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GR8677 #52 |
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Alternate Solutions |
DaveyClaus
2006-11-17 05:10:14 | One cannot use Gauss's Law to determine  within. Only spherical and cylindrical Gaussian surfaces (and "pillboxes") allow one to take E out of the integral; we cannot automatically assume that because  in this situation.
 Instead, we simply recall that Laplace's equation,  , with boundary conditions specified allows no maxima or minima within the boundary. So if the potential is V on the cube surface, potential cannot be greater than or less than V within.
Now we are in a position to state that within... |  |
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Comments |
phys2718
2008-10-16 16:29:33 | To elaborate on DaveyClaus:rnrnWe can "guess" a potential function of the form rnV(x,y,z) =  , where is the potential on the boundary of the cube. Now we simply ask (1) Does this function satisfy Laplace's equaton? Answer: yes (2) Does this function satisfy the boundary conditions? Trivially, yes. Then we are guaranteed that we have found the unique solution to the problem.rnrnBy the way, I have been posting a lot of comments on this site and I should thank Yosun for creating it. With a lack of study books and resources for this test, I believe this website is extremely useful and beneficial for any student in preparation for the Physics GRE.
phys2718
2008-10-16 16:30:58 |
And of course the formatting screws up when I decide to thank Yosun for the website
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|  | DaveyClaus
2006-11-17 05:10:14 | One cannot use Gauss's Law to determine within. Only spherical and cylindrical Gaussian surfaces (and "pillboxes") allow one to take E out of the integral; we cannot automatically assume that because in this situation.
Instead, we simply recall that Laplace's equation, , with boundary conditions specified allows no maxima or minima within the boundary. So if the potential is V on the cube surface, potential cannot be greater than or less than V within.
Now we are in a position to state that within... | |
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