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

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...

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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

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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...

Bare Basic LaTeX Rosetta Stone
LaTeX syntax supported through dollar sign wrappers $, ex., $\alpha^2_0$ produces $\alpha^2_0$ .
type this...	to get...
$\int_0^\infty$	$\int_0^\infty$
$\partial$	$\partial$
$\Rightarrow$	$\Rightarrow$
$\ddot{x},\dot{x}$	$\ddot{x},\dot{x}$
$\sqrt{z}$	$\sqrt{z}$
$\langle my \rangle$	$\langle my \rangle$
$\left( abacadabra \right)_{me}$	$\left( abacadabra \right)_{me}$
$\vec{E}$	$\vec{E}$
$\frac{a}{b}$	$\frac{a}{b}$

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