Increase rate of Er"

I saw an article related to high K materials. It states that increasing the effective electric permittivity will also increase the imaginary part of the dielectric constant. It further states that loss tangent will increase as well.

Any one could help explain why the imaginary part has to increase?

Since loss tangent=Er"/Er', to increase loss tangent, I suspect that the Er" has to increase at a faster rate for loss tangent to increase. If Er" is not increase as fast as Er', the loss tangent may even drop.

Please help to share your insight on this.

In any material, the real and imaginary components of permittivity are not directly linked; they often behave separately in a different way.
In High-K materials, the imaginary part (loss tangent) is often quite low; therefore such materials are e.g. used for dielectric resonators with the desired small dimensions (against cavity, air-filled ones) while the resonator Q is high.
Lossy materials like concrete have typically low real permittivity while the imaginary part is high. Both are frequency dependent.

Thank you for the reply!

Not sure why some people are assuming that increase Er' (real part) will increase the Er" (imaginary part) as well?

To answer this question may need knowledge on materials.

Any one can share more insight on this?

Thanx.

The best answer for a true discussion of this phenomenon lies in the Kramer-Kronig relationship. basically, if the imaginary part of the permittivity 'bumps' the imaginary part will 'wiggle'...think of a Lorentzian Oscillator. But also, the KK relationship shows that for a material to have a causal impulse response (material reaction to the incident EM field), the material properties have to be dispersive.

have Fun!