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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 |
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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