Question

A dielectric material has relative permittivity 18 and loss tangent ${10}^{-3}$ at 1 GHz. The distance over which the wave amplitude reduces to $\frac{1}{e}$ of its original value is __________m

• A. 22.5

Answer 1

The correct option is A 22.5
$Thelosstangenttan\delta =\frac{\sigma }{\omega \epsilon }$

$or\sigma =\omega {\epsilon }_0{\epsilon }_{r}tan\delta$

$=2\pi \times {10}^9\times \frac{1}{36\pi }\times {10}^{-9}\times 18\times {10}^{-3}$
$={10}^{-3}\mho /m$

$\because Losstangentisverythedielectricisalossdielectricandwecanapproximatelygettheattenuationconstantofthewaveas.$

$\alpha =\frac{\sigma }{2}\sqrt{\frac{\mu }{\epsilon }}=\frac{{10}^{-3}}{2}\times 120\pi \times \frac{1}{\sqrt{18}}=4.44\times {10}^{-2}N/m$

$\therefore Thedistanceoverwhichthewaveamplitudereducesto\frac{1}{e}ofitsoriginalvalueof\frac{1}{\alpha }=22.52m.$