If dielectric constant increase then how it will affect on the conductivity of the material? Everyone tries to increase conductivity of a dielectric material. Why? How it will help in practical application?
Dear Ratna sarkar,
Before answering your query I would say first you should be clear about the fact that "everyone and anyone is not synonymous" in that similar way " dielectric and insulator" and most important fact that semiconducting materials can show character of dielectric as well as metal in different physical condition(s). One need to understand or explore those physical conditions e.g., there are many ferroelctric materials which are fast ion conductor in high temperature range and may have some potential application in the SOFC. Mott insulator are another materials which shows insulator to metal transition.........
Additionally, dielectric and conductivity are independent phenomenon. However total number of charges in the system has to be constant, thus the sum of immobile and mobile charges; unless charges injected externally.
Hello Ratna,
I try to explain some complex properties, but the answer will not be generally valid.
The permittivity eps of a material is a complex magnitude (in mathematical sense): eps = epsR - j epsI
The real part relates to the polarization within an electric field and depends strongly on frequency. As an example water: At frequency zero we have epsR > 80 due to the strong electric dipole moment and for visible light epsR = n²=1,33² = 1,77 (here n refractive index). The inertia at high frequencies reduces eps.
EpsR is a measure for the stored energy in the medium. The imaginary part epsI is a measure for the dissipated energy and has a maximum. Example water: The maximum absorption of water is observed between 10 and 100 GHz. This is used in microwave ovens to heat soups and so on. The 2 components of eps are related by the so called loss factor tan(delta) = epsI/epsR = energy_dissipated/energy_stored.
If you want to optimize the dielectric for dissipation in heat, so you must look for a suitable frequency. If you want to store energy (in a capacitor), than you must look for a low loss factor.
The loss factor is to be completed by a second term (see attachment) which contains the static conductivity condS: condS/(2*pi*epsR). If the medium has a conductivity, than electron collisions enhance the loss. In the optical range you have an enhanced reflectivity.
Generally spoken, you must know, what you want to create. You can develop devices with different properties: high or low energy storage, heat production, reflectivity.
With Regard
R. Mitdank
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