dont have the same sign.
The interpretation that comes out of it is that when the charges have opposite sign, each charge acts as a deformation "sink" for the other charge deformations of opposite sign; that is the deformations generated by one particle are weakened by the deformations generated by the other. This deformation weakening effect is all the more important as the charges get closer and closer until they eventually overlap and yield (in principle) a zero deformation field. Since the universe seems to prefer low energy states, charges with opposite signs attract one another as a consequence.
The opposite is true of charges with the same sign whereby the deformations generated by one charge is simply enhanced by the presence of the other charge. Thus the EM field has more "curvature" energy to store than what it would have had if the charges had been accounted separately (or if they were infinitely far from one another). Since Nature again prefers low energy states, this implies that charges with the same sign will repel each other.
-
$\begingroup$ down voting seriously? Please put a comment to explain. $\endgroup$– gatsuCommented May 22, 2015 at 14:21
-
1$\begingroup$ nice answer, but perhaps too complex for this kind of questions $\endgroup$– user46925Commented May 25, 2015 at 4:43
-
1$\begingroup$ great answer, I just knew explanations from QFT books, but that here is much more insightful $\endgroup$ Commented Feb 28, 2023 at 7:38
$\begingroup$
$\endgroup$
you can draw feynman digrams and then calculate scattering amplitudes and it is in the non relativistic limit is proportinal to potential.so if the potential is positive it means they repel. this sort of claculation is done in peskin book and A.Zee book.in peskin book page no 125.
this is the most rigorous work to prove gravity is always attractive. by means of same thing assuming it follows field theory.
$\begingroup$
$\endgroup$
Because observations made by physicists have found that this is what nature does.
$\begingroup$
$\endgroup$
Firstly, I would like to say that there is no particular terminal separation between negative charges and positive charges. Actually you will understand it better if I would clarify in this way that scientists first saw that having even follow the same statistical distribution i.e. Fermi Dirac distribution some of them actually repel others and some do attract other. See an atom has electrons,protons and neutrons. A body having more electrons than protons is said to be negatively charged and one having excess of protons is said to be positively charged. This is a primary assumption they took, rather assigned it like that.This fact is a result of arbitrarily assigning positive charge to the proton and vice versa. Now consider a glass rod rubbed with silk. The electrons are transferred from the glass to the silk,and so the glass rod attains a positive charge. Now,if we bring a similarly rubbed glass rod next to it,they will repel. This is an observed fact and a law of nature. However they are called positive because we have assigned a positive charge to the proton.
