- #36
TSny
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voko's explanation is nice. I was thinking of a similar but cruder way to look at it by using ##U = \frac{k}{2}\displaystyle\sum\limits_{i\neq j} \frac{q_i q_j}{r_{ij}}## for a system of particles.
Thus, break the triangle into small patches of area ##\Delta A_i## so you have ##U = \frac{k\sigma^2}{2}\sum\limits_{i\neq j} \frac{\Delta A_i \Delta A_j}{r_{ij}}##.
Then scale the entire figure (including the grid) by a linear scaling factor so that you still have the same number of patches. See attached figure. You have a new energy
##U = \frac{k\sigma\,'\,^2}{2}\sum\limits_{i\neq j} \frac{\Delta A_i' \Delta A_j'}{r_{ij}'}## Then think about how the new primed quantities compare to the original unprimed quantities. But for the new ##\sigma\,'## you'll want to use the charge density that results from folding, not the charge density that would result from the shrinking of the figure.
[EDIT: You can see that this is just a "poor man's version" of voko's elegant derivation.]
Thus, break the triangle into small patches of area ##\Delta A_i## so you have ##U = \frac{k\sigma^2}{2}\sum\limits_{i\neq j} \frac{\Delta A_i \Delta A_j}{r_{ij}}##.
Then scale the entire figure (including the grid) by a linear scaling factor so that you still have the same number of patches. See attached figure. You have a new energy
##U = \frac{k\sigma\,'\,^2}{2}\sum\limits_{i\neq j} \frac{\Delta A_i' \Delta A_j'}{r_{ij}'}## Then think about how the new primed quantities compare to the original unprimed quantities. But for the new ##\sigma\,'## you'll want to use the charge density that results from folding, not the charge density that would result from the shrinking of the figure.
[EDIT: You can see that this is just a "poor man's version" of voko's elegant derivation.]
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