1. For dielectrics in alternating field, polarizability a e is a complex quantity. The imaginary part of a e is zero for





Ask Your Doubts Here

Type in
(Press Ctrl+g to toggle between English and the chosen language)

Comments

  • By: guest on 02 Jun 2017 12.59 am
    Imaginary part of a e is zero for ω = 0 and ω = ∞.
Show Similar Question And Answers
QA->What minimum number of non-zero non-collinear vectors is required to produce a zero vector?....
QA->The minimum number of non-zero non-collinear vectors required to produce a zero vector is?....
QA->The average of 20 numbers is zero. Of them , at the most ,how many may be greater than zero?....
QA->The average of 20 numbers is zero. Of them, at the most, how many may be greater than zero?....
QA->The average of 20 numbers is zero. Of them, How many of them may be greater than zero, at the most?....
MCQ->For dielectrics in alternating field, polarizability a e is a complex quantity. The imaginary part of a e is zero for....
MCQ->When a dielectric material is subjected to alternating field, the absorption of energy by the material from the field is given by the imaginary part of polarizability.....
MCQ->Assertion (A): For elemental dielectrics, P = Na eEi where N is number of atoms/m3, P is polarization, a e is electronic polarization and Ei is internal electric field.Reason (R): In elemental dielectrics there are no permanent dipoles or ions.

....
MCQ->When subjected to alternating stresses, an insulating material is characterised by complex dielectric constant ∈'r - j∈"r . In such materials, the dielectric losses under alternating stress is proportional....
MCQ->Assertion (A): In Figure, the intercept of the line on y-axis is equal to N (a e + a i), where N is number of atoms/m3, a e is electronic polarizability and a i is ionic polarizability.Reason (R): a e and a i are independent of temperature.

....
Terms And Service:We do not guarantee the accuracy of available data ..We Provide Information On Public Data.. Please consult an expert before using this data for commercial or personal use | Powered By:Omega Web Solutions
© 2002-2017 Omega Education PVT LTD...Privacy | Terms And Conditions
Question ANSWER With Solution