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Is charge of something for (e.g.) an electron related to electromagnetic space if it exists due to energy, due to which it may have mass? I don't know about quantum mechanics or advanced particle models. Can anyone just simply give an intuitive idea?

EDIT I want to mean what actually gives electron charge if it is not assumed fundamental but result of some other physical phenomenon or it is just the quantity defined to explain physical interactions?I think now it is clear

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closed as unclear what you're asking by John Rennie, Kyle Kanos, Brandon Enright, Jim, Emilio Pisanty Apr 3 at 23:57

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I'm unsure what you're after. Do you want to define it as your title says, or do you want an intuitive idea, as in your question text? –  BMS Apr 3 at 16:09
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See also: electronics.stackexchange.com/q/103489 –  The Photon Apr 3 at 20:34
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Duplicate of physics.stackexchange.com/q/57199 –  Kyle Kanos Apr 4 at 15:38

2 Answers 2

up vote 16 down vote accepted

Charge is a fundamental conserved property of particles. It is, if you like, a measure of how much a particle interacts with electromagnetic fields. A particle with charge can produce and be affected by electromagnetic fields. This is what we mean when we say a particle has charge. Its a simple quantised way to measure the coupling strength of particles with the appropriate force.

e.g:

charge for EM force, color charge to strong force, etc.

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Charge is a quantity which arises from Noether's theorem, due to continuuous global symmetries (up to a total derivative) of an action, and as such we have many types of charge, other than electric. For example, consider the Dirac Lagrangian,

$$\mathcal{L} = \bar{\psi}(i\gamma^{\mu}\partial_{\mu}-m)\psi$$

which describes fermions. It is invariant by a change of phase, i.e. $\psi \to e^{-i\alpha}\psi$, which has a conserved current, namely,

$$j^{\mu} = \bar{\psi}\gamma^\mu \psi$$

The conserved quantity, i.e. Noether charge, arising from the symmetry is the integral over all space of the zeroth component, therefore

$$Q = \int \mathrm{d}^3 x \, \, \bar{\psi} \gamma^0 \psi = \int \mathrm{d}^3 x \, \, \psi^{\dagger}\psi$$

The quantity $Q$ is indeed electric charge. In addition, when the field $\psi$ is quantized, and expanded as a plane wave with operators as Fourier coefficients, it can be shown that $Q$ also has the interpretation of particle number for fermions. In classical electromagnetism, charge also determines the magnitude of the effect of magnetic and electric fields on charged matter, via the Lorentz force relation,

$$\vec{F} = q\left(\vec{E}+ \vec{v} \times \vec{B}\right)$$

The elementary charge $e$ also plays the role of the coupling constant in quantum electrodynamics, which roughly determines the strength of an interaction term, namely,

$$\mathcal{L}_{\mathrm{int}} \sim e \bar{\psi}\gamma^\mu A_\mu \psi$$

corresponding to an interaction vertex involving a photon, positron and electron. In addition, the coupling $e$ actually depends on a scale, dictated by the beta function (to one loop order), $$\beta (e) = \frac{e^3}{12\pi^2}$$

So the idea that there is a single $e$ is wrong; from the beta function, assuming no additional corrections alter its nature, we may deduce the coupling increases with increasing energy scale.

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This is just sad. He is probably a high school kid, he has clearly written that he doesn't know much and here you are trying to show off by throwing this math on his face. –  Parth Vader Apr 3 at 19:39
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+1 -- After all, this answer provides an avenue towards understanding what's meant by "field (strength)" and "coupling" in the first place; and that "electromagnetic" charges and fields may be (systematically, reliably) distinguished from certain other kinds of charges and corresponding fields. –  user12262 Apr 3 at 19:52
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@ParthVader: personally, I find it awfully cynical to assume that JamalS is trying to show off by giving a thorough, high-level answer, and that the OP has no use for it because of his age and current level of understanding. Besides, there's the rest of the community to consider as a secondary audience. –  Nick Stauner Apr 3 at 20:26
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@NickStauner The answer is first and foremost for the person asking the question and then for the rest of the community. I said 'show off' because the OP very clearly mentioned that he wants a simple answer. If JamalS wanted to give such a high level answer, the least he could have done was explain briefly what the equations meant. –  Parth Vader Apr 3 at 20:45
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And frankly, I don't even consider this to be a good high level answer. When I say give me an explanation on some natural phenomenon and all the other person gives me is a mathematical equation, I'll say, but what does it suggest? Mathematics is a good way of describing something but it is the intuition that matters. The 'decoding' of the math if you will. –  Parth Vader Apr 3 at 20:50

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