Question

1. Drude model I The frequency dependent conductivity of an electron gas in a metal is to a very good approximation given by the Drude model, J(w) = σ(w)E(w) o(a) (a) Consider a channel shown below, made of A. When 60Hz AC power is applied, this channel has lk2 of resistance. If a material can be regarded as an insulator when its resistivity is over 10162 m, from what frequency A may work as an insulator? Calculate using the real part of the complex conductance. (τ- 1x10-5s) (b) The imaginary part of the complex conductance shows that the current lags behind the electric field. This lagging effect can be modeled as a function of inductance. The equivalent circuit model for the Drude conductivity in metal is schematically shown below. Find the vales of the conductance G and the inductance L in terms of the given parameters 2P A(Area)

The frequency dependent conductivity of an electron gas in a metal is to a very good approximation given by the Drude model

Answer 1

To solve this problem the details calculation are given below
Delta(w) = delta_DC/1 - I w z = delta_DC (1 + I w z)/1 + w^2 tau^2 delta(w) = delta_DC/1 + w^2 tau^2 + I delta_DC w_2/1 + w^2 tau^2 real part imaginary part \r\n Now, tau = 10^- 5 sec, w = 60 Hz: delta = 1/rho = 1/10^16 ohm^- 1 therefore delta(w)_real = delta_DC/1 + w^2 tau ohm = 10^- 16 ohm^- 1 = 10^- 16/1 + (60)^2 times (10^- 5)^2 = 9.999 times 10^- 17 ohm^- 1 rightarrow real part of conductivity Imaginary part of conductivity delta(w)_imag. = r_DC w_2/1 + w^2 tau^2 J = delta E