0
$\begingroup$

I am currently studying the textbook Microwave Engineering, fourth edition, by David Pozar. Section Fields at a General Material Interface of chapter 1.3 FIELDS IN MEDIA AND BOUNDARY CONDITIONS says the following:

The time-harmonic version of (1.4), where $S$ is the closed "pillbox"-shaped surface shown in Figure 1.6, can be written as $$\oint_S \bar{D} \cdot d \bar{s} = \int_V \rho \ dv. \tag{1.29}$$ enter image description here

(1.4) is written as follows:

$$\oint_S \bar{D} \cdot d \bar{s} = \int_V \rho \ dv = Q \tag{1.4}$$

$\bar{D}$ is the electric flux density, in coulombs per meter squared (Coul/$\text{m}^2$), $\rho$ is the electric charge density, in coulombs per meter cubed (Coul/$\text{m}^3$), and $Q$ is the total charge contained in the closed volume $V$.

So aren't (1.29) and (1.4) identical? What is the difference in the aforementioned "time-harmonic version"?

$\endgroup$

1 Answer 1

1
$\begingroup$

If you look carefully, Eq. 1.4 has a calligraphic symbol $\bar{\mathcal{D}}$, while Eq. 1.29 has a regular $\bar{D}$. They are related by factoring out a harmonic time dependence $\bar{\mathcal{D}}=\Re{[e^{j\omega t} \bar{D}]}$ which in the book is called "phasor notation". This is mostly useful when assuming the time dependence of the fields is purely harmonic with other properties being in a steady state. Thus they call it "time harmonic version".

$\endgroup$
1
  • 1
    $\begingroup$ Ahh, you're right: the textbook actually has written $$\oint_S \bar{\mathcal{D}} \cdot d\bar{s} = \int_V \rho \ dv = Q \tag{1.4}$$ This is my mistake. Thank you for the clarification. $\endgroup$ Commented Mar 15, 2022 at 14:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.