In reply to Jürg Wyttenbach's message of Fri, 30 Aug 2019 23:58:02 +0200: Hi, [snip] >The answer is simple > >q^2 --> rm. Charge square is proportional to rotating mass. In a proton >much more mass is needed to produce the same charge. Ergo adding an >electron can do nothing... > >J.W. Given that both mass & charge of the proton are known, what radius do you calculate?
Regards, Robin van Spaandonk local asymmetry = temporary success
