Determine whether germanium is a conductor at a certain frequency

Click For Summary

Discussion Overview

The discussion revolves around determining whether germanium can be classified as a conductor at a frequency of 2.4 GHz, focusing on the relationship between conductivity (sigma), frequency (omega), and permittivity (epsilon). The scope includes mathematical reasoning and application of relevant equations.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant states the relationship sigma = omega * epsilon and attempts to calculate epsilon based on given values.
  • Another participant suggests that calculating the time constant epsilon/sigma could indicate whether germanium is a good conductor, implying a short time constant would suggest good conductivity.
  • A participant calculates the relative permittivity (epsilon_r) to be 17.24, based on their computations involving epsilon and the permittivity of free space.
  • Further calculations are presented, confirming the value of omega and reiterating the calculated epsilon and epsilon_r.

Areas of Agreement / Disagreement

Participants present calculations and methods but do not reach a consensus on whether germanium is definitively a conductor at the specified frequency. The discussion remains unresolved regarding the implications of the calculated values.

Contextual Notes

Limitations include assumptions about the definitions of conductivity and the conditions under which germanium may be considered a conductor. The discussion does not resolve the implications of the calculated time constant or the significance of the relative permittivity value.

DODGEVIPER13
Messages
668
Reaction score
0

Homework Statement


Determine, whether germanium (sigma=2.3) is a conductor at the frequency of 2.4 GHz.
sigma=?(omega)(epsilon)


Homework Equations





The Attempt at a Solution


Well I know omega=2∏f and I have f. Thus I could find omega. Given sigma I could find epsilon but what excatly does that tell me about it being a conductor? hen this is done I get epsilon to be 1.526e-10
 
Physics news on Phys.org
I could find for a time constant epsilon/sigma and if it is short I will know it is a good conductor is that ok?
 
I got the relative epsilon to be 17.24
 
ω=2∏f=2∏(2.4x10^9)=1.507x10^10
σ=2.3
σ=ωε
ε=1.526x10^-10
ε0εr=1.526x10^-10
εr=(1.526x10^-10)/(8.85x10^-12)=17.24
my final answer is 17.24 am I good?
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
6K
  • · Replies 36 ·
2
Replies
36
Views
6K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
4
Views
2K
Replies
10
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K