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Loss tangent and dielectric power loss

Short answer

Loss tangent tan delta = epsilon_r'' / epsilon_r' quantifies dielectric loss. Power loss per unit volume is W_vol = omega epsilon_0 epsilon_r' E_rms^2 tan delta, showing it is proportional to angular frequency omega and tan delta.

What is loss tangent? Show that power loss in a dielectric material per unit volume is the function of frequency of the applied field and loss tangent.

Loss tangent

In an ideal dielectric the current leads the voltage by exactly 90°, so no power is lost. In a real dielectric, polarization cannot follow an AC field instantly, so part of the energy is turned into heat. This is described by a complex relative permittivity

εr=εr′−jεr′′\varepsilon_r = \varepsilon_r' - j\varepsilon_r''

where εr′\varepsilon_r' is the ordinary (storage) part and εr′′\varepsilon_r'' is the loss part. The loss tangent is

tan⁡δ=εr′′εr′\tan\delta = \frac{\varepsilon_r''}{\varepsilon_r'}

δ\delta is the angle by which the current falls short of leading the voltage by 90°. Good insulators have tan⁡δ∼10−4\tan\delta \sim 10^{-4} (polyethylene); lossy ones are 0.01–0.1.

      I_total
        ^   /
  I_cap |  /
  (wC'V)| /  delta
        |/________> V
          I_loss (wC''V)

Power loss per unit volume

Consider a parallel-plate capacitor with area AA, spacing dd, filled with the dielectric, and an applied voltage V=V0ejωtV = V_0 e^{j\omega t}.

Capacitance with complex permittivity:

C=ε0(εr′−jεr′′)AdC = \frac{\varepsilon_0(\varepsilon_r' - j\varepsilon_r'')A}{d}

Current:

I=jωCV=ωε0Ad(jεr′+εr′′)VI = j\omega C V = \frac{\omega\varepsilon_0 A}{d}\left(j\varepsilon_r' + \varepsilon_r''\right)V

The term ωε0εr′′AdV\frac{\omega\varepsilon_0\varepsilon_r'' A}{d}V is in phase with VV and causes power loss; the jεr′j\varepsilon_r' term is the lossless capacitive current. So the dielectric acts like a capacitor with a parallel conductance

Gp=ωε0εr′′AdG_p = \frac{\omega\varepsilon_0\varepsilon_r'' A}{d}

Power dissipated (using rms voltage VrmsV_{rms}):

W=GpVrms2=ωε0εr′′AdVrms2W = G_p V_{rms}^2 = \frac{\omega\varepsilon_0\varepsilon_r'' A}{d}V_{rms}^2

Divide by the volume AdAd and use Erms=Vrms/dE_{rms} = V_{rms}/d:

Wvol=WAd=ωε0εr′′Erms2W_{vol} = \frac{W}{Ad} = \omega\varepsilon_0\varepsilon_r'' E_{rms}^2

Since εr′′=εr′tan⁡δ\varepsilon_r'' = \varepsilon_r'\tan\delta:

Wvol=ω ε0 εr′ Erms2tan⁡δ\boxed{W_{vol} = \omega\, \varepsilon_0\, \varepsilon_r'\, E_{rms}^2 \tan\delta}

Conclusion

The dielectric loss per unit volume is directly proportional to:

  • the frequency ω=2πf\omega = 2\pi f of the applied field,
  • the loss tangent tan⁡δ\tan\delta (or loss factor εr′tan⁡δ\varepsilon_r'\tan\delta),
  • the square of the field strength.

So insulation for high-frequency or high-voltage use (cables, RF capacitors) must have a very low tan⁡δ\tan\delta. The same heating is used usefully in microwave ovens and dielectric heating of plastics and wood.

✓ Answer checked by Claude Opus 5.5 (high effort)

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