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Connecting with a wire makes the whole thing one conductor and thus an equipotential. Hence V/k = Q1/R1 = Q2/R2. -Carl
On May 2, 2017, at 8:29 PM, brian whatcott <betwys1@sbcglobal.net> wrote:
I lately took to writing responses on the Quora Q&A site for recreation. I rarely stray from electronics and one or two othersubject areas where I have some insight. But the other day, withinternal alarm bells ringing, I strayed into a basic physics question,where some high-schooler (?) transcribed a question directly from somequestion paper, as they sometimes do.
You will recognize it, I expect. Two spheres of 4 and 6 cm radius are charged to 80 and 60 uCoulomb respectively. How is the charge redistributed when the spheres are connected with a wire? I answered on general principle, in this way: the charge is redistributed in proportion to the spheres' surface area. 4^2 / (4^2 + 6^2) X (80 + 60) uC and 6^2 / (4^2 + 6^2 ) X (80 + 60) uC. about 43 : 97 uC I was surprised to find (when I finally checked) that the charge division (as given in online sources) is proportional to spheres' radii, not area. 56 : 84 uC How was I so easily falsified? Brian W
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Carl E. Mungan, Professor of Physics 410-293-6680 (O) -3729 (F)
Naval Academy Stop 9c, 572C Holloway Rd, Annapolis MD 21402-1363
mailto:mungan@usna.edu http://usna.edu/Users/physics/mungan/
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