# Why does the polarization point inwards for this problem?

• NucEngMajor
In summary, the problem deals with a thick spherical shell made of dielectric material with a frozen in polarization. The electric field inside and outside the shell is calculated and it is assumed that there is no free charge in the system. The polarization of the molecules creates a negative surface charge on the inner surface of the dielectric and a positive surface charge on the outer surface. The polarization vector is radially outward everywhere inside the dielectric, but at the inner surface it produces a negative surface charge density. This is an assumption for the problem and Griffiths defines "free charge" in a previous section.
NucEngMajor

## Homework Statement

Griffiths 4.15: A thick spherical shell of inner radius a and outer radius b is made up of dielectric material with a frozen in polarization. P = k / r in the r-hat direction. There is no free charge (why?). Find the electric field inside and out.

## The Attempt at a Solution

I have the solution, but the only part that I don't understand is why the polarization is negative at r=a (pointing inwards, why not radially out like at b?). This may be trivial, but a thorough explanation would be helpful, and perhaps and generalization to other cases. Also they tell us there is no free charge, but could someone explain why physically? Thank you

NucEngMajor said:
I have the solution, but the only part that I don't understand is why the polarization is negative at r=a (pointing inwards, why not radially out like at b?).
See if this helps: http://www.a-levelphysicstutor.com/field-capacit-1.php#dielectrics
You can see how the polarization of the molecules creates a negative surface charge on one of the surfaces of the dielectric and positive on another surface.

Also they tell us there is no free charge, but could someone explain why physically? Thank you
This is just one of the assumptions for this problem. It is assumed that the only macroscopic charge density in the system is due to the polarization of the molecules. Free charge would be extra charge added to the system. I think Griffiths defines "free charge" a couple of pages previous to this problem.

Additional comment: The polarization vector ##\vec{P}## is radially outward everywhere inside the dielectric. But, at the inner surface (r = a) the radially outward polarization produces a negative, bound surface charge density. It's not the polarization that's negative, it's the surface charge that's negative.

## 1. Why does the polarization point inwards for this problem?

The polarization of an electromagnetic wave depends on the direction of the electric field. In this problem, the electric field is pointing towards the center, causing the polarization to also point inwards.

## 2. Does the material being polarized affect the direction of polarization?

Yes, the material being polarized can affect the direction of polarization. Different materials have different arrangements of atoms and molecules, which can affect the direction of the electric field and thus the direction of polarization.

## 3. How does the polarization of light affect its properties?

The polarization of light affects its properties in various ways. For example, polarized light can be used to reduce glare, improve contrast, and enhance the 3D effect in certain types of glasses. It can also be used in communication systems, such as in polarized sunglasses to reduce glare from reflected light.

## 4. Can the direction of polarization be changed?

Yes, the direction of polarization can be changed through various means, such as passing the light through a polarizing filter or reflecting it off a surface at a certain angle. Additionally, the polarization of light can also be rotated by passing it through certain materials, such as liquid crystals.

## 5. How does polarization relate to the wave nature of light?

Polarization is a property of transverse waves, which includes electromagnetic waves like light. The direction of polarization is perpendicular to the direction of propagation of the wave and is determined by the direction of the electric field. This property helps to explain the behavior of light in various situations, such as when it is reflected or transmitted through different materials.

Replies
4
Views
2K
Replies
10
Views
3K
Replies
2
Views
2K
Replies
3
Views
3K
Replies
29
Views
3K
• Calculus and Beyond Homework Help
Replies
3
Views
432
Replies
4
Views
3K
• Atomic and Condensed Matter
Replies
0
Views
1K