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:
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'} $
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.
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.
Options A and B correctly represent aspects related to the dielectric loss tangent.
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:
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?
The wave length (λ) in meters of an electromagnetic wave is related to its frequency (f) in MHz as:
Bending of light wave as it passes between material of different optical density
The wave impedance of a medium is equal to: