Question

Which option(s) represents/represent the dielectric loss tangent of a substrate?

• A. Ratio of the real to imaginary parts of the total displacement current
• B. \(\left(\omega\epsilon''+\sigma\right)/\left(\omega\epsilon'\right)\)
• C. Ratio of the electric susceptibility to permittivity
• D. Ratio of the polarization vector \(\vec{P}\) to the displacement vector \(\vec{D}\)

Hint:

Loss tangent compares the in-phase (lossy) current to the out-of-phase (reactive) current: \(\tan\delta=(\omega\epsilon''+\sigma)/(\omega\epsilon')\).

Answer 1

The Correct Option is A, B

Step 1: Recall the standard definition of loss tangent.
For a lossy dielectric with complex permittivity \(\epsilon=\epsilon'-j\epsilon''\) and conductivity \(\sigma\), the loss tangent measures how much energy is dissipated compared to how much is stored. It is defined as
\[ \tan\delta=\frac{\epsilon''_{eff}}{\epsilon'} \]
where the effective imaginary part folds in the ohmic loss due to conductivity,
\[ \epsilon''_{eff}=\epsilon''+\frac{\sigma}{\omega} \]

Step 2: Write the loss tangent explicitly.
Substituting,
\[ \tan\delta=\frac{\epsilon''+\dfrac{\sigma}{\omega}}{\epsilon'}=\frac{\omega\epsilon''+\sigma}{\omega\epsilon'} \]
This is exactly option (B), so (B) is correct.

Step 3: Look at the total current in the dielectric.
The total current density (conduction plus displacement) in phasor form is
\[ J=\sigma E+j\omega\epsilon E=\sigma E+j\omega(\epsilon'-j\epsilon'')E=(\sigma+\omega\epsilon'')E+j\omega\epsilon'E \]
The real part, \((\sigma+\omega\epsilon'')E\), is in phase with the field and represents the dissipated (lossy) part. The imaginary part, \(\omega\epsilon'E\), is the reactive, lossless part that stores energy.

Step 4: Form the ratio of real to imaginary parts.
\[ \frac{\text{Real part}}{\text{Imaginary part}}=\frac{\sigma+\omega\epsilon''}{\omega\epsilon'} \]
This is the same expression as the loss tangent found in Step 2. So describing the loss tangent as the ratio of the real to imaginary parts of this total current is also a valid statement, and option (A) is correct too.

Step 5: Check option (C).
The electric susceptibility \(\chi\) relates polarization to field by \(P=\epsilon_0\chi E\), and permittivity is \(\epsilon=\epsilon_0(1+\chi)\). The ratio \(\chi/\epsilon\) is a plain real number and says nothing about any phase lag, so it cannot represent a loss. Option (C) is incorrect.

Step 6: Check option (D).
Since \(D=\epsilon_0E+P\), the ratio \(|P|/|D|\) is just another way of writing \(\chi/(1+\chi)\), again a real, lossless quantity with no imaginary part involved. Option (D) is incorrect.

Final Answer:
The loss tangent is correctly represented by
\[ \boxed{\text{options (A) and (B)}} \]