How Does Polarization Affect Electric Field Calculation in a Dielectric Disc?

[Saketh]

I'm trying to teach myself polarization and dielectrics by doing problems, but it's not going as well as I'd hoped. Here's the first problem that I got stuck on:

A round dielectric disc of radius R is statically polarized so that it gains the uniform polarization P, with the vector P lying in the plane of the disc. Find the strength E of the electric field at the center of the disc if d << R.​

I thought that since the disc is polarized along its plane, there would be a charge density $$\sigma$$ on one half of it, and a charge density of $$-\sigma$$ on the other half of it. I wasn't sure if $$\sigma = P$$, but I set them equal anyway. I also wasn't sure if $$\sigma$$ is uniform along each half-surface, but I did that anyway. Then I thought that a uniform polarization vector means that the whole surface is charged up unformly, which means no net electric field at the center.

I realize this is going to sound silly, but I have no idea how I'm supposed to solve this problem. Where am I supposed to start?

 

[Nate810]

The first thing you need to do is to determine the charge distribution on the dielectric disc. Since the polarization is uniform, this means that the same amount of charge will be distributed uniformly across the surface of the disc. To calculate this, you can use the equation \sigma = \epsilon_0 E P, where \epsilon_0 is the permittivity of free space and E is the electric field at the surface of the disc. Once you have this, you can then calculate the electric field at the center of the disc by using the equation E = \frac{1}{4 \pi \epsilon_0}\frac{\sigma}{r^2}, where r is the distance from the center of the disc to the point where the electric field is being measured.

 

[kyle8921]

I can understand your frustration with learning a new concept and facing difficulties in solving problems. First of all, it is important to have a clear understanding of the concept of polarization and dielectrics before attempting to solve problems. This will help you approach the problem with a better understanding and increase your chances of solving it successfully.

Now, let's break down the problem and try to understand it step by step. The problem states that a round dielectric disc is statically polarized, which means that the molecules within the dielectric material are aligned in a specific direction, resulting in a net dipole moment. This dipole moment is represented by the polarization vector P, which is lying in the plane of the disc.

The problem also mentions that the disc has a radius R and the distance between the center of the disc and any point on its surface is d, which is much smaller than R (d<<R). This information is important as it gives us an idea of the size of the disc and the distance at which we are trying to find the electric field.

Now, as you correctly pointed out, the polarization of the disc results in a charge density \sigma on one half of the disc and -\sigma on the other half. However, it is important to note that the charge density is not equal to the polarization vector P. The charge density is given by \sigma = P/(4\pi) where P is the magnitude of the polarization vector and 4\pi is a constant.

Next, we need to consider that the disc is in a dielectric material, which means that it has a dielectric constant \epsilon_r. This dielectric constant affects the electric field and is given by \epsilon_r = 1+\chi_e where \chi_e is the electric susceptibility of the material.

Now, let's focus on the electric field at the center of the disc. We know that the electric field is a vector quantity and is given by the equation E = \sigma/(\epsilon_0\epsilon_r) where \epsilon_0 is the permittivity of free space. However, since we are only interested in the electric field at the center of the disc, we can consider only the components of the electric field that are perpendicular to the plane of the disc. This means that the electric field at the center of the disc is given by E = \sigma/(2\epsilon_0\epsilon_r).

Using the information we have gathered so