Electromagnetism (Dielectrics)

[roam]

Homework Statement

A dielectric slab with a susceptibility $\chi_e$ rests on a conducting plate whose upper surface carries a free surface charge density $\sigma$. Show that the polarization surface charge density $\sigma_{pol}$ on the lower face of the dialectic slab is:

$\sigma_{pol} = - \sigma \frac{\chi_e}{1+\chi_e}$

Homework Equations

The polarization vector P is: $P=\chi \epsilon_0 E$

The electric field between the surfaces is

$E=\frac{\sigma - \sigma_{pol}}{\epsilon_0} = \frac{\sigma}{\epsilon_0} \frac{1}{1+\chi}$

The Attempt at a Solution

I believe in this case $\sigma_{pol} = P$, so combining the two equations I get

$\sigma_{pol}=P=\chi_e \epsilon_0 \frac{\sigma}{\epsilon_0} \frac{1}{1+\chi}$

$= \sigma \frac{\chi_e}{1+\chi_e}$

However I did not get the minus sign in front of my expression. What is wrong?

I know that $\sigma_{pol}$ and $\sigma$ must be of opposite sign, but mathematically how do I bring in the negative sign? Is my approach to the problem correct?

Any help would be appreciated.

 

[gabbagabbahey]

I believe in this case $\sigma_{pol} = P$

Justify your belief with some physics...

 

[roam]

Justify your belief with some physics...

I modeled the situation as a dielectric slab in a uniform field so that the lines of E pass through the slab. The field is generated by the conductor and so the free charges in it induce opposite charges on the lower face of the dialectic slab.

In this situation according to my textbooks we have bound charge of $\rho_{pol}=-\nabla .P$ within the dielectric and the magnitude of the surface charge density is $\sigma_{pol} = P. \hat{n}$.

Why is this not correct? How else can I approach this problem?

 

[gabbagabbahey]

I modeled the situation as a dielectric slab in a uniform field so that the lines of E pass through the slab.

How do you justify the assumption that the field is uniform? (It is, but you need to justify it in your solution)

the magnitude of the surface charge density is $\sigma_{pol} = P. \hat{n}$

In which direction does the electric field point (at the lower surface of the dielectric)? In which direction is the outward surface normal to the dielectric for its lower surface? What does that make \sigma_{pol}=\epsilon_o \chi \mathbf{E} \cdot \hat{\mathbf{n}}

 

[roam]

How do you justify the assumption that the field is uniform? (It is, but you need to justify it in your solution)

I don't know how to prove this assumption. But the course I'm doing is about electrostatics, so there are no time-varying electric fields, and we mostly assume the materials are uniform so the fields would have to be uniform.

In which direction does the electric field point (at the lower surface of the dielectric)? In which direction is the outward surface normal to the dielectric for its lower surface? What does that make \sigma_{pol}=\epsilon_o \chi \mathbf{E} \cdot \hat{\mathbf{n}}

Do you mean that should make the $\hat{n}$ part negative?

I believe the direction in which the electric field is pointing depends on the kind of charge it is carrying. The slab is sitting on the conducting plate so the surfaces are completely normal to each other. So we know the field lines are normal to the slab's lower surface.

Here is a diagram from a textbook about a parallel-plate capacitor with a dielectric:

http://img502.imageshack.us/img502/7273/dielectric.jpg​

I think in this problem the E lines are normal to the dielectric's surface similar to the picture above. How does that then really affect $\sigma_{pol}=\epsilon_o \chi \mathbf{E} \ \hat{\mathbf{n}}$? I'm really not sure...

 

[gabbagabbahey]

I don't know how to prove this assumption. But the course I'm doing is about electrostatics, so there are no time-varying electric fields, and we mostly assume the materials are uniform so the fields would have to be uniform.

The (infinite?) conducting plate has a free charge density on it, so it is free to redistribute itself in a non-uniform manner, if it is subject to a non-uniform external field. In order to justify the fact that the charge density will remain uniformly distributed here, I suggest you think about symmetry. Alternatively, what rgument was presented in the parallel plate example below? Can you adapt that argument to this situation?

Do you mean that should make the $\hat{n}$ part negative?

You tell me. Does the outward normal of the lower surface of the dielectric point upwards or downwards?

I believe the direction in which the electric field is pointing depends on the kind of charge it is carrying.

Doesn't the \sigma in your expression for E carry that information?

Here is a diagram from a textbook about a parallel-plate capacitor with a dielectric:

http://img502.imageshack.us/img502/7273/dielectric.jpg​

I think in this problem the E lines are normal to the dielectric's surface similar to the picture above. How does that then really affect $\sigma_{pol}=\epsilon_o \chi \mathbf{E} \ \hat{\mathbf{n}}$? I'm really not sure...

Look at the bottom half of that diagram. The \mathbf{E}-field of the conducting plate points upward (for positive \sigma_{\text{free}}). What about the outward surface normal for the dielectric's lower surface?

 

[roam]

The (infinite?) conducting plate has a free charge density on it, so it is free to redistribute itself in a non-uniform manner, if it is subject to a non-uniform external field. In order to justify the fact that the charge density will remain uniformly distributed here, I suggest you think about symmetry. Alternatively, what rgument was presented in the parallel plate example below? Can you adapt that argument to this situation?

I'm a little bit confused. When you say I think about symmetry, do you mean I have to use Gauss’s law for electric fields?

You tell me. Does the outward normal of the lower surface of the dielectric point upwards or downwards?

Thanks for clarifying. So, $\hat{n}$ is a unit vector normal to the surface, so I think it would be negative if it is pointing from the dielectric's lower surface toward the conductor (because it is pointing downward). Is this the correct idea?

Doesn't the \sigma in your expression for E carry that information?

How does the \sigma part carries that information? Charge comes in two varieties, plus or minus and the problem says there is a charge density \sigma on the conductor, but it doesn't say if it is positive or negative charge. But anyway I see now that this is not important for this problem.

Look at the bottom half of that diagram. The \mathbf{E}-field of the conducting plate points upward (for positive \sigma_{\text{free}}). What about the outward surface normal for the dielectric's lower surface?

It is pointing downward.