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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 ...

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Maxwell's equations do follow from the laws of electricity combined with the principles of special relativity. But this fact does not imply that the magnetic field at a given point is less real than the electric field. Quite on the contrary, relativity implies that these two fields have to be equally real. When the principles of special relativity are ...

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Yes, absolutely. In fact, Gauss's law is generally considered to be the fundamental law, and Coulomb's law is simply a consequence of it (and of the Lorentz force law). You can actually simulate a 2D world by using a line charge instead of a point charge, and taking a cross section perpendicular to the line. In this case, you find that the force (or ...

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Lubos Motl's answer is very good, but I think it's worth saying one or two additional things. You can regard magnetism as simply a byproduct of electricity, in the following sense: if you assume that Coulomb's Law is correct, and that special relativity is correct, and that charge is a Lorentz scalar (so that charge and current density form a 4-vector), ...

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This is not paradoxical and it is not necessary for any physical phenomenon to a priori have to obey any particular law. Some phenomena do have to obey inverse-square laws (such as, particularly, the light intensity from a point source) but they are relatively limited (more on them below). Even worse, gravity and electricity don't even follow this in ...

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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 ...

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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.

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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: $$\vec{E} = -\frac{q}{4\,\pi\,\epsilon_0}\left(\frac{\vec{R}}{R^3}+\frac{R}{c}\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\vec{R}}{... 14 I suppose you mean k_e=\frac1{4\pi\epsilon_0}. That comes from the fact that Coulomb's law can be stated as :$$F= \frac1{\epsilon_0}\frac1{4\pi r^2}q_1q_2 $$Now, \epsilon_0 is the electric constant, or the permittivity of free space, and it essentially scales the force. The 4\pi r^2 comes from the surface ... 13 Short Answer You've hit upon the quirk that the SI and CGS systems not only measure electric charge with different units, but also assign them different dimensionality. In SI, the Ampere is a base unit. Amperes are not made out of anything else - they are primitive, like meters, kilograms, and seconds. One Ampere is one Coulomb per second, so the unit of ... 13 The short answer is yes, and in fact you only need one single Maxwell equation, Gauss's law, together with the Lorentz force, to get Coulomb's law. More specifically, you need Gauss's law in its integral form, which is equivalent to the differential form for well-behaved fields because of Gauss's theorem. Thus, you use the law$$ \nabla\cdot\mathbf{E}=\rho/\...

12

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 ...

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Of course Coulomb's law has to be adapted! And it is therefore fortunate that there exist manifestly covariant formulations of electromagnetism that do not care how spacetime is curved. However, we should first briefly observe that Coulomb's law is not one of the fundamental laws of electromagnetism, though it has played a great role in its inception: ...

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James Clerk Maxwell thought about this one and showed the following. Suppose we have two concentric conducting spheres and we charge one up to a potential $\Phi$ relative to some grounding plane. Then the voltage of the inner sphere relative to the same ground is: $$\Phi_{inner} = \Phi \,q\, \left(\frac{\rho}{2}\log\left(\frac{\rho+1}{\rho-1}\right)-\frac{1}... 11 You are correct when you concluded that two classical point electrons could never touch each other. It would take infinite energy. 10 Not a direct answer to your question but still a surprising derivation of Maxwells equations: Feynman's proof of the Maxwell equations (FJ Dyson - Phys. Rev. A, 1989) shows, that it is possible to derive Maxwells equations from Newtons second law of motion and the uncertainty principle. 10 The mistake you made is in the way you stated Coloumb's law. It's either$$ \vec{F} = K \frac{q_1 q_2}{r^\color{red}3} \color{red}{\vec{r}} $$OR$$ \vec{F} = K \frac{q_1 q_2}{r^\color{red}2} \color{red}{\hat{r}} $$but definitely NOT$$ \vec{F} = K \frac{q_1 q_2}{r^\color{red}3} \color{red}{\hat{r}} $$10 If you want to avoid factors of \pi in the more fundamental equations like \nabla . E = \rho / \epsilon_0, you have to accept them where they belong, for instance in: E = \frac{1}{\epsilon_0} \frac{Q}{4 \pi r^2}. As remarked by others, Newton failed to put a factor 4 \pi into his gravitation equation (he stipulated g = G \frac{M}{r^2}, instead of ... 10 Concerning the factor \frac{1}{2}: It seems that OP in his classical reasoning only accounted for the Coulomb potential energy$$\tag{1}\langle U\rangle ~=~-k_e e^2 \langle \frac{1}{r} \rangle ~=~-\frac{k_e e^2}{a_0} ~<~0.$$Here k_e is Coulomb's constant and a_0 is the Bohr radius.^1 However we should also take the kinetic energy \langle T\... 10 Coulomb's law becomes invalid at distances of the order of the electron Compton wavelength and smaller, due to vacuum polarization. To first order in the fine structure constant, the electric potential due to a charge q at the origin is given by:$$V(r) = \frac{q u(r)}{r}$$where$$u(r) = 1 +\frac{2\alpha}{3\pi}\int_1^{\infty}du \exp(-2mru)\left(1+\frac{1}...

