133

As you said, it's probably not magnetism if the balls are free to rotate; there is no reason they wouldn't just flip over and stick together, north to south. You can test this by buying some of those toy magnetic balls. The repulsive configurations are highly unstable and turn attractive with the slightest touch. I'm going to go out on a limb and say it's ...


113

Well it has nothing to do with the Higgs, but it is due to some deep facts in special relativity and quantum mechanics that are known about. Unfortunately I don't know how to make the explanation really simple apart from relating some more basic facts. Maybe this will help you, maybe not, but this is currently the most fundamental explanation known. It's ...


45

I really appreciate the physical explanations made in other answers, but I want to add that Fourier transform of the Coulomb potential makes mathematical sense, too. This answer is meant to clarify on what sense the standard calculation is valid mathematically. Firstly, and maybe more importantly, I want to emphasize that The Fourier transform of f is not ...


41

Yes, potential energy can be negative: consider Newton’s law of gravitation $$V = -\frac{GMm}{r}$$ Where $G$ is Newton’s constant, $M$ and $m$ are masses, and $r$ is the distance between them. It can clearly be seen that this is always negative. The key thing is that the absolute value of potential energy is not observable; there is no measurement that can ...


35

The classical Coulomb potential can be recovered in the non-relativistic limit of the tree-level Feynman diagram between two charged particles. Applying the Born approximation to QM scattering, we find that the scattering amplitude for a process with interaction potential $V(x)$ is $$\mathcal{A}(\lvert p \rangle \to \lvert p'\rangle) - 1 = 2\pi \delta(E_p -...


30

Great question. First of all, you're absolutely right that until we find a universe with a different number of dimensions in the lab, there's no single "right" way to generalize the laws of physics to different numbers of dimensions - we need to be guided by physical intuition or philosophical preference. But there are solid theoretical reasons for ...


28

Permittivity $\varepsilon$ is what characterizes the amount of polarization $\mathbf{P}$ which occurs when an external electric field $\mathbf{E}$ is applied to a certain dielectric medium. The relation of the three quantities is given by $$\mathbf{P}=\varepsilon\mathbf{E},$$ where permittivity can also be a (rank-two) tensor: this is the case in an ...


28

You need to multiply by four not by two. To see why let's draw the situation: You are assuming the situation is as shown in the top diagram. So the two $M_1$s attract each other and the two $M_2$s attract each other. Those are the forces shown by the red lines. But you also need to include the force between $M_1$ on one side and $M_2$ on the other. Those ...


26

If $E = (m+M)c^2+k\frac{Q^2}{r}-G \frac{mM}{r} < 0$, then $(m+M)c^2-G \frac{mM}{r}$ must also be less than zero. Thus we must have $r < \frac{GmM}{(m+M)c^2}$. This is less than the Schwarzschild radius of a particle of mass $m+M$, and this gets to the heart of the matter: You are using Newtonian gravitation in a field where it does not apply.


26

It's true that a point particle with finite charge is problematic in electromagnetism because of the infinite field and associated energy near such a particle. However, we don't need that concept in order to make a defining statement about the electric field. Rather, we can use $$ {\bf E} = \lim_{r \rightarrow 0} \frac{\bf f}{q} $$ where $\bf f$ is the force ...


23

Yes, they can. Both interactions can be modeled using perturbative quantum field theory, where their strength is parametrized by a dimensionless coupling constant. Electromagnetic repulsion between two electrons can be written as a power series in $\alpha$, the fine structure constant, which is dimensionless and has a value of roughly 1/137. Meanwhile, the ...


22

Your confusion lies in the way you configured the problem. Let two charged particles revolve around the center of the system. It's quite clear in that viewpoint that any change in linear momentum of one particle is matched by a corresponding change of linear momentum in the second particle. Thus the linear momentum of the whole system remains constant. ...


22

For two electrons separated by distance $r$, we have $$F_g = \frac{Gm^2}{r^2}$$ and $$F_e = \frac{1}{4\pi\epsilon_0}\frac{e^2}{r^2}$$ The ratio is $$\frac{F_e}{F_g} = \frac{e^2}{4\pi\epsilon_0 G m^2}$$ Now choose a unit system in which $4\pi\epsilon_0 G = 1$, yielding $$\frac{F_e}{F_g} = \frac{e^2}{m^2}$$ So the constant Feynman is referring to is the charge-...


21

An experimentalists answer: Why do same/opposite electric charges repel/attract each other, respectively? Because careful physicists have made an innumerable number of observations and have found that this is what nature does. There is a long history of observations before any theory could be solidified. They observed the behavior of attraction with some ...


