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

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

Prologue
2009-11-06 18:57:04

If you like you can break this problem up into parts. First look at what the E field would be without the dielectric, namely

$E_{i}=\frac{\sigma}{\epsilon_{0}}$

Then with the dielectric in place

$E_{f}=\frac{\sigma}{K \epsilon_{0}}$

Now the difference in the two must be due to an electric field generated by the bound charges. Keep in mind that the only effect the dielectric has is to generate these bound charges nothing else, so the E field generated by the bound changes is not subject to the dielectric constant. The bound E field is given by

$E_{f}-E_{i}=E_{b}=\frac{\sigma _{b}}{\epsilon_{0}}$

By doing this you get

$E_{f}-E_{i}=\frac{\sigma}{K \epsilon_{0}}-\frac{\sigma}{\epsilon_{0}}$

So

$E_{b}=\frac{\sigma _{b}}{\epsilon_{0}}=\frac{\sigma}{\epsilon_{0}} \left (\frac{1}{K}-1 \right) = \frac{\sigma}{\epsilon_{0}} \left (\frac{1-K}{K} \right)$

This means the bound surface charge is given by

$\sigma _{b}=\sigma \left (\frac{1-K}{K} \right)$

Reply to this comment

abacus
2009-10-06 23:29:13

Here's another soln:

The bound charges in a dielectric plate redistribute so that $E_{inside} = E_{outside}/K$ (This is why capacitance is proportional to K)

Inside the dielectric, it looks like there's a surface with charge density is $\sigma/K$

$\sigma + \sigma_b = \sigma / K$
$\sigma_b = \sigma(1-1/K)$

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

nitin
2006-11-16 08:34:36

Let $\vec{D}=\epsilon_0\vec{E}+\vec{P}$ .

Take the divergence of this expression:

$\vec{\nabla}.\vec{D}=\epsilon_0\vec{\nabla}.\vec{E}+\vec{\nabla}.\vec{P}$ .

Let the permittivity of the dielectric be $\epsilon_D$ . Then, $\epsilon_0=\frac{\epsilon_D}{K}$ .

Now,

$\vec{\nabla}.\vec{D}=\sigma$ ,

$\vec{\nabla}.\vec{E}=\frac{\sigma}{\epsilon_D}$ , and

$\vec{\nabla}.\vec{P}=-\sigma_P$ .

Substitute these in the divergence equation, to get:

$\sigma_P=\sigma\frac{(1-K)}{K}$ , which is choice (E).

Reply to this comment

bterranova
2006-10-19 14:08:52

The divergence of the polarization is related to the total charge density, not the surface charge density.

CMSmonkey
2009-11-06 13:09:27

I agree!

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.

flutefreek
2007-10-02 02:46:27

$DA=\sigma A$
$D=\sigma$

$D=k \epsilon_0 E$ (Griff. eq 4.31)
$E=\frac{\sigma}{k \epsilon_0}$

$D=P+\epsilon_0 E$ (Griff. eq 4.21)
$P=\frac{\sigma (k-1)}{k} \hat{z}$

Jeremy
2007-10-24 11:25:22

Minor correction: $\hat{n}=-\hat{z}$ , since this is the "outward" direction for the dielectric. Therefore, you don't have to hand wave the sign!

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.

thebigshow500
2008-10-14 01:35:04

I will try your method if I were you.

However, by fitting k -> $\infty$ to (E), we get -Sigma. Is this making any sense?

gotfork
2008-10-16 15:01:48

I think this makes sense, -Sigma effectively cancels the surface charge since we don't expect to see any total surface charge since both materials are conductors.

a19grey2
2008-11-03 22:22:55

We expect to see -(Sigma) because the opposite charge will go to towards the charge on the conductor (like charges are repelled).
Also, both materials are not conductors. When k-> $\infty$ THEN the dielectric acts as a conductor. When k->1 is when the slab labeled "dielectric" is acting like vacuum would.... and since there can be no bound charges in vacuum, the bound charge $\sigma_p$ ->0.

Reply to this comment

StudyTime
2005-11-09 21:02:38

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Post A Comment!

You are replying to:

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.

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