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Incorrect Claim

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The claim in the text: "From a classical calculation,[3] the electric potential energy of one Planck charge on the surface of a sphere that is one Planck length in diameter is one Planck energy." appears incorrect to me. I did the substitution as a check, and found that one Planck charge uniformly distributed on a charged sphere is one halph a plank energy (since there is a factor of one halph in the electrostatic energy). can someone verify this for me? Pjbeierle (talk) 16:54, 1 August 2018 (UTC)[reply]

Other

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The equation for Planck charge was correct in the original version of the page, but someone changed it to an incorrect version. I've fixed it back to the original formula, and edited the text slightly.


It's easy to see which version is right from dimensional analysis. If you choose your units so that then Coulomb's law takes the form:

is , so this formula has the following dimensions:

has units of , and has units of , so has units of to match

The incorrect formula claimed the Planck charge squared was , but has units of which clearly doesn't match the units of --Tim314 16:30, 11 February 2007 (UTC)[reply]

you forget the very important coulomb constant Kc which dimension is the square of velocity. Then hbar divided by c becomens hbar times c or Planck charge is the squareroot of the Planck mass time Planck length!!!!!!!!!!!! -- unsigned comment added by 83.82.134.77
I didn't forget. can also be written as . But physicists sometimes use units where this constant is equal to 1, rather than the SI value of approximately . Of course, making equal to 1 (dimensionless) requires us to change the unit of charge. See the statcoulomb article for more details. -- Tim314 20:08, 26 September 2007 (UTC)[reply]
Also, please don't place your comment in the middle of one of mine. It makes it hard to know who said what. For this reason, I moved your comment to below mine. --Tim314 20:08, 26 September 2007 (UTC)[reply]


What is the physical interpretation of the planck charge? —Preceding unsigned comment added by 85.225.229.227 (talk) 19:48, 15 January 2008 (UTC)[reply]

Good question. It needs answering. Xxanthippe (talk) 02:01, 7 November 2008 (UTC).[reply]
To the best of my knowledge, this is one of the few Planck Units for which there is no real (useful) physical interpretation. Iameditingstuff (talk) 22:57, 23 May 2011 (UTC)[reply]

Quantum Electromagnetic Resonator

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Why this part was deleted?195.47.212.108 (talk) 07:47, 26 August 2009 (UTC)[reply]

Because you have been creating a large number of articles along these lines that appear to mostly be your own take on the subject, rather than material that has been published elsewhere. While the additions may be perfectly correct from a factual point of view, they'll still violate Wikipedia's "original research" policy if they're your own synthesis of available material. Wikipedia:Articles for creation is a rubber-stamp process, not an endorsement from Wikipedia. They mostly count on you to have already read the relevant policies on what kind of articles Wikipedia is trying to include, and the WP:V and WP:RS policies on how to include appropriate references to show that what you're including is appropriate (via citing appropriate published sources). I respect your efforts, but the type of article you've been creating might be more appropriately hosted on your own web site than in the encyclopedia (the encyclopedia's strictly a "secondary source", meaning every idea in it has to have been published elsewhere first). --Christopher Thomas (talk) 07:14, 29 August 2009 (UTC)[reply]
Actually wikipedia is a tertiary source WP:WITS, incorporating widely accepted results, typically those that have been published in review articles or books for the subfield in question. So some primary and secondary source material, even if published in reputable journals, may represent a contentious position not appropriate for wikipedia. Dark Formal (talk) 03:25, 31 August 2009 (UTC)[reply]

Coulomb constant

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In that nice big 3-or-4-equals sign equation at the top, could we add an expression showing the Planck charge explicitly in terms of ke, in addition to the expressions we're already showing there? (I.e. we already show the Planck charge in terms of ε0 and α, so why not also show it in terms of ke?) --Rogermw (talk) 02:17, 1 June 2013 (UTC)[reply]