21

Just to elaborate on the symmetry argument - Let's suppose in your first diagram you are observing the two charges from the side and we assume, as you have done, that the repulsion direction is vertical and to the right. If we now observe the two charges from the top looking down, we are presented with the exact same situation as before and we would say the ...


20

As with all derivations, it depends on what you want to treat as fundamental. Typically we would derive Coulomb's law from the Maxwell equations, so we're trying to solve $$\nabla\cdot \mathbf{E} = -\nabla^2 \varphi = q\delta(\mathbf{x})/\epsilon_0\qquad (1)$$ In $n$ spatial dimensions and in Cartesian coordinates $(x_1,\ldots,x^n)$, this becomes $$\sum_{k=1}...


19

You very well can. But the coordinates are used with the required symmetries in mind. In a line charge, the system is cylindrically symmetric about the axis of the line. This makes the vector equations to solve for the fields way easier. Plus, you can also invoke physical arguments to determine the direction of the fields. If you still have any doubt, just ...


19

those two laws look similar because they both describe the propagation of a long-range field through three-dimensional space which produces a force that acts (in the simplest example) between pairs of objects, and in which the strength of the resulting force depends on some extensive property of each (charge in one case, mass in the other). the long-range-...


19

Following R.W. Bird answer: Consider an isolated system of two particles: Since system is isolated, angular momentum (and linear) is conserved. I.e. $\vec\tau_{net}=\Sigma\space \vec r\times\vec F=0$ But clearly from figure $\vec\tau_{net}=\vec r\space\times\space\vec F_y\ne 0$ Thus as R.W. Bird noted, the system violates the conservation of angular ...


18

You are correct that Gauss's law alone cannot be used to derive Coulomb's law. Instead, you need to supplement it with the hypothesis that space is isotropic, but nothing more. The thing is really in the language. You start with Gauss's law, and then you say "consider a point charge...", without actually saying much about what you mean by that. In ...


18

Here is one line of reasoning: E&M is supposed to be a fundamental theory. Having an action principle may facilitate developing a consistent quantum theory. The structure of the Maxwell Lagrangian density $$ {\cal L}~=~-\frac{1}{4}F_{\mu\nu}F^{\mu\nu} + j^{\mu}A_{\mu}$$ does not depend on the spacetime dimension $n$. It possesses Lorentz symmetry. It ...


18

Unless I miscalculated, this does not seem possible for the following reason. Let us assess the electric field required to suspend a man with a mass of $m=70kg$. For simplicity, let us consider a plane capacitor with the plate area of $A=1m^2$, which seems reasonable for "one side of a human body". The force $F$ between the plates of the capacitor ...


17

The force does not change instantaneously, the correct way the electromagnetic field of (and thus the force exerted by) a moving electric charge is given by the Liénard-Wiechert potential, where one can see that the effect of the charge does not travel faster than light.


17

To add to ACuriousMind's answer on the Liénard-Weichert potentials, you can put these formulas into an even more wonderfully descriptive form since you can derive Feynman's formula from them for the radiation from a moving charge: $$\vec{E} = -\frac{q}{4\,\pi\,\epsilon_0}\left(\frac{\vec{R}}{R^3}+\frac{R}{c}\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\vec{R}}{...


15

There aren't that many repulsive forces out there which could be at play: magnetism electrostatic forces compressed material (solid, fluid or gas) pushing bodies apart Magnetism would have worked if the bodies would not be able to rotate, or if we could produce magnetic monopoles. Compressed springs are extremely hard to make completely invisible. Liquid ...


15

You're forgetting one thing: a particle cannot feel its own electric field, so a point charge that generates a $1/r^2$ field doesn't do anything unless acted upon by an external field. You also can't place a particle at $r=0$ of another particle's $1/r^2$ electric field, because, well, there's already a particle there. (Also, how are you going to get it ...


14

When the electrostatic force was originally being studied, force, mass, distance and time were all fairly well understood, but the electrostatic force and electric charge were new and exotic. In the cgs system, the charge was defined in relation to the resulting electrostatic force (it's called a Franklin (Fr) an "electrostatic unit" (esu or) sometimes a ...


14

We use cylindrical coordinates because they're convenient and because they allow us to solve the problem cleanly and effectively. You could attempt to use rectangular or spherical coordinates to formulate the problem (which will be doable enough) and to attempt to solve it (which will be much harder), but generally speaking, if your problem has a definite ...


14

Coulomb's law is only valid in Electrostatics. In other words, you cannot ask questions like "What would happen if one of the charges is moved (or disappears)?" and hope to find a sensible answer using Coulomb's law. Making a charge move or "disappear" violates Electrostatics. (This is the same reason that Coulomb's law does not hold to ...


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