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. |  | 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.
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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|* r^2d =  r^3d
(d is the length of our Gaussian cylinder)
|E| = r
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)?
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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 = E normal to the surface) not satisfy this question, making answer (E) correct?
panos85 2007-10-31 05:12:32 |
It says , not . The electric field near the surface of a conductor is constant, while the field in this problem is not.
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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.
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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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|  |
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$\int_0^\infty$ |
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$\partial$ |
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$\Rightarrow$ |
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$\ddot{x},\dot{x}$ |
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$\sqrt{z}$ |
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$\langle my \rangle$ |
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$\left( abacadabra \right)_{me}$ |
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$\vec{E}$ |
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$\frac{a}{b}$ |
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