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1

Coulomb's law is valid at all energies we have probed experimentally. There is a subtlety here because the fine structure constant changes with energy so the electrostatic force gets stronger with increasing energy, but it is still an inverse square law force. However a proton is not a point particle, so the force between for example a proton and electron ...


1

Yes, however the predominant forces are much, much stronger than electromagnetism. That is, there is a "strong nuclear force" which will routinely hold protons side-by-side in a tiny space, and a "weak nuclear force" which will routinely turn a neutron into a proton plus an electron, and those two forces completely violate what you'd expect from ...


4

It does give "Coulomb's law" with $\frac{1}{r^3}$, it gives it in its proper vectorial form $$ \vec E \propto \frac{\vec r}{r^3}$$ which, when taking the absolute values, yields the form you are probably more familiar with $$ E \propto \frac{1}{r^2}$$ since $\lvert \vec r \rvert = r$.


2

The colour Coulomb interaction is another name for the one-gluon exchange potential between (typically heavy) quarks or other sources having colour. It is a straightforward generalisation of the Abelian (i.e. standard) Coulomb potential stemming from one-photon exchange, $$ V(r) = -\alpha/r \; ,$$ where $r$ is the distance between the sources taken to be ...


0

If the shell is perfectly spherically symmetric, and the charge is perfectly evenly distributed on it, then the total field due to every singel charge is zero at every single point inside the sphere. This also happens with gravity, there is no gravitational force inside a uniformly distributed shell of mass due to the shell of mass. While each piece of ...


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The statement means that the net electric field at any given point inside the sphere adds up to zero due to all the varying contributions by the charges on the surface. They exactly cancel out, and hence for any point inside the sphere, the value of electric field is exactly zero.


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This question is answered in the Wikipedia article on the shell theorem. The gist is that it is not the total. But going beyond that, I think your question actually reflects a misunderstanding of what "electric field" means. The electric field is something which has a value at every point in space. If you try to calculate a total, e.g. by integrating the ...


0

As far as I know, Coulomb's law of electrostatic force is applicable on two different charges situated in same medium. Only in fluids (gas, liquid). In solids, Coulomb's law cannot be directly applied because the solid contributes unknown amount of force itself (it maintains mechanical stress). One needs to carefully analyze forces in solids; usually ...


0

Let me ask you: how are you characterizing the force between the two plates? Typically, the two plates are connected to a voltage source, and the force is found by multiplying the charge by the voltage, and dividing by the plate distance. The voltage already takes into account the charge on both plates, so there's no need to worry. If you're doing a more ...


2

To reach the Lienard-Wiechert potentials or to prove Feynman's equation (exposed in his lectures without proof), it's necessary to begin with the so-called retarded potentials expressed here conveniently by the following. \begin{equation} \phi\left(\mathbf{r},t\right)=\dfrac{1}{4\pi\varepsilon_{o}}\iiint ...


16

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} = ...


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.


3

The force is not propagated instantly. It takes time for the information to get from one point to another. You can treat that as an instant if you are working with small enough distances and velocities, but it's not. If you'll ever study field theory you'll meet retarded potentials that are just this: the field propagates at the speed of light and it's no ...



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