Polarisation of dielectric materials

[lys04]

Homework Statement

Is this correct?
For a uniformly polarised material i,e divergence of polarisation is 0, there will only be bound surface charges and so the dipole moment only comes from these bound surface charges.

However, for a non uniformly polarised material, the divergence is non 0 so the material will have both bound volume charge density and bound surface charge density. And thus both of these contribute to the dipole moment.

Relevant Equations

Uniformly polarised material:
$\vec{p}(\vec{x_0}) =\int_{surface} (\vec{x’}-\vec{x_0}) \sigma (\vec{x’}) dA’$

$=\vec{p}(\vec{x_0}) =\int_{surface} (\vec{x’}-\vec{x_0}) \vec{P} \cdot \hat{n} dA’$

Non uniformly polarised material:
$\vec{p}(\vec{x_0}) =\int_{surface} (\vec{x’}-\vec{x_0}) \sigma (\vec{x’}) dA’ + \int_{volume} (\vec{x’}-\vec{x_0}) \rho (\vec{x’}) dV’$

$=\vec{p}(\vec{x_0}) =\int_{surface} (\vec{x’}-\vec{x_0}) \vec{P} \cdot \hat{n} dA’ - \int_{volume} (\vec{x’}-\vec{x_0}) \nabla \cdot \vec{P} dV’$

^^

 

[TSny]

However, for a non uniformly polarised material, the divergence is non 0 so the material will have both bound volume charge density and bound surface charge density.

This statement is not necessarily true. It's possible for the polarization $\vec P$ within a dielectric material to be nonuniform and yet have $\vec \nabla \cdot \vec P = 0$ at every point inside the dielectric material. [Edited to remove additional comments that probably gives away too much information for a homework question.]

I'm not clear on the homework assignment. Are you given the statements listed in the "Homework Statement" and your assignment is to decide if the statements are correct?

 

[lys04]

This statement is not necessarily true. It's possible for the polarization P→ within a dielectric material to be nonuniform and yet have ∇→⋅P→=0 at every point inside the dielectric material.

Ah yeah I can see how that's possible.
But given a non-zero divergence, would the dipole moment be given by
$\vec{p}(\vec{x_0}) =\int_{surface} (\vec{x’}-\vec{x_0}) \sigma (\vec{x’}) dA’ + \int_{volume} (\vec{x’}-\vec{x_0}) \rho (\vec{x’}) dV’$?

I'm not clear on the homework assignment. Are you given the statements listed in the "Homework Statement" and your assignment is to decide if the statements are correct?

No, this is not a homework problem.

 

[TSny]

Ah yeah I can see how that's possible.
But given a non-zero divergence, would the dipole moment be given by
$\vec{p}(\vec{x_0}) =\int_{surface} (\vec{x’}-\vec{x_0}) \sigma (\vec{x’}) dA’ + \int_{volume} (\vec{x’}-\vec{x_0}) \rho (\vec{x’}) dV’$?

That looks right. Just to be clear, what is the meaning of $\vec x_0$? If the net charge of the dielectric is zero, does the dipole moment $\vec p$ depend on $\vec x_0$?

 

[lys04]

That looks right. Just to be clear, what is the meaning of $\vec x_0$? If the net charge of the dielectric is zero, does the dipole moment $\vec p$ depend on $\vec x_0$?

If the charge distribution has non zero total charge then the dipole moment depends on where you measure it. But it doesn’t matter if the charge distribution has 0 net charge

 

[TSny]

If the charge distribution has non zero total charge then the dipole moment depends on where you measure it.
But it doesn’t matter if the charge distribution has 0 net charge

Ok.
$\vec x_0$ in the integrals represents the choice of origin for calculating the dipole moment. If the net charge of the system is zero, then the dipole moment will not depend on $\vec x_0$..