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7. Defining the symbol k k in Coulomb's law, F = kq1q2 r2, F = k q 1 q 2 r 2, to be k = 1/4πϵ0 k = 1 / 4 π ϵ 0, is perfectly allowed when one understands it simply as a definition of ϵ0 ϵ 0. The motivation for this definition is that when you work out the forces between two oppositely charged plates of area A A and charge Q Q a distance d ...
Upon replacing in the expression for ΔE, one finds that: ΔE ≈ϵ1 +ϵ2 +Vcoul. where. ϵ = ∫d3k q2 2ε0k2. is the self interaction energy of the charges with themselves (can be interpreted as the emission and absorption of a scalar photon by the same charge) and. Vcoul = ∫d3kq1q2eik⋅(r1−r2) 2ε0k2 = q1q2 4πε0|r1 −r2|.
Closed last year. We can notice that in the Coulomb's law equation, F = 1 4πϵ ⋅ q1q2 r2 (1) (1) F = 1 4 π ϵ ⋅ q 1 q 2 r 2. 4πr2 4 π r 2 factor in the denominator expresses directly the surface of a virtual sphere with radius r r. Actually we can look at this equation as it was for 3 3 dimensional objects. If we suppose want to ...
The "traditional" form of Coulomb's law, explicitly the force between two point charges. To establish a similar relationship, you can use the integral form for a continuous charge distribution and calculate the field strength at a given point.
6. Coulomb's law is not precisely true when charges are moving-the electrical forces depend also on the motions of the charges in a complicated way. One part of the force between moving charges we call the magnetic force. It is really one aspect of an electrical effect. That is why we call the subject "electromagnetism."
The modern interpretation of the interaction of two charged particles is by means of Quantum Electrodynamics, where the resulting force is due to an exchange of photons between two fermions. When you go through the formalities of quantum field theory, you can see quite easily that Coulomb's force law is only an approximation of the interaction ...
Given the Gauss law AND the Lorentz force, yes it's possible to derive Coulomb's law as it already been answered. So I think the question is if it's possible to derive it given ONLY the four Maxwell equations (and not the Lorentz force). The answer is still yes since the Lorentz force is equivalent to the Faraday law and it can be derive from it.
Converting CGS centimeters and grams to SI meters and kilograms, then equating the two expressions for the force in Coulomb's law, we find the conversion that one Coulomb is the same amount of charge as 2.99792458 ×109 2.99792458 × 10 9 esu. Reference. This answer is a summary of an appendix to Purcell's Electricity and Magnetism.
The vector version of Coulomb's law is: F 1 = kq1q2 r2 r^21 F → 1 = k q 1 q 2 r 2 r ^ 21. Note the difference in notation from your expression: F 1 F → 1 is the force felt by charge 1. r^21 r ^ 21 is the unit vector from charge 2 towards 1. Now, like charges (same sign) repel, so the force will point in the same direction as r^21 r ^ 21 ...
Here is a fairly recent paper on this topic titled Upper bounds on the photon mass. As @count-iblis (correctly) points out, you can think of the Compton wavelength of an electron (about 2 ×10−12 m) as a lower bound for the validity of Coulomb's law. However, quantum field theory predicts the nature of the corrections.