Skip to main content

I think if you can measure the permittivity, then you must have an LCR bridge. using the latter: 1- measure the impedance of your sample, if it is >100kohm then>100 $k\,\Omega$ then go to 3 2- if your impedance is <100 kohm$k\,\Omega$ then measure the following parameters, using your LCR bridge, Cs, D or Q, or better theta. 3- measure Cp, D or Q, or better theta. Now you have some data for certain frequency range, you can start calculation.

4- if your D or Q values are within the limits of the bridge, it is easier, so I'll assume this for now & we'll call option A: using the cell of your measured sample,measure the same Capacitancecapacitance for the air, Cs$C_s$ or Cp$C_p$. Or you can calculate it using your sample cell dimensions.

Now

  • divide all the measured Cs or Cp by the value for the air. you can get the permittivity
  • multiply the value of D or 1/Q by all the permittivity values you have. you you have the Loss (Imaginary part of the permittivity).
  • Apply the relation conductivity=2PiF EoE''. then you're there. hope this is easy to follow. Kind regards

I think if you can measure the permittivity, then you must have an LCR bridge. using the latter: 1- measure the impedance of your sample, if it is >100kohm then go to 3 2- if your impedance is <100 kohm then measure the following parameters, using your LCR bridge, Cs, D or Q, or better theta. 3- measure Cp, D or Q, or better theta. Now you have some data for certain frequency range, you can start calculation.

4- if your D or Q values are within the limits of the bridge, it is easier, so I'll assume this for now & we'll call option A: using the cell of your measured sample,measure the same Capacitance for the air, Cs or Cp. Or you can calculate it using your sample cell dimensions.

Now

  • divide all the measured Cs or Cp by the value for the air. you can get the permittivity
  • multiply the value of D or 1/Q by all the permittivity values you have. you you have the Loss (Imaginary part of the permittivity).
  • Apply the relation conductivity=2PiF EoE''. then you're there. hope this is easy to follow. Kind regards

I think if you can measure the permittivity, then you must have an LCR bridge. using the latter: 1- measure the impedance of your sample, if it is >100 $k\,\Omega$ then go to 3 2- if your impedance is <100 $k\,\Omega$ then measure the following parameters, using your LCR bridge, Cs, D or Q, or better theta. 3- measure Cp, D or Q, or better theta. Now you have some data for certain frequency range, you can start calculation.

4- if your D or Q values are within the limits of the bridge, it is easier, so I'll assume this for now & we'll call option A: using the cell of your measured sample,measure the same capacitance for the air, $C_s$ or $C_p$. Or you can calculate it using your sample cell dimensions.

Now

  • divide all the measured Cs or Cp by the value for the air. you can get the permittivity
  • multiply the value of D or 1/Q by all the permittivity values you have. you you have the Loss (Imaginary part of the permittivity).
  • Apply the relation conductivity=2PiF EoE''. then you're there. hope this is easy to follow. Kind regards
Source Link

I think if you can measure the permittivity, then you must have an LCR bridge. using the latter: 1- measure the impedance of your sample, if it is >100kohm then go to 3 2- if your impedance is <100 kohm then measure the following parameters, using your LCR bridge, Cs, D or Q, or better theta. 3- measure Cp, D or Q, or better theta. Now you have some data for certain frequency range, you can start calculation.

4- if your D or Q values are within the limits of the bridge, it is easier, so I'll assume this for now & we'll call option A: using the cell of your measured sample,measure the same Capacitance for the air, Cs or Cp. Or you can calculate it using your sample cell dimensions.

Now

  • divide all the measured Cs or Cp by the value for the air. you can get the permittivity
  • multiply the value of D or 1/Q by all the permittivity values you have. you you have the Loss (Imaginary part of the permittivity).
  • Apply the relation conductivity=2PiF EoE''. then you're there. hope this is easy to follow. Kind regards