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GR8677 #10
Problem
 GREPhysics.NET Official Solution Alternate Solutions

Electromagnetism$\Rightarrow$}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!

Alternate Solutions
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2019-08-10 03:09:48
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varshakr
2016-12-27 10:46:29
depends on units
Giubenez
2014-10-22 02:55:55
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.
flyboy621
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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nmast
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.
 kroner2009-11-27 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.
wittensdog
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?
Andresito
2006-03-14 22:37:41
Gauss' law E*A = Q/epsilon

Yours is an approximation.

Thanks Yosun.
 lxanderl2009-06-28 17:23:43 Depends on if he's using cgs or SI

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}\$ $\frac{a}{b}$