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Power loss in a glass dielectric

Short answer

Power loss per unit volume is calculated as ω ε0 εr tanδ E_rms². With f=1 MHz, εr=2.6, tanδ=7×10⁻⁵, E_peak=0.71 V/m, the loss is 2.55×10⁻⁹ W/m³.

A glass dielectric has dielectric constant 2.6 and loss tangent 7×10⁻⁵ at 1 MHz. Calculate the loss of power per unit volume if the signal applied to the glass has peak amplitude of 0.71 V/m.

Power lost per unit volume in a lossy dielectric under AC:

Wvol=ω ε0 εr tan⁡δ  Erms2W_{vol} = \omega\,\varepsilon_0\,\varepsilon_r\,\tan\delta\;E_{rms}^2

Data: εr=2.6\varepsilon_r = 2.6, tan⁡δ=7×10−5\tan\delta = 7\times10^{-5}, f=1f = 1 MHz, Epeak=0.71E_{peak} = 0.71 V/m.

ω=2πf=2π×106=6.283×106 rad/sErms=0.712=0.502 V/m,Erms2=0.252Wvol=6.283×106×8.854×10−12×2.6×7×10−5×0.252=2.55×10−9 W/m3\begin{aligned} \omega &= 2\pi f = 2\pi\times10^6 = 6.283\times10^6\ \text{rad/s} \\ E_{rms} &= \frac{0.71}{\sqrt2} = 0.502\ \text{V/m}, \quad E_{rms}^2 = 0.252 \\ W_{vol} &= 6.283\times10^6\times8.854\times10^{-12}\times2.6\times7\times10^{-5}\times0.252 \\ &= 2.55\times10^{-9}\ \text{W/m}^3 \end{aligned}

Answer: Power loss ≈ 2.55×10−92.55\times10^{-9} W/m³.

(If the peak value is used directly in place of the rms value, the result is 5.10×10−95.10\times10^{-9} W/m³, i.e. twice as large.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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