# What are the actual physical effects if the relative permittivity of a dielectric increases?

I learn some basic information about the relative permittivity. I can understand that when the relative permittivity $$\epsilon_r \to \infty$$ that material is a perfect conductor. On the other hand, for any finite value of $$\epsilon_r$$, that material is a perfect dielectric.

However, I need help to obtain an intuitional or qualitative understanding of a specific physical setting, that is, two-layer different materials, a dielectric (diel_A) and a vacuum, between two plates (electrodes) with a fixed external electric field. In this case, what is the actual physical effect, or what happens if the relative permittivity of "diel_A" increases? Thank you in advance!

• A perfect conductor has permittivity of negative infinity. Feb 15 at 11:30

For a parallel plate capacitor without any dielectric material $$C_0=\frac{\epsilon_0 A}{s}$$

Now consider a simple situation where you have a capacitor with a half-filled dielectric with dielectric constant $$\kappa$$ (same as relative permittivity). $$\frac{1}{C_s}=\frac{1}{C_v}+\frac{1}{C_d}$$ Putting $$C_v=A\epsilon_0/(s/2)$$ and $$C_d=A\kappa \epsilon_0/(s/2)$$ will give $$C_s=\frac{2\kappa }{\kappa +1}C_0$$ $$\lim_{\kappa\rightarrow -\infty}C_s=2C_0$$ The explanation for this can be understood from here. It's clear that if $$s\rightarrow s/2$$ so that $$C_0\rightarrow 2C_0$$.

Edit: The notation is as follows

• $$s$$ stands for the distance between the plate.
• $$C_s$$ is the combined capacitance.
• $$C_v$$ is the capacitance of the vacuum part.
• $$C_d$$ is the capacitance of the dielectric part.
• $$\epsilon_0$$ is the permittivity of free space.
• thank you for the help, could you explain the notations? For example, what is $\epsilon_0$ and $s$? Feb 15 at 10:20
• I added notations. :) Feb 15 at 11:46