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The electric charge on baryons (e.g. protons) and leptons (e.g. electrons) are the same in magnitude. Baryons are made of quarks but Leptons are not. So where is the connection which makes the electric charge EXACTLY the same in magnitude? If these charges were not equal the whole atomic structure would be different, and atoms possibly unstable. Put more crudely, how does an electron "know" what the charge is on a proton if they have no structure in common?

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marked as duplicate by John Rennie, user36790, AccidentalFourierTransform, fffred, Jon Custer Dec 26 '16 at 4:42

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  • $\begingroup$ Its wrong to say both baryons and leptons have same electric charge in general. e.g. $\Delta^{++}$ has charge has +2, which has no leptonic counterpart. $\endgroup$ – AMS Dec 25 '16 at 8:31
  • $\begingroup$ Also note that the question, "How does one charge know what another charge is?" has a simple answer: the electric field surrounding the other charge acts on the first charge with the appropriate magnitude. Another side note: it's dangerous to anthropomorphize things like elementary particles. Do too much of that and you start straying out of physics and into mysticism. $\endgroup$ – probably_someone Dec 25 '16 at 10:16
  • $\begingroup$ There are a few questions being asked here that I see; the first is, "Why is charge quantized at all?" This question does not yet have a satisfactory answer that does not involve undiscovered things like magnetic monopoles (see Dirac's argument, en.wikipedia.org/wiki/Magnetic_monopole#Dirac.27s_quantization). $\endgroup$ – probably_someone Dec 25 '16 at 10:19
  • $\begingroup$ The second is "Why do protons and electrons have the same (opposite) integer multiple of the fundamental charge?" Is this the question you were actually meaning to ask? $\endgroup$ – probably_someone Dec 25 '16 at 10:21
  • $\begingroup$ Yes, I agree with your reply probably_somebody. (My anthropomorphisation was only to clarify the question). I think the exactness of the charge will be ensured by the quantisation.; there cannot be a small discrepancy. Sorry about the duplicate question John R, but it is my first question post! $\endgroup$ – user140203 Dec 26 '16 at 19:14