GREPhysics.NET: GR9277 #64

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NoPhysicist3
2017-03-23 12:31:21

The words \"certain region\" are VERY confusing. However, when choosing between B and E, one should keep in mind that it is unlikely for ETS to consider a correct answer containing ultimate statements. Therefore B is the only correct.

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Naismith
2011-10-10 04:30:08

What do they mean by "in a certain region" ? In my opinion, it is always possible to find a region small enough so that it doesn't contain any charges, therefore charge density. The charge then will be outside the region...

h.fei10
2012-11-04 07:42:12

That's not possible. The electric field pevades this certain region, so does the charge.

calcuttj
2014-09-03 17:57:04

Think about the field inside a cylinder of constant charge density.

The cylinder has radius R, constrain r < R such that

|E|* $\pi$ r^2d = $\frac{4}{3}$ $\pi$ r^3d $\frac{\rho}{\epsilon_0}$

(d is the length of our Gaussian cylinder)

|E| = $\frac{4}{3}$ r $\frac{\rho}{\epsilon_0}$

The region could be the z axis inside the cylinder

Not necessarily the only charge distribution to create E =kz, and this definitely doesn't prove there is ALWAYS a distribution to create a field like this, but it definitely disproves that the field is impossible

Now think about this. If there wasn't a charge density in the region. shouldn't the field be decreasing (i.e. E=k/z)?

calcuttj
2014-09-10 16:36:07

I made a mistake in my last comment, ignore it.

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r10101
2007-10-27 16:32:37

Why does a small region of vacuum near the surface of an infinite charged plate (with constant $\vec{E}$ = E $\hat{n}$ normal to the surface) not satisfy this question, making answer (E) correct?

panos85
2007-10-31 05:12:32

It says $E_z=kz$ , not $E_z=k\hat{z}$ . The electric field near the surface of a conductor is constant, while the field in this problem is not.

tonyhong
2008-10-25 01:54:24

this is a trap...

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sharpstones
2006-12-01 10:25:11

how could you possibly construct a charge density that would make such an E field?

mhas035
2007-04-04 23:58:56

Remember that it says that the field is only in a certain region. We just need the region to be small with a relatively large charged plane of constant charge density.

evanb
2008-06-24 11:56:52

How about a uniform-density infinite-plane slab. So, it would be thick and the region of interest would be from the middle of the slab to the edge of the slab.

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Why does a small region of vacuum near the surface of an infinite charged plate (with constant $\vec{E}$ = E $\hat{n}$ normal to the surface) not satisfy this question, making answer (E) correct?

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$\int_0^\infty$	$\int_0^\infty$
$\partial$	$\partial$
$\Rightarrow$	$\Rightarrow$
$\ddot{x},\dot{x}$	$\ddot{x},\dot{x}$
$\sqrt{z}$	$\sqrt{z}$
$\langle my \rangle$	$\langle my \rangle$
$\left( abacadabra \right)_{me}$	$\left( abacadabra \right)_{me}$
$\vec{E}$	$\vec{E}$
$\frac{a}{b}$	$\frac{a}{b}$

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