# Polarization and Bound Charges

1. May 31, 2010

### manofphysics

Consider a uniform, isotropic , homogeneous solid dielectric slab.
We know, induced surface charge=$$\overline{P}.\widehat{n}$$
and $$\overline{P} \alpha$$ $$\overline{E}$$
So, as applied electric field increases, polarization per unit volume increases.
which implies that surface charge also increases.

But how can that be?Refer to the figure attached (similar fig. is given in Griffiths). This fig. tells us that the effective bound charge is the net charge at the surfaces which should be constant as the no. of atoms in a dielectric is constant.
Can someone explain this apparent anomaly to me?

#### Attached Files:

• ###### pol_and_area_charge.gif
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2. May 31, 2010

### Antiphon

Polarzation is not just counting charges in a unit volume but includes the degree of separation between the dipoles. The dipoles separate more as the applied field increases.

3. May 31, 2010

### manofphysics

I know that distance of dipole increases as applied electric field increases and hence dipole moment increases. But $$\overline{P}$$ is dipole moment per unit volume.And that should remain constant as volume also increases as dipole moment increases.

4. May 31, 2010

### gabbagabbahey

No, the volume of a dielectric slab stays the same (to a very good approximation at least) as the dipoles get stretched, squeezed and rotated. Hence, the Polarization changes.

5. May 31, 2010

### manofphysics

Ok, if volume remains approx. constant and P increases, then as higher electric field is applied surface charge of the conductor = P.n also increases?How can that be, as charges at the surface are constant as shown in the figure( attached it in my first post).

6. Jun 1, 2010

### Born2bwire

The material, like a dipole, is quasi-neutral. If we look at the dipole, we note that after a certain distance far enough away, the dipole looks like a neutral object because the fields cancel out. If we increase the distance of separation between the charges, we have to move accordingly farther away to get the same effect. Thus, for a given constant distance of observation away from a dipole, the dipole looks less and less like a neutral body the more the two charges are separated.

This is the same thing that is happening here. The bound charges are not true charges, they are the equivalent charges that represent the net field that you see due to the separation of charges within the volume due to the applied field. Because the applied field increases the distance between the dipoles, we see a more pronounced net field from the volume due to the discontinuity at the surface (there is no dipole above the surface dipoles to cancel out the distended charge).

So, if the bound charge is mearly a mathematically equivalent charge to represent the distortion from neutrality of the volume due to the applied field, then we would expect that the bound charge must increase when we increase the polarization of the dipole moments to accordingly account for the more pronounced net fields that we would observe.

7. Jun 1, 2010

### manofphysics

I have understood the point, born2bwire. But in Griffiths, he empathically says that bound charges are true accumulations of charges and not imaginary equivalent charges or mathematical tools to facilitate calculation.

8. Jun 1, 2010

### Born2bwire

They are constructed of true charges, but the magnitude of these charges, or rather the charge density, is taken to be an effective charge density. Griffiths explains, as I stated previously, that the surface charge arises because of the discontinuity of the surface. If we were to observe a given dipole within the volume, there is always an adjacent dipole that has an opposite charge that gets drawn near the charges of our dipole to effectively cancel out the fields of the dipole. At the surface, there are no dipoles above the dipoles whose charges congregate on the surface. Thus, close to the surface we will see a net field from these dipoles. The physicality of how the field arises is as Griffiths states, from true charges. The question then becomes what are the magnitudes of these charges.

Take, for instance, the figure that Griffiths uses to demonstrate this in his section on the physicality of the charges. If we have a line of dipoles, say:

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This is roughly equivalent, from a given distance away, to:

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Obviously though, the actual fields between differing dipole moments must be different. So the two cases,

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Will be seen roughly as:

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with the difference being the magnitude of the equivalent surface charges here. We can see obviously in the latter case where the moment is stronger that, due to the constant lattice spacing, the interior charges cancel out even more than before and that the surface charges are even further separated from each other and from the interior charges. So while the number of charges on the surface does not change (we will assume that we cannot create such a large moment such that the displacement is on the order of the lattice constant) the quasi-neutrality of the volume grows weaker.

9. Jun 1, 2010

### manofphysics

Thanks a lot, born2bwire...I really appreciate your help.It may sound funny but this was one thing which I had not understood for a long time even while using it for solving many problems.