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Question

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

Defining Dielectric Loss Tangent ($\tan \delta$)

The dielectric loss tangent, often denoted as $\tan \delta$ or the dissipation factor, measures the energy dissipated as heat within a dielectric material when subjected to an alternating electric field. It quantizes the inefficiency of the dielectric in storing electrical energy.

For a dielectric material, the complex permittivity is defined as $\epsilon^* = \epsilon' - j\epsilon''$. Here:

  • $\epsilon'$ (epsilon prime) is the real part, representing the dielectric constant (energy storage).
  • $\epsilon''$ (epsilon double prime) is the imaginary part, representing the dielectric loss factor (energy dissipation).

The loss tangent is fundamentally defined as the ratio of the imaginary part to the real part of the complex permittivity:

$ \tan \delta = \frac{\epsilon''}{\epsilon'} $

Analyzing Option A: Displacement Current Components

Consider the displacement current density $\vec{J}_D$ in a dielectric medium under an AC field $ \vec{E} = \vec{E}_0 e^{j\omega t} $. We have $\vec{D} = \epsilon^* \vec{E} = (\epsilon' - j\epsilon'') \vec{E}$.

The displacement current density is $\vec{J}_D = \frac{\partial \vec{D}}{\partial t} = j\omega \vec{D} = j\omega \epsilon^* \vec{E}$.

Substituting $\epsilon^*$: $ \vec{J}_D = j\omega (\epsilon' - j\epsilon'') \vec{E} = (\omega\epsilon'' + j\omega\epsilon') \vec{E} $

This expression shows that the current has a real component proportional to $\omega\epsilon''$ (related to loss) and an imaginary component proportional to $\omega\epsilon'$ (related to energy storage).

Option A states: "Ratio of the real to imaginary parts of the total displacement current". Interpreting "total displacement current" as $j\omega\vec{D}$, the ratio of its real part ($\omega\epsilon''$) to its imaginary part ($\omega\epsilon'$) is:

$ \frac{\text{Real Part}}{\text{Imaginary Part}} = \frac{\omega\epsilon''}{\omega\epsilon'} = \frac{\epsilon''}{\epsilon'} = \tan \delta $

Therefore, Option A represents the dielectric loss tangent.

Analyzing Option B: Loss in Conductive Dielectrics

In a dielectric material that also exhibits conductivity ($\sigma$), the total current density is $\vec{J}_{total} = \sigma \vec{E} + \vec{J}_D$.

$ \vec{J}_{total} = \sigma \vec{E} + j\omega \vec{D} = \sigma \vec{E} + j\omega (\epsilon' - j\epsilon'') \vec{E} $

$ \vec{J}_{total} = \left[ \sigma + j\omega(\epsilon' - j\epsilon'') \right] \vec{E} = \left[ (\sigma + \omega\epsilon'') + j\omega\epsilon' \right] \vec{E} $

Here, $(\sigma + \omega\epsilon'')$ represents the total AC conductivity (sum of conduction loss and dielectric loss components), and $\omega\epsilon'$ represents the energy storage component (scaled by frequency). Option B gives the ratio:

$ \frac{\omega\epsilon'' + \sigma}{\omega\epsilon'} $

This represents the ratio of the total AC conductivity loss term to the dielectric storage term, often considered the effective loss tangent for a lossy, conductive dielectric. Therefore, Option B is also a valid representation in this context.

Evaluating Other Options

  • Option C: The ratio of electric susceptibility ($\chi_e$) to permittivity ($\epsilon'$) is incorrect. $\chi_e = \epsilon' - \epsilon_0$, so $\chi_e/\epsilon'$ does not equal $\tan \delta$.
  • Option D: The ratio of the polarization vector ($\vec{P}$) to the displacement vector ($\vec{D}$) ($\vec{P}/\vec{D}$) is related to susceptibility but is not the dielectric loss tangent.

Options A and B correctly represent aspects related to the dielectric loss tangent.

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Important Questions from Electromagnetic Wave Propagation

  1. For sky waves, following statements are given:

    (A) n > 1, this shows 81 \(\rm\frac{N}{f^2}\) positive

    (B) n > 1, show 81 \(\rm\frac{N}{f^2}\)  Negative

    (C) n < 1 shows 81 \(\rm\frac{N}{f^2}\)  < 1

    (D) v g x v p= c 2

    (E) n = 0 shows 81 \(\rm\frac{N}{f^2}\)  = 1, f = f c

    Choose the correct answer from the options given below:

  2. If the Polarization vector is given as N and the Direction of propagation is given as K then which one of the following relations is correct?

  3. The wave length (λ) in meters of an electromagnetic wave is related to its frequency (f) in MHz as:

  4. Bending of light wave as it passes between material of different optical density

  5. The wave impedance of a medium is equal to:

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