---
title: "Power loss in a glass dielectric"
subject: "Electrical Engineering Material"
chapter: 3
key: "semester_3/electrical_engineering_material:glass-dielectric-power-loss"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/glass-dielectric-power-loss
site: BCT Engineering
---

# Power loss in a glass dielectric

Electrical Engineering Material · Chapter 3: Dielectric Materials

- Asked in: 2081 Chaitra (new course), 2 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/glass-dielectric-power-loss
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-3

## Question

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.

## 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.

## Answer

Power lost per unit volume in a lossy dielectric under AC:
$$
W_{vol} = \omega\,\varepsilon_0\,\varepsilon_r\,\tan\delta\;E_{rms}^2
$$

**Data**: $\varepsilon_r = 2.6$, $\tan\delta = 7\times10^{-5}$, $f = 1$ MHz, $E_{peak} = 0.71$ V/m.

$$
\begin{aligned}
\omega &= 2\pi f = 2\pi\times10^6 = 6.283\times10^6\ \text{rad/s} \\
E_{rms} &= \frac{0.71}{\sqrt2} = 0.502\ \text{V/m}, \quad E_{rms}^2 = 0.252 \\
W_{vol} &= 6.283\times10^6\times8.854\times10^{-12}\times2.6\times7\times10^{-5}\times0.252 \\
&= 2.55\times10^{-9}\ \text{W/m}^3
\end{aligned}
$$

**Answer: Power loss ≈ $2.55\times10^{-9}$ W/m³.**

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

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. The power loss per unit volume used is:
   A. ω ε0 εr tanδ E_rms^2
   B. ε0 εr E_rms/tanδ
   C. ω ε0 εr E_rms
   D. ω ε0 tanδ E_peak
   Answer: A. W = ω ε0 εr tanδ E_rms^2. (Key point: Formula)

2. With peak E = 0.71 V/m, E_rms^2 is about:
   A. 0.252
   B. 0.71
   C. 0.504 x 2
   D. 0.50
   Answer: A. E_rms = 0.71/sqrt2 = 0.502, squared 0.252. (Key point: RMS conversion)

3. The power loss is approximately:
   A. 2.55e-6 W/m^3
   B. 2.55e-9 W/m^3
   C. 5.10e-9 W/m^3
   D. 6.3e-9 W/m^3
   Answer: B. 6.283e6 x 8.854e-12 x 2.6 x 7e-5 x 0.252 = 2.55e-9 W/m^3; using the peak gives twice that. (Key point: Result)

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