Talk:Planck charge

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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 4πε0 = 1 then Coulomb's law takes the form: F = \frac{q_1 q_2}{r^2}

Force is \frac{energy}{length}, so this formula has the following dimensions:

\frac{energy}{length} = \frac{charge^2}{length^2}

\rightarrow energy \times length = charge^2

\hbar has units of energy \times time, and c has units of \frac{length}{time}, so \hbar c has units of energy \times length to match charge2

The incorrect formula claimed the Planck charge squared was \frac{\hbar}{c}, but \frac{\hbar}{c} has units of \frac{energy \times time^2}{length} which clearly doesn't match the units of charge2 --Tim314 16:30, 11 February 2007 (UTC)

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. kc can also be written as \frac{1}{4 \pi \epsilon_0}. But physicists sometimes use units where this constant is equal to 1, rather than the SI value of approximately 9 \times 10^9. Of course, making kc 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)
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)


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)