GREPhysics.NET: GR8677 #52

GR8677 #52

Problem

GREPhysics.NET Official Solution

Electromagnetism $\Rightarrow$ }Potential

By Gauss Law, if there are no charges inside the cube, then the electric field inside would be 0. The potential $\phi$ is related to the field $\vec{E}$ by $\vec{E} = \nabla \phi$ , and thus since $\vec{E}=0$ , one infers that $\phi$ is constant. Since the potential function has to remain continuous its value everywhere inside of the cube is the same as that at the surface of the cube, which is given as $V$ .

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). Reply to this comment
DaveyClaus 2006-11-17 05:10:14	One cannot use Gauss's Law to determine $E=0$ within. Only spherical and cylindrical Gaussian surfaces (and "pillboxes") allow one to take E out of the integral; we cannot automatically assume that $E=0$ because $\rho=0$ in this situation. $\hspace{15pt}$ Instead, we simply recall that Laplace's equation, $\nabla^{2}V=0$ , 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. $\hspace{15pt}$ Now we are in a position to state that $E=0$ within... Reply to this comment

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

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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:

$\Phi_{center}=\frac{1}{n}\sum_{i=1}^n V_i$

In our particular problem, $\Phi_{center}=\frac{1}{6}\sum_{i=1}^n V_i=V$ , since $V_i=V$ on the six sides of the cube.

Herminso
2009-09-21 18:54:48

In LaTeX:

$\Phi_{center}=\frac{1}{n}\sum_{i=1}^n V_i$

In our particular problem, $\Phi_{center}=\frac{1}{6}\sum_{i=1}^6 V_i=V$ , since $V_i=V$ on the six sides of the cube.

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Herminso
2009-08-30 15:52:35

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phys2718
2008-10-16 16:29:33

To elaborate on DaveyClaus:rnrnWe can "guess" a potential function of the form rnV(x,y,z) = $V_0$ , where $V_0$ 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 $E=0$ within. Only spherical and cylindrical Gaussian surfaces (and "pillboxes") allow one to take E out of the integral; we cannot automatically assume that $E=0$ because $\rho=0$ in this situation.

$\hspace{15pt}$ Instead, we simply recall that Laplace's equation, $\nabla^{2}V=0$ , 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.

$\hspace{15pt}$ Now we are in a position to state that $E=0$ within...

spacemanERAU
2009-10-15 18:33:00

is a cube not a "pillbox"?

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LaTeX syntax supported through dollar sign wrappers $, ex., $\alpha^2_0$ produces $\alpha^2_0$ .
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$\partial$	$\partial$
$\Rightarrow$	$\Rightarrow$
$\ddot{x},\dot{x}$	$\ddot{x},\dot{x}$
$\sqrt{z}$	$\sqrt{z}$
$\langle my \rangle$	$\langle my \rangle$
$\left( abacadabra \right)_{me}$	$\left( abacadabra \right)_{me}$
$\vec{E}$	$\vec{E}$
$\frac{a}{b}$	$\frac{a}{b}$

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