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GR9277 #64
Problem
 GREPhysics.NET Official Solution Alternate Solutions
\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$\Rightarrow$}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

Alternate Solutions
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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.fei102012-11-04 07:42:12 That's not possible. The electric field pevades this certain region, so does the charge.
 calcuttj2014-09-03 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)?
 calcuttj2014-09-10 16:36:07 I made a mistake in my last comment, ignore it.
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?
 panos852007-10-31 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.
 tonyhong2008-10-25 01:54:24 this is a trap...
sharpstones
2006-12-01 10:25:11
how could you possibly construct a charge density that would make such an E field?
 mhas0352007-04-04 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.
 evanb2008-06-24 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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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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