GREPhysics.NET: GR8677 #54

GR8677 #54

Problem

GREPhysics.NET Official Solution

Alternate Solutions

Electromagnetism $\Rightarrow$ }Dielectric

Recall the following equations,

$\nabla \cdot \vec{D} = \rho_f /; /; /; \vec{D}=\epsilon_0 \vec{E} + \vec{P}=\kappa \epsilon_0 \vec{E}.<br />$

Also, recall the relation between bound charge and polarization,

$\nabla \cdot \vec{P} = -\sigma_b.<br />$

Use the divergence theorem on the above equations to apply the elementary Gauss' Law to the region,

$\oint \vec{D} \cdot dA = \int \sigma dA \Rightarrow D=\sigma. <br />$

$\oint \vec{P} \cdot dA = -\int \sigma_p dA \Rightarrow P=-\sigma_p. <br />$

But, since $\vec{D}=\epsilon_0 \vec{E} + \vec{P}=\kappa \epsilon_0 \vec{E}$ , one has $D=\sigma=\kappa \epsilon_0 E\Rightarrow E=\frac{\sigma}{\kappa \epsilon_0}$ . Plug that into the other relation for D (and use the result for P from above) to get, $\sigma = \frac{\sigma}{\kappa} - \sigma_p$ . Thus, $\sigma_p = \sigma \frac{1 - \kappa}{\kappa}$ , as in choice (E).

Alternate Solutions

hooverbm 2012-10-01 09:32:47	Super fast way to do this. This is exactly what a capacitor is (one side of it), with a dielectric wedged inbetween the plates. Compare capacitor in vacuum with one with dielectric. More field lines pass through the capacitor in vacuum. However, with the dielectric, some field lines are "intercepted" by negative charges that are induced. The induced charge that accumulates is never going to exceed the surface charge density of the conductor. For most dielectric materials, the induced charge density must be lower. Therefore, we know that the solution must be negative and the induced charge density be smaller than the surface charge density of the conductor. Pick a number for K, find the one that gives a fraction of the amount and is negative. Thus E. Reply to this comment
flyboy621 2010-11-14 21:27:59	As K approaches 1, the dielectric does not conduct at all, and the charge approaches 0. As K gets very large, the dielectric acts like a good conductor, and the charge approaches minus sigma. The only solution that matches both criteria is (E). Reply to this comment
Herminso 2009-09-21 19:02:57	The charge induced on the dielectric must be negative since $\sigma$ >0 on the conductor (remember images method), in addition that induced charge must be less than $\sigma$ for a dielectric which have $k>1$, thus the only answer choice is (E). Reply to this comment
sevensixtwo 2006-07-31 22:25:33	Whoa, dude! This is actually nice and simple. No divergence theorem needed. use gauss law for dielectrics with cylindrical surface. because E=0 in the conductor there is only flux through one end of the cylinder go gauss law becomes: D A = sigma A, so D = sigma then, D = k e0 E (Griff. eq 4.31) so E = sigma / (k e0) plug those values into: e0 E + P = D (Griff. eq. 4.21) simplify to get P = [sigma (k-1)/k] (z-hat) then dot by n-hat = z-hat to get the bound surface charge in the conductor. then the bound charge in the dielectric just has the opposite charge so its choice E. also, if you're not sure that we were solving for the bound charge in the conductor remember that the field in the dielectric has the same sign as the free charge. since we said the field was positive, the bound charge in the dielectric has to be opposite to the positive sign of the bound charge in the conductor. Reply to this comment
StudyTime 2005-11-09 21:06:49	Alternatively, you can check the limits. SigmaBound->0 for k->1, SigmaBound->SigmaSurface for k->infinity. Only choice (E) survives. Reply to this comment

Comments

fredluis
2019-09-17 02:05:02

The solution given above is wrong. Purely qualitatively, by Ampere\'s Law, B field will be in the +/- Z direction at point P. The energy flux vector S points outward (obviously). Thus E is in the XY plane (Orthogonal Triad). tree removal

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