What is the experimental value of the ratio between the proton and the electron charge? Or more generally, is there a table that lists the ratio of the different nuclei charges to that of the electron?
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4Have you looked in the Review of Particle Physics? That is always the first place to check for these kinds of things. – dmckee --- ex-moderator kitten Nov 10 '14 at 23:38
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1Theoretical analogue: http://physics.stackexchange.com/q/21753/2451 – Qmechanic Nov 11 '14 at 00:08
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Sry, on phone that's why brief. Yes I looked there quickly and there they quoted something like $(q_p+q_e)/e$ without defining these quantities, (or I missed them) – Physics_maths Nov 11 '14 at 00:08
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Duplicate? http://physics.stackexchange.com/q/55513/ – dmckee --- ex-moderator kitten Nov 11 '14 at 04:02
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@dmckee: looks pretty close, though this seems to be more of the ratio than the sum (simple algebra swap though). – Kyle Kanos Nov 11 '14 at 04:03
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I'm with @dmckee - but for simple math, this is the same question. – Floris Nov 11 '14 at 07:16
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[PDG] quoted something like $(q_p+q_e)/e$ without defining these quantities,
That is exactly what you asked for. Recall that the charge on the electron $q_e$ is negative and that on the proton $q_p$ is positive, so the sum there is exactly the difference in their magnitudes. Taking it as a fraction of the defined base charge $e$ makes it a dimensionless value that does not depend on units, and $$ \frac{\left| q_p + q_e\right|}{e} \le 1 \times 10^{-21}$$ is a pretty good measurement.
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So here both $q$'s are measured while $e$ is the reference charge? Why not divide by $q_e$? – Physics_maths Nov 11 '14 at 11:06
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1No doubt because it complicates the uncertainty expression, and with the difference being valued at $10^{-21}$ or less it simply doesn't matter to the result. – dmckee --- ex-moderator kitten Nov 11 '14 at 19:30