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\prob{64}
If an electric field is given in a certain region by $E_x=0,E_y=0,E_z=kz$, where k is a nonzero constant, which of the following is true?

  1. There is a time-varying magnetic field.
  2. There is charge density in the region.
  3. The electric field cannot be constant in time.
  4. The electric field is impossible under any circumstances.
  5. None of the above.
Electromagnetism}Gauss Law

Gauss Law gives \nabla \cdot \vec{E} = \rho/\epsilon_0. Since the divergence of E in Cartesian coordinates is non-zero, there is a charge density in the region. QED
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Comments
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.NEC
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|*\pir^2d = \frac{4}{3}\pir^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.
NEC
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...
NEC
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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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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