9

It's a good observation that the electric and gravitational fields both satisfy Poisson's equation $$\nabla^2\Phi_G = 4\pi\rho_G, \qquad \nabla^2\Phi_E = -\frac{\rho_E}{\epsilon_0}$$ where $\Phi_G, \Phi_E$ are the gravitational and electric potentials and $\rho_G,\rho_E$ are the mass and charge densities. It would seem from the perspective of Newtonian ...

9

Pure convention. There is no reason alternative conventions couldn't be used, apart from the need to avoid confusion. Newton introduced the constant to make the force law simple, whereas the electrostatic definition with the $4\pi$ is designed to make Poisson's equation (one of the equations for the electric field) look simple. You can write a Poisson ...

9

+1, Good question,. While I don't think your idea has much of a physical implications, it is a good analogy (in my opinion, at least). A fair approximation to General Relativity is Newtonian Gravity. A better one is Newtonian Gravity with some special relativistic corrections (I mean a modification to Newton's gravity where the masses $m$ are replaced ...

9

A short answer, is that to estimate interaction energy (which says if same charges attract or repel), you use propagators. Propagators come from the expression of Lagrangians. Finally, the time derivative part for dynamical freedom degrees in the action must be positive, and this has a consequence on the sign of the Lagrangian. Choose a metrics $(1,-1,-1,-1)... 8 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 ... 8 Systems of units are in some sense flexible and optional. The relationship $$\text{Electrostatic force twixt two point-like charges} \propto \frac{(\text{one charge})\times (\text{the other charge})}{(\text{distance between them})^2} \tag{1}$$ is an experimental fact. In SI, we have units for Force, distance and charge such that (1) is not ... 8 There is another 'infinity' (among others) lurking in classical electrodynamics which is evident when one calculates the electrostatic energy$W$of a uniform spherical charge distribution of radius$a$and total charge$Q$$$W = \frac{3}{5}\frac{Q^2}{4\pi \epsilon_0 a}$$ Thus, by this result, a point (zero radius) particle of charge Q has 'infinite' self-... 8 The energy in a level$n$is given by $$E = - \frac{Z^2 R_E}{n^2}$$ where$R_E$is the Rydberg energy ($R_E = 13.6\mathrm{eV}$). Therefore, greater$n$means lower energy (in absolute value), i.e., the electron is less bounded. 7 The length scale$L$has to be present in the denominator for dimensional reasons – only logarithms of dimensionless quantities are really "well-defined" unless one wants to introduce bizarre units such as the "logarithm of a meter". On the other hand, the dependence on$L$is largely trivial and unphysical for most purposes. Replace$L$by$K$and you will ... 7 Yes the dimension is different. In SI the current (A) a base unit independent from length (m), mass (kg) and time (s) because we choose to, but in CGS Gaussian unit this is not (1 unit of current = 1 g1/2 cm3/2 s-2), by setting$\epsilon_{0,SI} = \frac1{4\pi}\$. This also leads to some perhaps unintuitive results, like the unit capacitance in CGS Gaussian is "...

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