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GR8677 #52
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Alternate Solutions |
Herminso
2009-08-30 15:58:04 | An easier & faster way to do this problem is find the potential at the center of the cube as the average of the potential on the six sides, then it is V.
That can be generalized to any regular polyhedron with n sides (n=4,6,8,12,20), the n surfaces can be at different potentials Vi. The potential at the center of the polyhedron is the average of the potential on the n sides (see Jackson 3ed pag 94). |  | 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 |
alemsalem
2009-10-02 09:41:35 | the easiest way is to find a solution since we know it must be unique,, obviously a constant V in the cube satisfies both Laplace equation and the boundary condition
|  | Herminso
2009-08-30 15:58:04 | An easier & faster way to do this problem is find the potential at the center of the cube as the average of the potential on the six sides, then it is V.
That can be generalized to any regular polyhedron with n sides (n=4,6,8,12,20), the n surfaces can be at different potentials Vi. The potential at the center of the polyhedron is the average of the potential on the n sides (see Jackson 3ed pag 94).
Herminso
2009-09-21 18:50:47 |
In LaTeX:
In our particular problem,  , since  on the six sides of the cube.
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Herminso
2009-09-21 18:54:48 |
In LaTeX:
In our particular problem,  , since on the six sides of the cube.
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| | Herminso
2009-08-30 15:52:35 | An easier & faster way to do this problem is find the potential at the center of the cube as the average of the potential on the six sides, then it is V.
That can be generalized to any regular polyhedron with n sides (n=4,6,8,12,20), the n surfaces can be at different potentials Vi. The potential at the center of the polyhedron is the average of the potential on the n sides (see Jackson 3ed pag 94). | | 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...
spacemanERAU
2009-10-15 18:33:00 |
is a cube not a "pillbox"?
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keyboarder
2010-05-07 18:35:08 |
spacemanERAU asked 'is a cube not a "pillbox"?'
A "pillbox" is most often used as a Gaussian surface at an interface between two different media. Since Gauss's Law applies for any arbitrary size of surface, the "pillbox" is made to shrink until it is of negligible size (infinitesimal): Either flat in one dimension or very small in multiple dimensions. Once it is small enough--especially near an interface--it is reasonable to assume that the E-field inside is symmetrical and/or constant throughout the pillbox.
The cube described in the problem is of finite size. Thus, it is not a pillbox in the context meant by DaveyClaus.
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keyboarder
2010-05-07 18:49:19 |
It is worth clarifying that it is not only the choice of Gaussian surface which allows one to "take E out of the integral". The Gaussian surface is chosen to "match" the *charge distribution*. For example, choose a sphere with a spherically symmetric charge distribution, or a cylinder with a wire or coaxial (all with a common axis).
In other words, the symmetry of the Gaussian surface should match the symmetry of the charge distribution. This is what allows the flux integral to be simplified.
A spherical or cylindrical Gaussian surface will not help at all if the charge distribution (and hence, the E-field going through the Gaussian surface) has some other configuration.
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