Understanding Dielectric Materials: E Field and Coulomb Force Explained

[Fibo112]

My basic understanding of a dielectric material is that the coulomb force between charges in a dielectric must be reduced by some factor since some of the electric field energy gets stored in the medium. The following thought experiment is confusing me and I am wondering where my mistake lies.

Lets consider the border area between two different dielectrics. Now consider a rectangular path where two opposing sides are each in a different dielectric. All sides are short enough that any E field would be constant if it were the same medium and the sides travellng through the border are very short compared to the others.

Now the work done by the E field along this path must be zero. This seems to imply that the component of the E field along the path is equal in the different dielectrics. But arent all components of the E field in the dielectric smaller by some factor?

 

[kuruman]

My basic understanding of a dielectric material is that the coulomb force between charges in a dielectric must be reduced by some factor since some of the electric field energy gets stored in the medium.

That is correct. The electric field is reduced by the polarization of the dipoles in the dielectric.

Now the work done by the E field along this path must be zero.

What work done? Perhaps you mean the integral $\int \vec E \cdot d \vec l$?

But arent all components of the E field in the dielectric smaller by some factor?

Yes they are, but the factor is not the same in the two media. I think what you are talking about is the proof that the tangential component of the E-field is continuous across the boundary. That is usually shown by using $\oint_C \vec E \cdot d \vec l=0$ around the closed loop $C$ that you mentioned in the limit that the "sides" of the loop shrink to zero.

 

[Fibo112]

That is correct. The electric field is reduced by the polarization of the dipoles in the dielectric.

What work done? Perhaps you mean the integral $\int \vec E \cdot d \vec l$?

Yes they are, but the factor is not the same in the two media. I think what you are talking about is the proof that the tangential component of the E-field is continuous across the boundary. That is usually shown by using $\oint_C \vec E \cdot d \vec l=0$ around the closed loop $C$ that you mentioned.

Thanks for your answer! I am still a bit confused though. If a have some charge in a dielectric and look at the field it produces it will be smaller than the field in a vacuum by the factor of the dielectric constant. Let's say I have a charge surrounded by vacuum and the vacuum is surrounded by a dielectric. Will the field in the dielectric be reduced by the same amount as if the space between the point in question and the charge was filled entirely with the dielectric?

 

[Charles Link]

The reduction in the electric field $E$ that the dielectric experiences is due to the polarization charges that form on the boundary of the material. Surface polarization charge per unit area $\sigma_p=\vec{P} \cdot \hat{n}$. $\\$ This effect on the electric field from this surface polarization charge, that is responsible for the reduced electric field experienced inside a dielectric with an applied external electric field, will depend upon the geometric shape of the dielectric material. $\\$ A uniform polarization $P$ in the dielectric does not cause any electric field to develop=at least in a macroscopic sense. $\\$ A sphere will cause an $E_p=-\frac{1}{3} \frac{P}{\epsilon_o}$. A long cylinder has nearly zero $E_p$. Meanwhile a cylinder turned sideways has $E_p=-\frac{1}{2} \frac{P}{\epsilon_o}$.$\\$ $\\$ A flat slab has $E_p=-\frac{P}{\epsilon_o}$. (Outside a flat slab, $E=E_o$). $\\$ Here $E_{total}=E_o+E_p$ inside the material. $\\$ In the case of the flat slab, this makes $D$ continuous across the material, and in the material $E_{total}=\frac{E_o}{\epsilon_r}$ , where $E_o$ is the applied field, and $\epsilon=\epsilon_r \epsilon_o$ defines the dielectric constants.