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  GR8677 #10
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Verbatim question for GR8677 #10
Electromagnetism}Gauss' Law

Only choices C, D, and E make sense, since the field at r=0 has to be 0. Moreover, since a uniform charge distribution is applied, the field has to increases linearly within the volume of the sphere. Choice D and E are both out, leaving just C.

Direct verification via applying Gauss' Law for the inside of the sphere: E(4\pi r^2)=Q=\rho \frac{4}{3}\pi r^3\Rightarrow E=\frac{\rho r}{3}\propto k r.

Outside, all five choices converge to an inverse-square decreasing field---so, no worry there!

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2016-12-27 10:46:29
depends on unitsNEC
2014-10-22 02:55:55
Nice bad pics:
the trend for x>R isn't an inverse square (if you estimate for E(x=2R)/E(R) you find 1/2 and not 1/4.
Furthermore it seems to have an horizontal asymptote > 0......

And so, all the graph are wrong.
2010-11-09 20:54:39
A minor correction: the field increases linearly with the radius of the sphere, not the volume. Hence the straight line of graph C.
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2009-11-24 09:16:37
Since the charge density is uniform, shouldn't E go like Q enclosed? Being a sphere, this would be like r^3, making D correct.
2009-11-27 19:08:54
No. Q is proportional to r^3, but E is proportional to Q/r^2 which means E is linear in r, as in choice C.
2009-09-24 21:45:16
ETS sure does love to throw in unnecessary details - who cares if the charge is positive or negative when it comes to the field magnitude?NEC
2006-03-14 22:37:41
Gauss' law E*A = Q/epsilon

Yours is an approximation.

Thanks Yosun.
2009-06-28 17:23:43
Depends on if he's using cgs or SI
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Since the charge density is uniform, shouldn't E go like Q enclosed? Being a sphere, this would be like r^3, making D correct.

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